MYP
Mathematics
A
concept-based
approach
4&5
Extended
Rose Harrison • Clara Huizink
Aidan Sproat-Clements • Marlene Torres-Skoumal
3
Great
Clarendon
Oxford
It
Street,
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in
How
to
use
this
book
Chapters
Chapte rs
For m,
e ach
focu s
Relat ion ship s
Chapters
for
1–3
4 –15
each
on
a nd
focus
one
of
t he
key
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L ogic.
on
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concept s
Mathematic s:
Repre sentation,
Change,
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unit s.
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ABCD
Objective
by
using
8).
BD
part
written
20).
Objective boxes
T his
show
method.
6 + 18
Practice
work
questions.
.
M
(6,
inquir y
1
midpoint
3 + 9
the
generalization?
Worked
Example
and
predict?
shows
where
Geometr y
should
not
student s
Soft ware
use
could
(DGS)
technolog y
use
or
are
Graphical
Computer
indicated
Display
Algebra
with
a
Calculator s
Systems
(GDC),
(CA S).
cros sed- out
Place s
where
icon.
The notation used throughout this book is largely that required in the DP IB programs.
v
Each
topic
ends
with:
Summary
The
standard
dispersion
deviation
that
gives
an
is
a
measure
idea
of
how
of
Using
close
data
original
how
A
data
values
representative
small
standard
are
the
to
the
mean
deviation
mean,
is
of
and
the
values
are
close
to
shows
the
that
mean.
The
of
values
all
the
of
the
x
deviation
original
means
Using
Σ
the
this
are
the
same
as
the
in
x f
the
formula
sum
of
all
the x
of
all
the
mean
the
values
of
a
set
of
divided
values
by
frequency
table
divided
by
is
the
,
total
the
sum
frequency.
to
data
calculate
the
presented
in
standard
a
deviation
frequency
table
for
is:
Summary
of
the
points
The
normal
distribution
with
most
is
a
symmetric
values
close
to
the
mean
the
values.
tailing
frequency
o
evenly
graph
is
a
in
either
direction.
bell-shaped
Its
curve.
practice
Number
questions
1
and
2
the
data
provided
is
of
strawberries
on
each
plant
fed
with
for
special
the
discrete
values.
and
In
of
key
notation,
Mixed
set
data.
sum
of
a
of
n
number
a
values
distribution,
,
of
units
x
is
mean
the
units
discrete
standard
the
so
data.
A
data
notation,
the
strawberry
plant
food:
Mixed
population.
14
15
17
17
19
19
12
14
15
practice
–
15
summative
1
Find
the
each
data
mean
and
the
standard
deviation
of
Analyze
2,
b
21
3,
3,
4,
4,
5,
5,
6,
6,
21
kg,
24
kg,
of
using
to
the
nd
whether
special
there
strawberry
is
an
plant
25
facts
kg,
27
kg,
29
kg
Bernie
recorded
hamster
ate
the
each
weight
day.
Here
of
food
are
his
in
of
the
food.
6
3
kg,
data
assessment
eect
a
the
set:
grams
and
skills
his
results:
learned, including
c
x
f
3
2
4
3
5
2
Day
Food
a
(g)
Find
the
the
1
2
3
4
5
6
7
8
9
55
64
45
54
60
50
59
61
49
mean
weight
of
and
food
standard
the
deviation
hamster
ate
b
Interval
Bernie
range
assumes
that
his
hamster’s
in
a
the
9days.
d
questions, and
questions
of
over
problem-solving
of
contexts.
food
Frequency
−
Review
in
The
art
ancient
culture
to
and
create
context
of
origami
involves
models.
has
folding
Some
roots
at
models
in
sheets
are
Japanese
of
The
resulting
paper
designed
creases
are
A
illustrated
in
this
B
diagram.
C
to
E
resemble
just
animals,
celebrate
the
owers
beauty
or
of
buildings;
geometric
others
design.
Review in context
D
–
summative
F
assessment
within
the
questions
global
H
context
G
The
artist
for
the
topic.
Z
thinks
lengths
FH
and
DY
are
equal.
1
b
Explain
why
EX
EG.
2
Reection
Reect
and
statement
How
have
v i
you
on
discuss
explored
the
statement
of
inquiry?
Give
specic
examples.
inquir y
of
the
Table
of
contents
Access
your
suppor t
website:
www.oxfordsecondar y.com/myp-mathematics
Form
E1.1
Space
What
if
we
Dierent
all
had
eight
ngers?
•
E9.1
bases
2
Another
3D
E9.2
dimension
•
Orientation
Mapping
the
167
world
•
Relationships
Oblique
E2.1
Multiple
doorways
Composite
triangles
183
•
functions
21
Change
E2.2
Doing
and
Invserse
undoing
•
E10.1
functions
35
Time
Log
E10.2
for
a
change
•
functions
Meet
the
202
transformers
•
Logic
Rational
E3.1
How
can
Vectors
we
and
get
there?
vector
functions
216
•
spaces
54
Equivalence
E11.1
Unmistaken
identities
•
Representation
Trigonometric
E4.1
How
to
stand
out
Data
inferences
from
the
crowd
identities
233
•
76
Generalization
E12.1
Go
ahead
and
log
in
•
Simplication
Laws
E5.1
Super
powers
Fractional
of
logarithms
249
•
exponents
98
Justication
E13.1
Are
we
very
similar?
•
Quantity
Problems
E6.1
Ideal
work
Evaluating
for
lumberjacks
involving
triangles
264
•
logs
108
Models
E14.1
A
world
of
dierence
•
Measurement
Nonlinear
E7.1
Slices
The
of
pi
unit
inequalities
282
•
circle
and
Systems
trignometric
functions
E15.1
Branching
out
•
124
Conditional
probability
299
Patterns
E8.1
Making
it
all
Arithmetic
add
and
up
Answers
315
Index
350
•
geometric
sequences
145
v i i
What
if
we
all
had
eight
E1.1
ngers?
Global
context:
Objectives
●
Understanding
Scientic
Inquiry
the
concept
of
a
●
number
F
system
●
Counting
●
Converting
in
dierent
numbers
bases
from
one
base
to
MROF
Using
How
have
What
●
How
operations
is
a
can
number
you
in
dierent
How
other
●
D
are
written
in
base?
write
numbers
mathematical
bases
Communication
understanding
to
interpret
Would
base
●
2
been
in
other
bases
operations
intercultural
numbers
bases?
●
Use
innovation
questions
●
C
ATL
technical
history?
another
●
and
communication
similar
in
you
base
be
to
operations
and
dierent
in
from
10?
better
o
counting
in
2?
How
does
form
inuence
function?
N U M B E R
Y
ou
●
should
use
the
already
operations
of
know
how
addition,
1
Calculate
subtraction and long multiplication
in
base
10,
without
a
to:
a
10 442
understand
place
value
2
27
●
How
●
What
●
How
Humans
record
of
language,
these
but
or
all
cultures
you
the
nd
the
values
Egyptian
of
symbol
total
all
numerals
a
a
10 887
d
14 078
7891
in
system
of
down
the
value
of
the
5
×
71
in:
b
511 002
d
1.5
bases
in
in
history?
other
bases?
to
down
a
or
in
.
the
forms
was
the
word
dierent
Each
dierent
of
way,
of
notation
Egyptian
an
additive
dierent
symbols
the
a
quantity.
ancient
a
5,
word
tally:
same
has
ways
write
symbol
a
43
15
written
numbers
you
dierent
use
b
diagram?
used
The
value
the
this
the
been
dierent
number
number
Each
in
even
use
hand:
base?
write
would
have
number.
hieroglyphic
number
written
maybe
dierent
numbers
many
bugs
the
in
you
represent
represent
system.
can
could
represents
they
Dierent
to
have
You
a
How
green
might
“ve”.
is
used
numbers.
number
You
have
have
762
351
c
Numbers
×
Write
a
F
+
by
calculator
c
●
these
value
number
by
and
adding
together.
these
symbols:
Green
are
called
stroke
heelbone
1
10
coiled
shield
bugs
sometimes
green
stink
rope
bugs,
as
they
100
produce
odor
if
a
pungent
handled
disturbed.
lotus
ower
pointed
1000
The
number
nger
10 000
11
is
written
,
and
36
is
tadpole
scribe
100 000
1 000 000
written
.
E1.1 What if we all had eight ngers?
3
or
Reect
and
discuss
1
ATL
●
Write
each
number
in
Egyptian
numerals:
99
●
Write
these
●
What
are
the
advantages
●
What
are
the
disadvantages?
Our
number
The
position
The
two
numbers
system,
of
a
numbers
represent
the
often
digit
below
same
as
ordinary
of
the
called
tells
you
have
the
Egyptian
in
Moving
The
the
left,
third
furthest
the
right
second
column
number
numerals,
system?
is
a
place
value
system
same
four
digits,
but
digits
do
not
always
amount.
3
digit
numbers:
value.
3642
A
(decimal)
Arabic
its
10 240
4623
thousands
column
column
represents
and
3
represents
represents
collections
of
units
individual
collections
ten
tens:
of
the
objects:
ten
units
units:
hundreds
the
tens
column
column
The pattern continues: each column is worth 10 times more than the column to
The
number
4
364
actually
represents
a
collection
of
that
many
its right.
objects:
hundreds
tens
units
3
6
4
E1 Form
N U M B E R
Reect
●
●
and
Does
the
Does
it
The
Our
symbol
represent
Roman
does
the
column
the
to
have
system
of
dierent
is
do
known
each
values
in
the
numbers
have
need
as
place
a
a
symbol
symbol
decimal
value
•
•
10
thousands
×
to
for
(also
column
10
×
hundreds
as
is
10
tens
2
10
In
make
binary
represent
base
ten
×
a
(base
place
two)
•
value
ten,
times
•
or
the
so
why
denary)
value
of
the
each
column
•
•
2
worth
×
sixteens
2
eights
4
number
is
twos
2
a
set
the
×
number
column
base.
to
its
right.
2
units
1
2
represents
point
natural
2
fours
3
•
10
twice
×
2
10 111
any
•
1
10
with
•
tenths
0
10
system
2
binary
zero,
10
units
1
10
×
The
2051?
zero?
decimal
can
and
right.
3
You
205
dierent?
not
system
×
•
2
something
value
value
its
0
numerals
place
number
because
discuss
•
•
•
0
2
2
containing:
2
sixteens
eights
fours
twos
units
1
0
1
1
1
The
subscript
2
means
that
the
number
10 111
is
written
in
base
2.
Tip
2
Base
3
-
ternary,
or
trinary
Computer
lots
8
of
GB
20
logical
contexts
(gigabytes),
GB,
etc.
systems
relating
Also,
16
use
to
GB,
base
2.
computers.
32
whereas
GB,
the
64
prex
This
For
GB
kilo
is
why
you
example,
and
so
on,
usually
see
SD
rather
means
powers
cards
than
1000
of
which
10
(e.g.
2
in
Base
4
-
quaternary
Base
8
-
octal
Base
12
store
GB,
there
are
-
duodecimal
10
1000
so
meters
there
are
in
a
1024
kilometer),
bytes
in
a
in
computing,
kilobyte.
the
term
kilo
means
1024
(2
),
Base 16 - hexadecimal
(widely
used
in
computing)
E1.1 What if we all had eight ngers?
5
The
base
number
Base
2
of
has
Example
Find
a
number
system
the
two
has.
system
Base
unique
10
tells
has
symbols,
you
ten
0
how
many
unique
and
unique
number
symbols
symbols,
0
the
through
9.
1.
1
value
of
12 011
in
base
10.
3
4
3
3
2
3
1
3
0
3
3
Write
81
27
9
3
the
powers
of
3
above
the
digits,
1
0
starting
1
81
2
+
2
×
27
=
12 011
0
+
a
1
=
1
at
the
right-hand
=
139
Add
up
the
parts
Use
that
make
up
the
value
of
each
10 111
subscripts
to
show
binary
b
number
in
base
11 001
c
each
number
to
base
b
21 002
2
is
the
22 101
c
412
3
human
332
e
64
5
f
9
being
is
these
numbers
1010
in
ascending
100 011
4
The
number
Describe
1111
3
1 101 011 000
the
order:
1011
2
=
5
856
2
a
10
relationship
between
1 101 011 000
and 110 101 100
2
b
Find
the
value
of
2
1 101 011
2
Exploration
1
Use
this
table
7
1
to
explore
6
3
5
3
2187
4
3
729
questions
3
3
243
a
g
2
3
81
to
1
3
27
0
3
9
3
3
1
7
a
Explain
how
the
value
of
3
tells
you
that
the
number
1038
will
10
have
b
7
1038
digits
=
in
729
10
base
+
how
3.
309
10
Explain
this
.
10
sum
tells
you
that
the
rst
digit
of
1038
in
base
3
10
will
be
1.
Continued
6
aware
that
counting.
solving
–
4
from
without
77
8
Problem
pleasure
mind
experiences
5
counting,
Write
base.
10.
3
3
the
1 101 101
2
the
d
number.
10.
Music
Convert
a
the
1
2
2
side.
1
10
Practice
Find
+
3
139
3
1
3
1
with
E1 Form
on
next
page
Gottfried
Leibniz
it
N U M B E R
c
309
=
243
10
+
66
10
Explain
10
how
this
sum
tells
you
that
the
second
digit
of
1038
10
in
d
base
66
<
3
will
also
be
1.
81
10
10
Explain
how
this
inequality
tells
you
that
the
third
digit
of
1038
10
in
e
base
=
66
3
2
will
×
be
27
10
0.
+
12.
10
Explain
how
this
sum
tells
you
that
the
fourth
digit
of
1038
10
in
f
base
3
will
Continue
be
the
2.
process
to
nd
the
value
of
1038
in
base
3.
10
g
2
Check
Use
a
the
your
solution
method
1000
to
in
base
step
by
1
converting
to
2
The
●
Start
the
b
513
with
the
number,
remainder
If
●
Repeat
>
new
4
units
●
Here
down
0,
base
10.
base
3
c
673
the
replace
from
digit
and
n.
the
then
Divide
to
base
4.
10
digits
of
working
n
by
b
a
to
and
number
the
nd
n
in
base
b,
left.
the
quotient
q,
and
r
value
n
the
remainders
the
to
produces
the
Write
q
algorithm
with
●
into
10
following
starting
back
convert:
10
3
it
algorithm
with
of
r
the
value
beginning
to
is
the
left
used
to
with
of
of
q ;
the
any
new
you
convert
otherwise
value
have
1038
stop.
of
n,
already
into
base
and
record
written
down.
3.
10
n ÷ b
n
q
r
1038
346
0
346
115
1
115
38
1
38
12
2
12
4
0
1038
÷
3
346
=
÷
346
3
=
r
0.
115
The
r
1.
units
The
digit
3s
is
digit
is
0.
1.
Change
Change
n
to
n
346.
to
115.
2
115
4
1
1
1
0
1
÷
3
=
38
r
1.
The
3
38
÷
digit
is
1.
Change
n
to
38.
3
3
=
12
r
2.
The
3
digit
is
2.
4
Read
the
12
÷
3
=
4
r
0.
The
3
4
÷
3
=
1
r
1.
The
3
digit
is
0.
digit
is
1.
q =
0.
number
5
in
base
b
from
upwards
here:
1 102 110
6
1
Use
a
the
algorithm
1000
10
to
base
2
to
÷
3
=
0
r
3.
The
3
digit
is
0.
Stop
because
convert:
b
513
10
to
base
3
c
673
to
base
4.
10
E1.1 What if we all had eight ngers?
7
Practice
1
Convert
2
999
to:
10
Use
a
base
2
b
base
3
c
base
4
d
base
the
prefer
2
a
Find
the
value
of
472
in
base
bases.
Hence
nd
the
value
of
472
in
base
5.
8
3
By
rst
given
a
converting
to
base
10,
nd
in
base
7
b
431
5
d
the
in
1011
in
8868
base
in
6
e
base
3
h
write
the
1331
Determine
you
know.
Use
your
in
101 101
same
Benito:
2
c
base
8
which
students’
Determine
d
Find
the
values
of
to
the
part
numbers,
of
your
in
number
A
bottle
A
shipping
2
base
4
in
dierent
3
Claudio:
a,
to
a,
b,
c
and
students
and
list
any
the
such
other
value
that
the
to
a,
for
1331
b,
c
and
=
2061
d
to
nd
the
value
of
the
d
=
you
in
Explain
can
gain
8
1000
how
from
ascending
3213
=
order.
1000
c
number
packs
12
bottles
will
hold
to
12
a
box.
There
are
12
in
d
base
10.
in
each
crate.
bottles
in
each
shipping
orders
600
bottles.
boxes
in
a
crate.
crates.
bottles
customer
base.
b
of
Find
the
number
container.
of
crates
and
boxes
to
order.
customer
single
orders
bottles
on
page
price
81
for
9
Calculate
basic
per
bottles.
this
Determine
shows
four
appropriate
bottle
the
number
of
complete
E1 Form
boxes
order.
is
dierent
values
orders,
for
the
with
shaded
some
information
cells.
€1.40.
Continued
8
base
b
of
The
in
2
container
table
9
d
largest
number
missing.
base
Donatello:
information
numbers
possible
d
used
number
The
in
c
the
A
2468
3213
the
and
4
factory
this
2
bases:
Determine
full
110 213
i
Determine
A
base
9
2061
four
minimum
answer
Exploration
1
in
4
a
Use
214
f
b
answer
c
e
the
solving
students
the
base
110 111
a
b
in
2
Alberto:
a
numbers
2
Problem
Four
these
6
9
4
of
8
2
g
value
base.
223
to
numbers
10.
8
b
method
on
you
5
next
page
convert
to
dierent
N U M B E R
Customer
Total
name
ordered
Total
Notes
Containers
Crates
Boxes
Singles
cost
Mr
a
n/a
0
5
0
0
b
c
d
e
f
g
4
4
1
0
€8400
j
k
l
m
n
Antinoro
10%
discount
Mr
6000
on
Drouhin
complete
containers
i
(%
Herr
discount
h
Müller
on
the
whole
order)
Mrs
1500
n/a
Symington
5
An
employee
containers,
repeat
the
0
singles
a
Write
this
b
c
could
Mr
Why
●
When
1,
Base
2,
do
because
since
and
every
written
as
all
orders
singles,
time,
so
the
an
are
made
company
order
of
3
of
a
does
crates,
number
not
7
of
need
boxes
to
and
370.
Mr
notating
Drouhin’s
is
the
12
4,
5,
6,
problem
and
Herr
Müller’s
orders
using
these
you
12
uses
7,
8
write
ten
and
writing
most
involve
number
Mrs
in
to
relates
Symington’s
used
to
writing
numbers
order
in
this
way.
as
use
a
base
base?
10
to
What
count?
makes
the
use?
symbols
twelve
two
numbers
way
commonly
Similarly,
requires
this
3
dierent
9.
in
commonly
convenient
both
symbols
in
people
number
orders
bases.
discuss
think
a
the
number
(duodecimal)
When
extra
you
(decimal)
3,
12
the
and
number
0,
how
Describe
10
boxes
Antinoro’s,
non-denary
●
Base
be
that
convention.
Reect
ATL
crates,
headings
Explain
in
suggests
base
to
2
describe
uses
symbols,
two
but
the
whole
symbols:
you
cannot
0
numbers:
and
use
1.
‘10’
or
‘11’
digits.
bases
greater
than
10,
letters
are
used
needed.
for
the
Sometimes
lowercase
a
In
base
12,
the
symbols
are
0,
1,
2,
3,
4,
5,
6,
7,
8,
9,
A
and
and
b
letters
are
used
B.
instead
A
and
E1.1 What if we all had eight ngers?
of
uppercase
B.
9
Example
Find
the
2
value
of
A3B
in
base
10.
12
2
1
12
0
12
12
144
12
1
A
3
B
=
A3B
(10
×
The
144
+
3
×
12
+
11
×
letter
A
represents
10,
and
B
represents
11.
1)
12
10
=
1487
10
Example
Find
the
3
value
of
500
in
base
16.
n ÷ b
n
q
r
500
31
4
31
1
F
1
0
1
500
31
500
=
Convert
1
to
base
a
b
301
c
the
value
in
base
190
b
2766
e
c
The
symbol
÷
16
=
0
r
1.
The
and
(base
other
Explain
d
Give
why
10
the
20
10
51 966
16)
digital
there
number
your
digit
for
digit
G0
f
64 206
10
codes
are
often
used
as
passwords
for
home
Wi-Fi
services.
are
256
possible
2-character
passwords
in
hexadecimal.
10
Find
16
solving
Hexadecimal
b
units
47 806
10
Problem
a
The
16:
10
48 879
hubs
15.
4.
12
10
3
r
BAA
16
10
d
1
r
10:
210
Find
=
31
3
12
2
16
=
is
15
4.
is
Change
F.
n
Change
to
n
31.
to
1.
2
16
Practice
a
16
1F4
10
1
÷
÷
of
answer
possible
to
a
E1 Form
10-character
suitable
degree
of
hexadecimal
accuracy.
passcodes.
is
1.
Stop
because
q
=
0.
N U M B E R
Counting
US
in
President
Abraham
“Four
score
a
nation,
new
men
A
is
are
the
and
seven
a
group
number
‘quatre-vingt-dix’
Objective
iii.
move
Dierent
C:
number
our
liberty,
or
set
system,
or
‘four
Mayans
dierent
bases
show
and
that
Exploration
The
ago
in
of
20,
80
is
used
up
19.
to
had
the
For
Write
The
For
the
twenties
forms
symbols
you
a
can
of
base
20
tell
the
and
1s
dierent
to
forms
and
or
continent
that
all
seven’
‘four
is
87.
twenties’
and
90
representation
of
dierent
system,
but
representation.
symbols
and
In
this
number
bases.
only
three
and
symbols:
5
in
combination
leave
bars
4,
0–19
do
in
10,
full
base
wrote
enough
one
and
the
in
the
19
in
Mayan
purpose
Eectively,
digits
say,
=
same
10.
space
between,
with
14
6
to
create
numbers
vertically
between
a
10
units
(two
11
in
the
=
in
Mayan
are
the
ascending
digits
so
horizontal
place
and
16
numerals.
they
place
one
place
place
1
20
s
1
20s
1
1s
order.
that
bars)
in
numerals
Mayan
the
It
they
and
as
the
digits.
was
could
105
twenties
(two
place).
×
20
)
place
5
place
0
place
2
(1
this
2
s
20s
on
proposition
ten’.
2
20
score
1
=
they
dierence
horizontal
‘four
between
number
3
0–9
value,
important
so
mathematical
are
move
symbols
from
from
place
forth
the
3
numbers
numbers
numbers
to
began:
example:
=
1
brought
dedicated
‘quatre-vingt’
0
They
fathers
and
Address”
Communicating
between
Exploration,
years
“Gettysburg
equal.”
means
French
Lincoln’s
conceived
created
‘score’
In
ATL
20s
13
2
+
(1
×
20)
+
1
=
421
(5
×
20
)
+
(0
×
20)
+
13
=
Continued
2013
on
next
page
E1.1 What if we all had eight ngers?
11
2
Find
the
value
of
these
Mayan
a
3
Write
4
numerals:
b
these
numbers
in
c
d
Mayan:
a
806
b
2005
c
10 145
d
16 125
e
43 487
f
562 677
g
100 000
h
2
Find
the
value
of
this
sum,
showing
all
your
million
steps:
+
The
Mayan
people
kept stingless bees for
honey.
Technically,
stingless
have
Express
your
5
Explain
why
6
Compare
answer
in
Mayan
notation
as
well
as
in
base
Mayan
system
is
part
additive
and
part
place
value.
a
are
sting
only
our
base
10
number
system
with
the
Mayan
system
and
any
advantages
Per forming
C
●
How
are
dierent
In
or
a
calculation
+
8 9
2
3
7
4
you
add
you
a
digit
the
units,
number
and
one
has
over
the
other.
operations
operations
in
base
in
other
bases
similar
to
and
10?
this:
then
the
that
operations
mathematical
from
greater
‘carry’
12
like
disadvantages
the
than
10
to
tens,
or
the
E1 Form
then
equal
next
the
to
10
hundreds.
(the
column.
base)
When
you
the
write
addition
down
so
small
from
about
one
as
that
is
painful
write
as
down
do
but
10.
they
the
bees
stingers
the
gives
units
a
mosquito
bite.
N U M B E R
You
can
add
‘carrying’
when
Example
Find
the
numbers
a
in
other
number
is
bases
greater
in
the
than
same
or
way
equal
to
as
the
decimal
base
numbers,
number.
4
value
of
465
+
326
7
7
Units
column:
5
+
6
=
11
.
In
base
7
this
is
1
seven
and
4
units,
or
14
10
Write
4
in
the
units
column
7
and
carry
a
1
Sevens
into
the
column:
sevens
column.
6
1
+
2
+
=
9
=
12
10
7
2
7
column:
4
+
3
+
1
=
8
=
11
10
So,
465
+
326
7
1124
7
Reect
You
=
could
7
and
work
discuss
out
465
+
4
326
7
●
converting
465
●
adding
●
converting
For
you
method
the
and
‘borrow’
Example
Calculate
a
326
result
you
10
in
adding,
as
in
Example
4,
or
by:
into
base
10
7
would
subtraction,
by
7
7
Which
7
back
you
can
to
prefer?
‘borrow’
base
base
7.
Explain.
the
base
number,
in
the
same
way
that
10.
5
783
9
267
9
7
>
3,
so
you
cannot
subtract
7
from
3.
1
Rewrite
3
units,
same
7
8
as
8
3
as
nines
13
7
7
13
7
9
Continued
on
next
7
=
because
eighty-ones
nines
7
9
13
3,
and
units,
Find
7
by
and
7
writing
is
the
eighty-ones.
it
in
base
9
12
9
10
7
10
=
5
10
=
5
9
page
E1.1 What if we all had eight ngers?
13
10:
7
6
=
1
in
base
9
and
in
base
7
So,
783
267
9
4
124
+
321
8
453
+
261
c
563
8
+
241
8
+
757
8
8
b
341
6
153
6
c
1231
6
402
6
6
Calculate:
a
115
23
8
+
45
8
b
2231
8
125
7
216
7
c
8463
7
+
728
9
541
9
+
18
9
9
Calculate:
a
92A1
+
4436
12
Vorbelar
b
10 000
12
Problem
5
77
8
231
6
4
b
8
Calculate:
a
3
9
Calculate:
a
2
=
515
9
Practice
1
=
2
10.
123
12
12
solving
the
alien
does
not
count
in
base
10,
but
in
order
to
make
things
If
easy
for
us
to
understand,
she
uses
our
base
10
symbols
in
the
usual
you
can’t
She
calculates
216
+
165
and
obtains
the
answer
think
count,
the
value
of
216
165,
giving
your
answer
using
Vorbelar’s
three
pocket
giving
two
and
discuss
performing
added
or
subtracted
Exploration
There
are
a
many
calculation,
are
in
the
is
it
same
important
base?
Why
that
or
the
why
two
numbers
not?
of
setting
out
the
calculation
742
×
36
10
Here
is
one
being
4
ways
10
approach.
First
multiply
with
6
×
2
the
=
2
12,
in
so
the
the
top
row
units
write
2
by
6,
one
column
at
a
time,
starting
column.
in
the
units
column
and
1
in
the
tens
column.
Continued
14
E1 Form
of
in
your
and
Fido
then
only
them.
5
–
When
dog
base.
biscuits
Reect
try
403.
putting
Find
dogs
order.
on
next
page
Phil
Pastoret
5
N U M B E R
6
×
4
tens
write
6
×
7
2
in
=
calculate
tens
×
2
=
3
tens
×
4
tens
tens
×
7
the
742
tens,
1
=
in
×
×
completes
Add
together
above
the
to
to
write
4
the
so
in
3
1
add
the
2
Try
times
to
use
write
6
in
the
Here
Use
is
a
the
so
tens
column
and
to
the
2
in
the
column.
write
column.
2
in
the
hundreds
thousands.
=
21
and
×
3
two
thousands,
write
2
in
so
the
add
ten
1
to
the
thousands
1
in
the
column.
tens.
parts
calculate
already
727
×
found.
736
.
Explain
how
you
have
10
tables.
the
same
method
to
calculate
727
multiplication
table
to
help
table
you
in
base
calculate
×
736
.
Describe
any
diculties.
9
9.
727
×
9
4
2
tens
9
3
the
thousands
10
used
in
tens.
hundreds,
the
742
4
column.
6.
so
12
column
This
method
add
hundreds,
and
hundreds
thousands
Use
6
and
so
42
742
3
column
1
=
column
completes
Now
3
tens,
hundreds
hundreds
hundreds
This
24
the
736
9
×
1
2
3
4
5
6
7
8
10
1
1
2
3
4
5
6
7
8
10
2
2
4
6
8
11
13
15
17
20
3
3
6
10
13
16
20
23
26
30
4
4
8
13
17
22
26
31
35
40
5
5
11
16
22
27
33
38
44
50
6
6
13
20
26
33
40
46
53
60
7
7
15
23
31
38
46
54
62
70
8
8
17
26
35
44
53
62
71
80
10
10
20
30
40
50
60
70
80
100
Explain
why
you
need
this
table
for
long
multiplication
in
base
9.
E1.1 What if we all had eight ngers?
15
Practice
5
1
multiplication
Here
is
a
in
base
5:
×
1
2
3
4
10
1
1
2
3
4
10
2
2
4
11
13
20
3
3
11
14
22
30
4
4
13
22
31
40
10
10
20
30
40
100
Use
a
it
to
complete
×
123
Copy
these
multiplications:
24
5
2
table
b
401
5
and
×
133
5
complete
×
1
2
1
1
2
this
multiplication
3
c
213
5
× 1310
5
table
for
4
5
10
2
4
5
10
2
4
12
4
4
12
5
5
10
10
base
5
6.
20
3
Hence
a
20
complete
×
155
these
multiplications:
b
212
6
Problem
Find
40
23
6
3
40
two
×
315
6
6
solving
numbers
(both
greater
than
1)
whose
product
is
1215
6
Because
long
multiplication
multiplication
multiplication
to
convert
convert
the
Division
easiest
the
or
that
How
use
the
16
to
to
you
base
2
its
10
E1 Form
o
for
most
table
write
cases
out
it
is
multiplication,
facts,
a
easier
and
then
base.
and
then
counting
table
facts,
divide
so
it
is
usually
them.
in
of
their
through
exists
system:
base
2?
function?
most
pass
memory
counting
in
the
can
numbers
inuence
that
but
You
multiplication
base
(binary)
signals
computer’s
for
need,
required
better
form
multiplication
tricky.
perform
binary
be
knowing
quite
you
10,
the
to
on
is
base
base
knowing
at
does
electrical
symbols
the
back
on
peek
Would
As
for
relies
bases
numbers
●
the
‘o ’.
two
relies
●
Computers
is
table
numbers
convert
A
D
dierent
numbers
also
to
in
0
in
and
operations.
the
two
1.
computer
states
(on
One
chips
or
reason
are
o )
it
for
either
uses
this
‘on’
just
N U M B E R
Reect
●
Create
tables’
●
Try
●
Can
You
and
and
long
with
multiplication
in
the
using
may
numbers
a
you
nd
it
new
to
work
bases,
Hexadecimal
represents
a
passwords
is
hand
2
binary.
Can
you
multiplications
in
4
in
a
to
or
base
because
steps.
‘learn
routers
in
so
in
your
times
2.
base
there
are
more
fewer
is
in
why
often
lot
use
of
digits,
symbols
In
have
in
numbers
fewer
digits.
hexadecimal
example,
That
example)
a
digits.
symbol
For
2?
are
numbers
each
digits.
binary.
for
base
There
have
and
because
binary
0 011 1111
and
of
2
symbols,
4
in
110 110 100
numbers
computers
bits,
×
lot
thus
more
3F
in
built-in
digits
0-9
and
the
you
Draw
10
show
represents
number
to
nd
nger
32
5
nger/thumb
diagram
Find
a
slow
and
are
modems
diagrams
extended
help
some
involves
converts
(in
in
A-F.
This
1
of
table
1 011 100 110
bases,
used
Exploration
An
out
there
string
hexadecimal
These
perform
multiplication
small
6
base?
multiplication
with
larger
letters
discuss
in
10
number
diagrams
b
way
of
counting
1
and
zero.
base
the
a
represents
to
show
68
10
in
binary
each
by
binary
curled
This
represented
in
a
diagram
the
using
one
your
ngers.
represents
represents
diagrams
below.
0.
one.
It
may
rst.
of
c
these
base
101
10
10
numbers
d
in
binary:
250
10
E1.1 What if we all had eight ngers?
17
Reect
●
Is
●
How
counting
in
●
and
is
in
base
counting
base
When
discuss
2
in
on
7
your
base
2
ngers
on
your
a
more
ngers
ecient
more
way
dicult
of
counting?
than
counting
10?
might
it
be
useful
to
try
to
count
in
base
2
on
your
ngers?
Summary
Our
number
system,
often
called
Arabic
numerals,
The
subscript
2
means
that
the
number
10 111
is
2
is
a
place
tells
you
value
its
system.
The
position
of
a
digit
written
The
Our
or
number
base
place
ten,
to
system
or
value
column
is
also
denary,
column
its
is
known
because
10
times
the
the
as
decimal,
value
value
of
of
base
unique
each
has
the
Base
2
10
×
10
×
10
×
thousands
hundreds
3
tens
2
10
10
•
units
1
10
•
tenths
0
•
10
In
10
A
unique
has
the
system
number
number
two
make
a
place
value
unique
tells
you
system
symbols:
symbols:
0
0
how
uses.
Base
through
and
many
10
9.
1.
base.
In
system
binary,
or
write
used
numbers
for
the
in
extra
bases
greater
symbols
than
10,
needed.
12
B
the
(or
symbols
are
sometimes
0,
1,
2,
3,
lowercase
4,
5,
letters
6,
a
7,
and
8,
9,
b).
point
with
can
base
add
way
as
numbers
decimal
in
other
bases
numbers,
in
the
‘carrying’
when
any
a
number
you
are
base
and
same
natural
number
•
You
can
a
1
decimal
You
of
10
letters
•
2.
symbols
ten
When
•
base
right.
×
•
in
value.
two,
number
is
greater
than
or
equal
to
the
base
each
number.
column
is
worth
×
•
•
twice
2
×
sixteens
•
the
column
2
eights
×
fours
to
2
its
×
right.
2
twos
units
•
•
For
division
it
usually
is
4
3
Mixed
1
Find
2
2
1
2
base
10,
the
value
in
base
10
of:
1101
b
answer
back
3
can
1 000 101
d
1221
You
45
3
12 000
f
4523
8
34 337
h
11
the
100
value
in
in
base
the
base
2
specied
b
100
in
base
4
d
255
in
base
3
f
in
base
2
in
1689
base
12
h
base
5
456
in
base
numbers
to
the
base
in
the
question.
measure
One
angles
degree
=
in
60
degrees,
minutes
minutes
or
60′,
and
and
minute
degrees,
=
60
15
seconds
minutes
angle
or
and
60″.
30
45° 15′
30″
=
seconds.
Convert
these
measures:
a
42.8°
into
b
90.15°
c
2.52°
d
25°15′
e
62° 30′
30″
into
base
10
(including
a
decimal)
f
58° 12′
20″
into
base
10
(including
a
decimal)
degrees,
minutes
and
seconds
in
base
12
j
degrees,
minutes
and
seconds
4
2785
in
into
degrees,
into
base
10
minutes
and
(including
a
seconds
decimal)
base
12
Calculate,
in
the
giving
your
answers
in
the
base
used
question:
10
in
base
16
10
l
15 342
in
base
16
a
1101
1011
2
E1 Form
+
2
10
c
18
into
12
10
10
1049
in
10
10
k
3
8162
10
i
base
10
1008
271
in
10
10
g
the
convert
of:
100
10
e
then
4EA8
Find
c
calculations,
18A9
16
a
the
2
8
2
convert
base,
110 110
3
i
dierent
practice
2
g
to
a
2
one
e
ecient
in
0
2
2
c
most
do
seconds.
a
multiplication
•
to
2
and
1001
b
11 101
2
+
+
2
10 111
2
+
1 101 101
2
101 101
2
N U M B E R
d
122
+
2101
3
f
e
2122
3
3012
+
10
+
20
012
3
Problem
3
solving
332
4
d
4
Ronnie
has
331 213
grams
in
weights.
4
g
33
010
+
312
212
4
+
Donnie
101
4
has
565
+
718
12
i
1A91
12
+
AB0
+
10B8
12
l
k
6E
12
F00D
+
+
wishes
Ronnie
and
the
out
weights.
the
Donnie
same
combined,
amount
using
16
the
weights
FACE
she
has
from
only
her
sets.
three
of
Remembering
each
type
of
that
weight,
16
Calculate,
in
weigh
AC
determine
5
to
12
16
16
in
3302
12
as
j
grams
4
4
Connie
h
231 232
giving
your
answers
in
the
base
the
weights
Connie
should
use.
used
e
question:
Ted
weighs
out
grams.
1330
Ed
weighs
out
4
twice
a
11 011
101
2
b
111 011
2
that
10 110
111
2
e
3011
2210
f
that
3
44 125
1045
h
B791
6
FEED
89A0
12
the
he
9
A
US
current
time
was
10:45,
use
addition
2
combined.
as
Remembering
only
the
three
of
weights
each
Ned
type
of
should
weight,
use.
committee
monetary
system.
decides
All
coins
to
streamline
will
be
phased
except
to
nd
the
time
that
minutes
from
for
1
cent,
5
cents
and
25
cents.
All
and
bills
(bank
notes)
will
be
phased
out
is:
and
b
Ted
16
subtraction
35
has
congressional
existing
a
and
12
out
If
Ed
BEEF
16
6
twice
6
the
i
as
determine
5150
6
6
out
102
4
50 112
weighs
2
3
231
4
g
d
2
Ned
11 101
2
much
c
amount.
replaced
with
bills
valued
$1,
$5,
$25,
$125
and$625.
now
1
hours
from
now
a
A
mathematician
suggests
that
the
dollar
2
should
c
3
hours
and
40
minutes
from
1
hour
and
e
6
hours
19
minutes
50
100
cents
minutes
Calculate,
giving
your
answers
in
the
base
the
worth
125
as
it
is
cents,
currently.
Suggest
reasons
this
why
might
be
the
a
mathematician
good
idea.
used
ii
in
be
ago
thinks
7
to
ago
i
and
revalued
now
not
d
be
Suggest
reasons
why
it
might
not
be
a
question.
good
b
Find
idea.
in
540
base
5.
10
Either
create
multiplication
tables
like
the
Hence
ones
in
Practice
5,
or
convert
the
base
10
and
then
convert
c
110
×
10
2
b
1011
2
×
11
2
c
1101
2
×
2101
×
21
3
e
14
3
×
23
5
A
set
of
weights
f
24
5
contains
×
Find
powers
of
4.
There
in
the
set.
The
three
lightest
heaviest
is
16 384
in
grams
Find
the
total
Rania
of
each
weight
the
one
of
the
is
1
g,
Write
create
there
down
of
these
weights,
size
a
the
total
are
three
above
weights
weight
only
Find
the
value
three
of
base
5
and
246
in
base
5.
$732
a
to
her
payment
credit
of
card
company.
$246.
the
total
amount
and
the
need
to
least
outstanding
number
of
(still
notes
she
repay
the
amount.
six
weights:
In
real
of
the
next
of
of
life,
bases,
but
systems
have
do
not
use
combinations
of
wanted
why
to
weights,
size
and
this
you
26
in
nd
would
the
total
one
of
the
next
two
of
the
size
notes
size
follow
irregular
patterns.
For
example,
in
that.
g.
each
2313
monetary
two
would
use
Remember
type
of
UK,
valued
to
that
there
£5,
Suggest
base
£10,
to
£1
£20
reasons
convenient
weight.
are
and
and
why
use
a
it
£2
coins,
then
be
perfectly
not
be
regular
base
above.
10.
relevant
weight
size,
after
of
if
you
three
three
of
1
the
notes
£50.
would
4
Explain
under
10
owes
suggested
c
$540
and
the
b
somebody
size
that
and
give
g.
weight
smallest
in
makes
regular
of
to
that
e
a
that
5
would
the
need
732
owed),
weight
notes
131
5
weights
are
of
system.
10
Find
are
number
2
She
8
would
new
101
2
d
d
smallest
back.
the
a
the
numbers
you
to
nd
g
next
that.
E1.1 What if we all had eight ngers?
19
as
Review
in
In
and
the
1960s
context
1970s,
NASA’s
Pioneer
and
3
The
scientists
also
encoded
some
information
ATL
Voyager
beyond
in
case
life
the
the
solar
craft
forms.
they
or
missions
system.
was
could
could
Each
tried
be
be
spacecraft
to
by
travel
by
messages
extra-terrestrial
choose
universally
measured
to
contained
intercepted
Scientists
hoped
which
launched
things
that
about
our
about
quantities
measured,
whereas
communicated,
can
aliens.
be
If
an
alien
culture
has
a
number
system,
there
good
chance
that
they
will
have
some
sort
value
system.
They
will
probably
have
of
whole
of
prime
to
include
numbers,
numbers.
One
and
maybe
suggestion
an
for
year.
per
The
prime
in
a
numbers
binary
Write
are:
b
List
c
Hence list
For
the
the
10,
to
communicate
binary.
11,
the
The
rst
111
and
101,
values
of
these
next
ten
the
prime
a
rst
15
prime
and
table
in
vertical
Here
these
the
the
—
some
symbols
our
Earth
many
times
while
orbiting
information
the
about
solar
the
system:
length
Day
length
Ear th
days
1408
Ear th
hours
1011.
Venus
224.68
Ear th
days
5832
Ear th
hours
numbers
in
Ear th
365.26
Ear th
days
24
Ear th
hours
Mars
686.98
Ear th
days
25
Ear th
hours
denary.
in
Jupiter
11.862
Ear th
years
10
Ear th
hours
Saturn
29.456
Ear th
years
11
Ear th
hours
Uranus
84.07
17
Ear th
hours
Neptune
164.81
16
Ear th
hours
binary.
Ear th
years
are
did
not
known
—
they
and
used
Calculate
and
in
year.
number
Give
your
of
Earth
answer
hours
in
in
a
denary.
and
b
given
the
to
|.
sequences
years
use
only
horizontal
Ear th
binary
using
Hence
Mars
nd
year.
the
number
Give
your
of
Mars
answer
in
days
in
a
denary.
|.
whether
—
or
|
represents
1.
Find
(to
the
nearest
whole
number)
the
Justify
of
Mars
days
in
a
Mars
year
in
binary.
answer.
Describe
and
365
how
axis
some
87.96
number
your
own
Mercur y
c
Determine
is
that
each
numbers
numbers
scientists
symbols
Instead,
lines:
are
its
gives
Year
Mars
humans.
(roughly)
that
in
solving
messages,
0s;
quantity
‘days’
rst
a
1s
of
than
units
message
the
in
One
rather
require
awareness
ve
numbers
not.
number
has
information
counted,
an
(decimal).
Problem
2
in
down
denary
be
does
the
because
around
Planet
might
is
be
measurements
Earth
year
rotates
planets
idea
counting
can
used
of
sun.
place
that
They
is
it
a
system.
because
counted
planet’s
days
1
solar
|,
each
the
next
sequence
three
and
predict,
using
—
d
terms.
Find,
for
in
each
binary,
of
the
the
number
other
outer
of
days
planets
in
a
year
(Jupiter
to
Neptune).
a
|
,
|—
b
|
,
|
,
c
|
,
||
,
—,
|—
—|
,
|—
—
—
—,
||—
—|
e
|—,
||
,
|—|
,
|—
—
—,
||—|
,
Describe
when
20
|—
—|
,
||—||
,
E1 Form
|—|—
—
any
problems
you
would
encounter
|—|—|
—|
for
trying
Venus.
to
perform
a
similar
calculation
Multiple
E2.1
Global
doorways
context:
Objectives
●
Performing
function
●
Using
Inquiry
operations
with
functions,
including
function
composition
to
solve
questions
●
How
●
What
●
How
C
to
do
you
add
and
subtract
functions?
is
function
composition?
can
model
you
use
function
real-world
composition
situations?
functions
●
Draw
innovation
real-world
D
ATL
technical
Does
a
composite
function
have
a
unique
decomposition?
Critical-thinking
reasonable
conclusions
and
generalizations
21
SPIHSNOITA LER
Decomposing
and
F
composition
problems
●
Scientic
Y
ou
●
should
draw
already
mapping
know
diagrams
how
for
1
to:
Draw
functions
a
mapping
represent
the
the
diagram
function
to
shown
on
graph.
y
2
0
6
4
x
2
2
4
6
2
4
6
●
nd
the
range
natural
of
a
domain
and
2
If
function
f
:  → ,
domain
a
f ( x )
nd
and
=
2x
the
range
+
natural
of :
3
b
f ( x ) =
x
2
1
c
f ( x ) =
d
f ( x ) =
x
x
2
2
e
●
evaluate
functions
f ( x ) =
3
a
Function
F
●
How
●
What
The
function
The
function
Example
do
is
the
you
add
function
is
is
and
subtract
− 3.
f (4)
x
Find:
b
f (
1
)
c
f (3 x )
functions?
composition?
equivalent
to
equivalent
.
to
1
function
.
given
by:
a
b
a
=
=

2 x
4 −
operations
and
Write
f ( x ) =

Substitute
=
b
2 2
E2 Relationships
f ( x ) and
g( x )
and
simplify.
A LG E B R A
ATL
Exploration
has
domain
has
has
Write
each
:
two
domain
function
a
rule
for
functions.
mapping
Example
a
State
b
Write
a
f
the
has
For
so
.
{x
:
and
the
x
≠
1}.
state
its
domain.
c
domain
Justify
your
3x
≠
has
function
with
a
formed
by
adding
mathematical
or
subtracting
explanation
or
a
,
domains
of
f,
g
and
and
.
h
function
2
⇒
and
state
its
domain.
.
≠
x
0,
≠
Exclude
values
of
x
that
make
the
denominator
Exclude
values
of
x
that
make
the
radicand
zero.
domain
h,
has
k ( x )
2
a
rule
2
the
g,
of
d
diagram.
domain
3x
For
b
0}.
.
,
h
≥
b
Find
g
x
domain
a
2
{x
domain
has
1
1
1
3x
≥
0,
so
1
≥
3x
⇒
≥
x
negative.
domain
=
( g
f
)( x )
2

=

=
k
has
−
(2x
+
1)
Write
with
a
common
denominator.
domain
E2.1 Multiple doorways
23
For
(
two
f
functions
g)( x )
is
Practice
the
f
( x )
and
g ( x ),
intersection
of
the
the
domain
domains
of
of
(
f
f
+
( x )
g)( x )
and
and
the
domain
of
g ( x ).
1
2
h (a )
1
=
2a
+ 1 and k ( a )
State
the
domains
Find
(h +
k )( a )
3
of
and
=
h
− ( a + 3).
and
state
k
its
domain.
2
2
h( x )
=
x
− 5x
3
h( x )
=
2x
and g ( x )
=
5 −
x .
Find
( g
–
h)( x )
and
state
its
domain.
2
4
State
the
Find
(h +
h (c )
=
Find
2c
−
x
and
domains
k )( x )
k(x )
of
and
h
state
− 5  and  k ( c )
(k
h )( c )
=
and
3x
and
its
=
+ 1 .
k.
domain.
2 − 3c .
state
its
domain.
1
5
g (x ) = 6x
− 8, h ( x ) =
and k ( x ) =
x
+ 4 .
Find
( g
+ h −
k )( x )
and
2x
(k
− h −
g )( x ) ,
and
state
1
6
f
( x )
and
g ( x )
=
x
a
State
b
Write
the
domain
down
Problem
of
f
and
domain
a
function
f
b
Write
a
function
g
c
Write
a
function
with
composition
mapping
maps
the
Domain
diagram
elements
of
the
of
f
+
of
with
So
the
with
is
a
domain {x
shows
A
:
domain {x
way
of
B,
:
x
:
x
x
≠
≠
g
3}.
≠
0
combining
three
onto
sets
and
g
A,
B
0}.
and
≠
3}.
functions.
and
maps
x
the
C
and
two
elements
functions
of
B
onto
1
1
2
4
3
3
9
13
f
=
{1, 2, 3}
Range
diagram,
f
maps
2
f
(2)
=
4
and
g (4)
=
−3
f
(1)
=
1
and
g (1)
=
3
f
(3)
=
9
and
g (9)
=
−13
24
of
g.
domain {x
=
From
domain
solving
Write
The
f
the
2
a
Function
domains.
1
=
x
7
their
onto
of
f
=
E2 Relationships
and
and
{1, 4, 9}
of
Range
of
g
=
g
then
g
maps
4
onto
3.
{3,
g
C :
3
Domain
4,
f
3,
−13}
B
is
the
range
of
f
C
is
the
range
of
g
A LG E B R A
Applying
gives
a
f
to
composite
two
elements
from
function
A
of
of
to
f
A
C.
and
and
The
g.
A
then
applying
function
that
composite
composite
You
From
write
the
this
domain
The
range
of
of
( x )
=
g (
=
f
a
f
of
as
g o
g o
Example
is
function
f
g o
f
is
( x ))
to
function
the
A
is
to
a
elements
C
is
of
called
B
the
combination
of
f
(1)
=
3
(2)
=
−3
g o
f
(3)
=
−13
the
domain
range
means
means
from
A
to
B
,
b
List
c
Find
d
State
is
say
f
the
g
of
‘g
of
g :
f
of
:
{3,
function
f
of
{1,
2,
3,
followed
by
g.
g
followed
by
f
and
f
is
a
mapping
diagram
the
elements
of
A,
B
for
and
the
can
also
by
function
write
g (
f
g.
( x ))
=
g (
f
(2))
=
g (4)
=
−3
the
domain
and
B
to
C.
,
composite
.
function
.
C
and
{1,
You
from
,
.
range
of
.
B
C
g
=
followed
3}.
function
and
A
A
x ’.
f
13}.
f
a
a
the
3
,
Draw
and
g o
is
( g ( x ))
function
a
f
and
diagram,
The
b
g
maps
functions.
The
g
the
mapping
f
1
2
7
3
4
5
5
6
3
3,
5},
B
=
{2,
4,
6},
C
=
{3,
5,
means
g
followed
7}
c
d
Domain
Range
of
of
is
is
{1,
{3,
3,
5,
5}.
7}.
E2.1 Multiple doorways
25
by f
Reect
ATL
and
●
Explain
●
For
the
belong
●
If
the
why
discuss
it
makes
sense
function
to
the
range
,
domain
of
f
1
is
the
of
to
write
explain
‘function
why
,
f
followed
the
1
f
f
2
is
a
(1)
Draw
b
List
c
Find
g o
d
State
the
For
a
entire
b
c
domain
The
of
f
(1), g o
composite
the
where
mapping
Find
b
State
f
h o
f
the
5
=
2x
a
Draw
b
State
For
two
(a )
of
g,
what
is
the
domain
of
?
g(f (x))
f
function
−2,
g (9 )
from
=
2,
composite
B
to
g (13)
C.
=
6
function
g o
f
C
f
of
(3)
g o
draw
f
a
(x )
mapping
diagram,
range.
2, h (3)
b ,
=
the
g o
= 1, g ( −2 )
=
a
=
(b )
here
=
4, g (0 )
5, h ( 4 )
=
c ,
f
=
(c )
represents
=
7,
=
2,
f
h (2)
(2)
=
d , g (d )
=
3,
=
−1, h ( 4 )
f
(5)
=
x , g ( c )
=
4,
=
−3, h (1)
f
f
domain
and
(8)
and
and
g
(h o
range
( x ) =
mapping
range
f
the
domain
●
the
range
of
f
●
the
range
of
g
of
x
of
− 5.
diagram
of
functions
the
(7 )
=
y , g (b )
two
=
5
6
=
z
h
f
h
●
2 6
.
must
f
6
36
4.5
8
64
8
10
100
)(10 )
h o
f
(x )
solving
the
Determine
=
( 6 ), h o
+ 3
a
and
range
and
is
for
function,
diagram
and
Problem
(x )
f
B
g
g (5)
and
and
g ( −5)
h (1)
A,
(2)
domain
where
,
f
domain
where
a
f
= 13 ,
elements
functions
4
and
the
g
f
(3)
B
diagram
o h ,
g o
f
to
mapping
state
f
A
a
each
h o
from
( 2 ) = 9,
f
a
then
3
5,
x,
as:
2
function
=
of
g’
g
(x)
Practice
image
by
g o
f
and
f
is
and
is
C
number
The
to
domain
represent
g o
of
f
f
is
{0,
1,
3,
5}.
( x ).
(x ).
g :
A
=
{1,
2,
domain
=
of
{7,
8,
3}
of
g
is
B
=
{4,
5,
6}
9}
dierent
E2 Relationships
composite
functions
g o
f
.
12.5
A LG E B R A
2
The
set
functions
of
real
f
(x )
numbers.
=
2x
and g ( x )
Applying
=
x
function
− 1 both
f
and
have
then
domain
function
g
and
to
range
input
x
∈,
value
3
the
gives:
2
x
f (x)
=
2x
g(x)
=
x
−
1
2
2
3
f
(3)
=
6
and
Applying
function
g (6)
=
function
g o
f
35
f
×
−
or
and
then
g o
function
g
f
(3)
to
=
1
35
input
or
value
x
gives
f (x)
g(x)
g
2
2x
2
ATL
f
(x )
=
(2 x )
Reect
●
For
●
Is
the
Example
and
4 x
=
35
composite
2x
(2x)
f (x)
2
−
1
4x
−
1
− 1
discuss
2
functions
equal
and
to
?
,
nd
Explain.
4
.
b
a
(3))
2
− 1 =
and
a
the
f
(x ):
x
g o
g (
Find:
c
d
Start
with
the
inside
function
f.
⇒
b
Start
with
the
inside
function
g
⇒
c
d
g ( x )
goes
in
here.
f ( x )
goes
in
here.
E2.1 Multiple doorways
27
Practice
3
2
1
f
(x )
a
f
e
=
g (x )
=
x
+ 1 and h ( x )
( g (3))
g (
f
i
3x ,
f
b
(0 ))
g (h (
f
o h (x )
f
j
Problem
(
g (
f
x
+ 2 .
Find:
2 ))
(
f
=
c
4 ))
f
g
( x ))
o h (2)
h o h (0 )
k
g o
g (x )
d
h
o
g (
h
h(
l
h ( g ( x ))
f
(
0.5)
1))
solving
2
2
A
composite
a
a
suitable
b
the
f
function
function
composite
( g ( x ))
=
2( x
− 5) + 1
and
f
(x )
=
2x
A
composite
Find:
g (x )
function
g (
f
( x ))
1
1
3
+ 1.
function
g( f
( x )) =
,
x
≠
±1
and
,
g (x ) =
x
≠ 1.
Find:
2
x
a
a
suitable
b
the
function
composite
f
and
function
f
state
o
x
1
its
domain
and
1
range
g (x )
2
4
f
(x )
=
7x
− 2
and
g (
f
( x ))
=
3x
− 70.
Find
g (33).
In
f
g
x
5
The
graphs
7x
show
the
functions
f
−
2
and
=
g.
33
g(33)
Both
y
have
domain
6
5
5
4
4
2
1
1
x
2
1
the
graphs
3
of
f
4
f
o
g (0 )
b
e
f
o
g (1)
f
sells
a
works
of
enough
Explain
to
in
Determine
hours
plus
get
what
represent
b
30
$500,
x
a
f
g,
o
f
g o
per
2%
the
g (x )
=
2 8
≤
6}.
f
x
0
6
2
1
3
4
5
6
evaluate:
(2)
c
(5)
g o
g
week
at
a
travel
commission
on
f
f
(3)
( g (
f
agency.
sales
d
4 x
+ 4
f
g (4 )
She
over
is
paid
$5000.
a
One
weekly
week
she
the
functions
f
( x )
=
0.02x
and
g ( x )
=
x
5000
context.
whether
and
g o
( 4 )))
f
o
g ( x )
or
g
o
f
( x )
represents
her
commission.
2
7
x
commission.
and
this
5
and
a
salary
≤
g(x)
2
0
Emily
0
3
f(x)
6
:
y
6
3
Using
{x
(x )
=
x
− 64.
E2 Relationships
Find
x
such
that
f
o
g (x )
=
0
4,
rst
nd
x
A LG E B R A
Modelling using compositions of functions
C
●
How
Objective
i.
identify
In
of
D:
2,
a
diver
increases
This
t
you
need
in
real-life
authentic
to
identify
to
real-world
situations?
contexts
real-life
the
model
situations
meaning
of
the
given
functions
in
light
Let
(0)
Let
deeper
the
into
pressure
the
diver’s
on
Depth
d
40
32
60
44
80
65
100
80
120
90
d
=
below,
pressure
pressure
is
(atm)
be
0
be
Explain
3
Use
the
4
Explain
the
and
the
f
what
tables
to
to
at
mass
body
dierent
of
water
above
the
times.
pressure
in
exerted
atmospheres
on
a
diver
(atm),
at
where
dierent
1
atm
is
30
40
50
60
70
80
90
1
2
3
4
5
6
7
8
9
10
connecting
nd
nd
Find
f
7
g
time
(60).
connecting
=
the
level.
20
15.
diver
increases.
10
=
g (60)
the
diver’s
0
function
how
sea
function
(20)
the
measured
at
(meters)
p
shows
ocean,
the
(meters)
15
table,
the
depth
20
f (t)
g
2
0
Pressure
2
of
0
The
Depth
so
shows
atmospheric
f
composition
mathematics
will
(seconds)
next
depths.
1
function
elements
descends
and
table
Time
The
use
problem.
Exploration
As
you
Applying
relevant
Exploration
the
can
t
and
Explain
the
depth
d
depth
what
and
f
d
(60)
means.
pressure
p.
means.
o
the
f
(120).
time
Explain
before
the
what
diver
g
is
o
at
f
(120)
a
means.
pressure
of
9
atm.
E2.1 Multiple doorways
2 9
Practice
4
Problem
1
The
price
solving
p
of
an
orange,
in
cents,
is
a
function
f
of
its
weight
w,
in
grams.
Oranges
are
domesticated
Weight
so
w
100
125
150
200
250
300
350
400
you
are
unlikely
nd
them
to
(grams)
Price
p
naturally
12
15
20
26
33
36
45
50
(cents)
The
weight
Radius
w
of
an
orange,
in
grams,
is
a
function
g
of
its
radius
r,
in
cm.
r
3.8
4.2
4.3
4.8
5.1
5.3
190
210
215
240
250
265
(cm)
Weight
w
(grams)
a
Find
f
(125)
b
Find
g (4.2)
c
Evaluate
d
Determine
between
2
A
f
as
a
increases
and
o
(350).
g (5.3).
and
in
a
explain
function
radius
of
computer
dot
at
f
g (5.1)
the
the
particular
starts
and
the
composition
an
orange
screen
middle
constant
what
and
saver
of
the
The
to
a
screen
expand
Copy
is
to
and
Time
t
c
The
radius
nd
the
The
area
Hence
3 0
r
is
A
of
f
with
top
this
given
the
simulates
and
in
an
the
relationship
cents.
expanding
expands
circle.
outward
a
and
height
table
for
0
1
0
2.5
by
the
of
bottom
the
20
of
cm.
the
=
The
its
The
circle
takes
4
seconds
screen.
radius
2
of
the
circle
3
at
time
t
seconds:
4
10
function
f
(t).
Use
the
values
in
your
table
(t).
circle
the
circle
radius
4
is
a
function
g (r)
of
the
radius
r.
Write
g (r).
express
so
cm
t
(cm)
function
function
d
the
complete
r
describes
price
0
rectangle
touch
means.
rate.
(seconds)
Radius
b
a
=
that
its
screen
20
t
this
area
A
of
the
E2 Relationships
circle
as
a
function
of
time
t.
the
to
growing
in
the
wild.
A LG E B R A
3
3
A
cylindrical
container
has
a
radius
of
50
cm
and
a
capacity
of
500
cm
.
3
It
is
being
lled
with
water
at
a
constant
50
rate
of
10
Write
t
b
Write
c
Find
d
Determine
A
of
a
a
function
for
the
second.
volume
v (t)
of
cm
water
in
the
container
after
seconds.
to
4
down
per
cm
h
a
cm
down
h o v (t )
reach
pebble
the
a
for
the
explain
long
it
height
what
takes
it
for
h
of
water
in
the
container
in
terms
of
v
means.
the
height
of
the
water
in
the
container
cm.
thrown
water
Write
and
how
50
is
formula
ripple
down
into
still
water
increases
the
radius
the
area
r
at
in
and
the
rate
terms
of
creates
of
2
a
cm
time
t
circular
per
after
ripple.
The
radius
r
second.
the
pebble
was
thrown.
2
b
A (r)
the
c
5
=
is
meaning
Find
For
πr
the
each
area
pair
a
write
b
describe
of
a
A
of
down
A o r
of
of
the
water
f
meaning
possible
water
ripple
in
terms
of
its
real
ripple
and
of
life
g
g
in
o
after
the
60
Fuel
consuption
vehicle
ii
Height
table
below:
f
use
for
the
composite
function.
iii
of
Radius
of
iv
is
a
function
of
water
of
of
in
a
tank
is
a
time.
a
Sum
of
Bank
time.
is
of
circle
is
of
angles
a
a
function
of
a
function
sides
savings
is
cost
Volume
is
a
function
of
fuel
a
of
regular
of
the
the
polygon.
function
of
of
a
Size
tank
is
a
function
is
a
function
of
the
radius
circle.
of
each
polygon
of
the
height.
Volume
of
the
number
Fuel
consumption.
time.
polygon
v
g
velocity.
function
Explain
seconds.
f
i
radius.
(t ) .
functions
the
the
the
is
a
angle
angles
Interest
amount
of
earned
of
of
a
function
the
is
a
regular
of
the
sum
polygon.
function
of
savings.
E2.1 Multiple doorways
31
You
can
graph
composite
functions
on
any
GDC
or
graphics
software.
For
2
example,
for
functions
f
( x )
=
x
+
2x
+
1
and
f
1
functions
f
( x ) =
f
3
(
f
1
( x )
=
3x
+
5,
the
two
composite
2
( x ))
and
f
2
( x )
=
f
4
(
f
2
1.1
( x ))
are
shown
in
the
graphs
below.
1
1.1
y
2
f
(x)
=
x
+
2x
+
y
1
1
8
8
2
f
(x)
=
3(x
=
(3x
+
2x
+
1)
+
5
+
2(3x
+
4
6
6
4
4
2
2
2
f
(x)
+
5)
5)
+
1
3
f
(x)
=
3x
+
5
2
0
0
4
4
●
Does
a
Exploration
1
Find
two
2
Expand
composite
functions
function
f
and
g
such
simplify
functions
when
Find
functions
two
4
have
a
unique
decomposition?
3
functions
and
x
2
2
4
Decomposing
D
3
x
2
2
the
and
composite
h
that
and
k
their
explain
function
such
composition
that
is
how
you
is:
might
nd
the
simplied.
their
composition
is:
The
has
word
composition
meanings
photography,
4
Simplify
,
and
nd
functions
h
and
k.
Are
the
functions
the
same
literature,
as
you
found
in
step
3?
music,
Explain.
chemistry,
5
Determine
its
whether
decomposition.
or
not
it
is
better
to
simplify
a
function
before
nding
Explain.
ower
design,
course
Does
and
every
Practice
composite
each
function
3
function
have
a
unique
solving
question,
h( x )
=
f
o
nd
two
functions
f
and
h( x )
=
(2 x
− 1)
g
whose
composition
is
the
g (x ).
3
1
decomposition?
5
Problem
For
discuss
3
2
h( x )
=
2x
− 1
3
h( x )
=
6x
3
4
h( x )
=
6
32
x
− 5
5
h( x )
E2 Relationships
=
3x
2x
− 1
6
h( x )
=
2
− 5
2
cooking,
arranging,
metallurgy,
printing,
Reect
in
art,
and
of
mathematics.
A LG E B R A
Reect
●
and
Compare
Did
●
you
Write
and
discuss
your
all
answers
nd
down
a
the
to
same
non-linear
decompose
function
4
means
the
in
Practice
function
function.
the
5
functions
with
g
and
that
Explain
context
of
the
others
f
in
your
class.
?
models
what
a
the
real
life
situation,
decomposition
of
the
problem.
Summary
The
function
is
the
same
The
as
f
The
function
is
the
same
followed
You
as
g
For
(
f
two
+
functions
g)( x )
and
the
f
( x )
and
domain
g ( x ),
of
(
f
of
f
the
domain
g)( x )
is
composite
of
of
can
f
Mixed
of
the
domains
( x )
and
(
x
g
is
the
function
g
g (
f
( x )),
and
say
=
g (
f
( x ))
means
f
followed
by
g
=
f
( g ( x ))
means
g
followed
by
f
practice
= 2 − 5 x , g
)
or
and
g ( x ).
2
Problem
2
f
function
f
the
x
1
of
x.’
( x )
intersection
the
write
of
( x )
by
function
x
(
=
)
x
− 4 x,
h
(
x
)
=
and
solving
4
x
6
2
k
(
x
= −
)
,
x
If
f
(x )
=
3x
+
k
and
4
g (x ) =
,
≠ 0.
nd
the
value
3
x
of
Write
down
these
functions
in
simplest
k
such
and
a
(
f
+
g )( x )
b
( g
f
)( x )
c
(
f
that
the
composite
functions
f
o
g
form:
g o
f
are
equal.
+ h )( x )
2
7
d
( g
g
(
h )( x )
e
(h +
k )( x )
f
(k
If
f
+
State
Find
the
f
2x
− 3
and
g (x )
=
,
2x
nd:
f
( g (x
1))
g (
b
f
(
3 x ))
k )( x )
8
2
=
h )( x )
a
f
(x )
o
domains
of
functions
g ( x ) and  g o
f
( x ),
f
and
and
g
state
If
f
(x )
=
x
− 1 and
g (x )
=
4 x
+ 3,
nd
2
below.
g (
f
(3 x
1))
their
9
The
notation
[x]
means
the
greatest
integer
domains.
not
a
f
(x )
=
x
+ 1,
g (x )
=
2x
exceeding
the
value
of
x.
If
f
(x )
= [ x ],
8
− 1
g (x )
= 15 x ,
and
h( x ) =
,
nd:
x
2
b
f
(x )
=
x
, g ( x )
=
x
− 1

a
f
c
f
(x )
=

x
+
x,
g (x )
=
Given
the
functions
p (t )
=
4t
+ 1,
q (t )
=
t
− 4
r (t ) =
, t
≠ 0,
nd
these
functions
and
The
a
the
b
number
certain
q ( r ( t ))
c
p (r (t ))
d
1
Given
the
functions
f
(x )
of
< 15,
food
B
and
g (x )
=
2x
f,
g
bacteria
is
(T
)
T
is
taken
=
B
in
can
20T
in
out
temperature
function T (t )
,
=
5))
c
h(
(π ))
f
food
be
refrigerated
modelled
− 80T
degrees
of
the
+ 500 ,
by
Celsius.
where
When
refrigerator,
the
r (q (t ))
food
4
(h (
temperature
function
1 < T
domains:
p (q (t ))
f
 
2
the
a
 
b
2
state
t
their
10
at
1
and
1

2 − 3x
2
3

g
2
=
4t
T
can
+ 2, 0
be
≤
t
modelled
≤
4,
by
where
the
t
is
the
+ 4
x
time
in
hours.
2
and
h( x )
=
x
− 2,
nd
in
terms
of
and
h
the
a
Find
the
composite
function
B
oT
(t )
and
functions:
explain
2
a
what
it
means
in
the
context
of
the
2
+ 4
b
4 x
+ 16 x
+ 14
problem.
x
x
5
Given
that
f
2
(x ) =
b
and
g (x )
=
2x
3
a
g (
f
( a ))
b
f
( g (
+ 1
Determine
how
count
5000.
long
it
takes
for
the
bacteria
nd:
to
be
b ))
E2.1 Multiple doorways
3 3
11
A
store
to
If
owner
decide
items
the
do
adds
50%
selling
not
sell
to
price
in
4
the
of
wholesale
items
weeks,
he
in
the
sells
price
ii
write
store.
them
20%
Write
down
selling
price
wholesale
a
s
function
of
price
an
that
item
in
represents
terms
of
the
write
for
Write
down
discounted
selling
c
Write
its
down
d
a
function
price
d
of
that
an
represents
item
in
terms
the
of
down
the
a
the
composite
discounted
the
From
function
an
price
online
discount
price
in
terms
of
b
You
for
see
price
was
that
a
25%
sale
function
that
your
of
an
item
sale.
a
x
much
you
then
you
rst
would
use
apply
the
the
25%
a
composite
the
how
of
represents
sale
shoes
and
if
then
function
much
you
use
you
rst
the
would
apply
You
store
go
favorite
the
the
discount
nd
a
pair
the
of
shoes
price
two
composite
a
and
of
you
the
like
shoes
functions
you
for
$50.
using
iii
iv,
and
state
the
the
found
in
dierence
in
price
you
receive
online
to
shoe
store
a
spend
is
them.
For
each
question,
nd
two
functions
f
it,
having
g
whose
composition
is
the
function
a
=
f
o g (x )
5
4
If
pay
discount
whose
h( x )
25%
if
$72.
shopping
coupon.
that
and
sale
Evaluate
and
and
of
coupon.
to
13
$5
eect
composite
how
shoes
down
between
12
a
represents
pay
price.
discount
wholesale
write
that
25%
wholesale
Find
the
coupon
price.
represent
the
cost
w
iv
b
g (x )
function
discount.
represents
a
a
the
at
iii
a
down
represents
original
cost
of
a
a
h( x )
=
−(x
+ 1)
c
h( x )
=
−5 x
+ 7
b
h( x )
=
d
h( x ) =
4 − 3x
pair
shoes:
3
2
i
write
down
represents
a
f
function
the
cost
eect
(x )
of
x
using
your
Review
1
An
oil
in
spill
meters.
At
4
x
e
discount
− 8
that
h( x )
= 10
coupon
context
is
modelled
time
t =
0
as
the
a
circle
radius
is
of
70
radius
m.
r
3
The
The
speed
hours,
of
can
a
be
car,
v
(in
modelled
km/h)
using
at
the
a
time
t
function
2
radius
increases
at
a
rate
of
0.8
meters
per
v (t )
minute.
r
a
Express
b
Find
the
radius
r
as
a
function
of
(in
=
40 + 3t
+ t
liters/km)
.
at
The
a
rate
speed
of
of
fuel
v
consumption,
km/h
can
be
time.
2
v

modelled
A(r)
the
is
composite
the
area
of
a
function
circle
of
A o r
(t )
what
it
radius
Determine
function

r (v ) =
+ 0.2
0.1

500
r.
Write
down
the
function
r (v (t )) ,
the
rate
of
means.
fuel
c
the

a
Explain
by
where
how
long
it
takes
for
the
area
consumption
as
a
function
of
time.
of
b
Find
the
rate
of
fuel
consumption
for
a
trip
2
the
oil
spill
to
exceed
100 000
m
that
2
The
formula
K (C )
=
C + 273
converts
4
temperature
into
Kelvin
temperature.
lasts
4
hours.
Celsius
Composite
functions
can
also
be
used
to
create
The
and/or
explain
number
puzzles.
Ask
a
friend
5
formula
C
(F)
=
(F
− 32
)
converts
to
Fahrenheit
think
of
any
number,
then:
9
temperature
a
Write
into
down
convert
a
Celsius
composite
Fahrenheit
●
multiply
●
divide
temperature.
function
temperature
into
a
Kelvin
Write
boiling
composite
and
freezing
temperatures
are
212°
and
32°
in
the
boiling
and
freezing
and
and
then
then
add
add
6
3.
the
the
steps
function
steps
are
as
that
functions,
shows
the
and
the
order
in
executed.
Simplify
the
Make
can
E2 Relationships
your
function,
friend’s
and
explain
how
to
number.
temperatures
5
Kelvin.
34
2
4
respectively.
nd
Find
by
by
in
b
Fahrenheit
result
down
which
The
the
number
to
temperature.
b
the
up
solve
your
own
using
number
composite
puzzles
which
functions.
you
Doing
E2.2
Global
and
context:
Objectives
●
undoing
Scientic
Inquiry
Understanding
onto
and
one-to-one
functions
and
technical
innovation
questions
●
What
●
How
are
onto
and
one-to-one
functions?
F
●
Finding
the
a
function
algebraically
do
you
nd
the
inverse
of
a
●
Restricting
the
necessary,
Justifying
each
inverse
the
so
a
domain
that
that
of
its
two
function
of
a
graphically
function,
inverse
functions
is
also
are
a
●
if
function
inverses
C
are
●
of
that
Is
do
you
inverses
the
know
of
inverse
when
each
of
a
two
functions
other?
function
always
a
function?
other
Justifying
How
a
function
is
self-inverse
●
Are
●
Can
there
functions
that
are
self-inverses?
D
ATL
Test
everything
be
undone?
Critical-thinking
generalizations
and
conclusions
3 5
SPIHSNOITA LER
Finding
●
of
function?
●
●
inverse
Y
ou
●
should
decide
whether
diagram
pairs
already
or
a
set
represents
a
mapping
of
a
know
1
ordered
function
how
to:
Decide
which
If
function,
state
a
it
does
reason
a
a
why.
B
f
p
b
q
c
r
g
A
B
a
p
b
q
c
r
h
M
=
it
a
x
b
y
c
z
{1,
each
a
represents
represent
a
c
If
not
the
A
b
2
diagram
function.
3,
4}.
relation
is
R
2,
not
=
a
{(2,
Determine
is
a
function
function,
3),
(1,
whether
4),
state
(2,
1),
from
the
(3,
M
or
to
M.
reason
2),
(4,
not
why.
4)}
1
b
R
=
{(2,
1),
(3,
4),
(1,
4),
(4,
4)}
2
●
use
to
the
ver tical
decide
if
a
line
test
relation
is
3
a
Use
the
each
ver tical
relation
line
here
is
test
a
to
determine
if
function.
function
y
y
a
b
x
x
●
form
the
composition
of
4
For
the
functions
f ( x )
=
2x
+
1
and
2
two
functions
3 6
E2 Relationships
g ( x )
=
a
g( x )
f

x
+
1,
nd:
b
g

f ( x )
A LG E B R A
Types
F
For
a
must
●
What
●
How
relation
map
For
a
to
of
are
do
functions
onto
you
between
one
and
function
f
:
nd
two
only
A
→
the
inverse
to
be
a
element
functions?
of
a
function?
function,
in
the
element
in
the
●
Each
element
in
A
domain
can
ordered
have
pairs
A
must
only
(a,
b
)
have
an
one
image
and
(a,
b
1
Onto
This
each
element
from
the
domain
range.
B :
Each
two
one-to-one
sets
one
●
contains
and
image
in
),
B.
in
So
then
b
2
the
if
=
a
range
B
function
b
1
2
functions
mapping
diagram
shows
a
function
A
f
:
A
→
B
B
f
a
p
b
q
The
is
Each
element
The
element
f
→
:
A
A
B
is
in
not
function
element
in
s
in
f
:
the
B,
an
A
c
r
d
s
domain
A
however,
onto
→
B
maps
is
not
to
the
one,
and
image
of
any
only
any
one,
element
element
in
A.
in
element
not
the
s
in
image
element
in
B
of
A
B
Therefore,
function.
is
onto
if
every
element
in
B
is
an
image
of
an
A
1
In
this
graph
of
y
=
− 1) ,
(x
if
you
draw
horizontal
lines
through
every
1
possible
y
y
2
=
(x
−
1)
2
2
y-value
in
the
then
range
these
lines
( y-values)
will
is
all
the
intersect
image
of
the
an
graph
element
line,
in
because
the
every
domain
element
(x-values).
1
0
1
3
1
The
only
horizontal
if
its
Reect
●
Do
●
Draw
the
all
line
graph
and
graphs
three
domain
test
for
intersects
discuss
of
straight
dierent
and
onto
any
of
a
line
function
drawn
on
f
is
it
onto,
at
least
if
and
once.
1
lines
graphs
range
functions:
horizontal
represent
that
the
onto
represent
functions?
onto
Explain.
functions,
and
specify
functions.
E2.2 Doing and undoing
37
4
5
x
f(x)
2
Top
As
right
x
the
can
set
is
the
graph
assume
of
real
numbers,
any
of
value,
numbers.
hence
this
the
function
the
The
(x ) :
domain
range
function
f
is
is
not
of
the
x
,
the
set
of
f
:

→
function
 .
is
non-negative
real
onto.
x
f(x)
2
If,
however,
then
f
( x )
passes
is
the
we
an
dene
onto
f
(x )
=
function,
horizontal
line
+
x
such
and
that
the
f
graph
: 
→
for
∪ {0},

these
y-values
test.
x
One -to-one
This
mapping
function
and
f
is
Y
a
A
is
=
f
:
X
{A,
diagram
→
B,
Y,
C,
one-to-one
function
the
f
image
function
f
is
:
X
functions
X
=
shows
{1,
2,
Y
f
the
3,
1
D
2
B
3
C
4
A
4}
D}.
function.
A
of
→
B
is
a
exactly
one-to-one
one
one-to-one
if
element
f
(a
)
=
f
function
in
(a
1
the
)
if
every
domain.
implies
element
In
that
a
2
other
=
in
the
words,
range
a
a
1
2
1
Looking
again
at
the
graph
of
y
1
=
(x
− 1) ,
every
y-value
is
the
image
of
y
y
=
(x
−
1)
2
2
2
exactly
line
one
x-value,
through
once
every
so
y
is
a
possible
one-to-one
y-value,
function.
these
lines
If
will
you
all
draw
a
intersect
horizontal
the
graph
line
1
only.
0
1
1
The
if
horizontal
and
one
only
its
line
test
graph
for
one-to-one
intersects
any
functions:
horizontal
a
line
function
drawn
f
on
is
it
one-to-one,
at
exactly
point.
Reect
●
if
How
and
does
discuss
the
horizontal
2
horizontal
line
test
for
line
test
one-to-one
for
●
Is
every
●
Is
a
●
Draw
●
How
can
a
function
be
onto
●
How
can
a
function
be
one-to-one
●
Can
straight
quadratic
3 8
a
three
line
a
one-to-one
function
dierent
function
be
neither
that
but
onto
E2 Relationships
functions
function?
one-to-one?
graphs
onto
dier
from
the
functions?
Explain.
represent
not
nor
Explain.
one-to-one
functions.
one-to-one?
but
not
onto?
one-to-one?
Explain.
3
4
5
x
A LG E B R A
Example
Determine
1
which
of
these
functions
are
onto,
one-to-one,
both,
y
y
a
or
neither.
y
b
c
x
0
0
x
0
x
y
a
0
x
Use
Horizontal
is
onto.
lines
Some
function
is
intersect
horizontal
not
the
graph
lines
in
at
intersect
least
at
one
more
point,
than
so
one
the
point,
the
horizontal
line
tests.
function
so
the
one-to-one.
y
b
0
x
Some
onto.
so
horizontal
Horizontal
the
function
is
lines
lines
do
do
not
not
intersect
intersect
the
the
graph,
graph
so
in
the
function
more
than
is
one
not
point,
one-to-one.
y
c
0
The
not
range
x
is
the
one-to-one
set
of
since
non-positive
horizontal
real
lines
numbers,
intersect
at
hence
more
it
is
than
onto.
one
It
is
point.
E2.2 Doing and undoing
3 9
Practice
1
1
Determine
a
f
: 
→
which
of
these
functions
are
onto,
one-to-one,
b

f
:
neither
A
→
B ;
or
A
both.
=
{x
:
5
≤
x
<
5},
B
=
{y
:
3
y
y
5
4
4
3
3
2
2
1
1
x
0
5
0
5
c
f
:
4
{x
3
:
-
2
2
1
1
≤
x
4
3
2
1
1
2
3
4
5
3
4
5
6
1
2
2
3
≤
2
x
8}
→
{y
:
-
1
≤
y
≤
1}
d
f
: 
→

y
y
1
6
4
0.5
2
x
0
5
4
2
1
1
2
3
4
1
2
3
4
5
2
x
0
2
4
1
e
f
: 
→
f

f
: 
→ { y
:
y
≥
−3}
y
y
5
4
4
3
3
2
2
1
1
x
0
5
4
3
2
1
1
x
0
5
3
2
1
1
1
2
3
3
4
4
4 0
E2 Relationships
6
≤
y
<
4}
A LG E B R A
f
g
: 
→

h
f
: {x
:
x ∈ , −3
y
≤
x
≤
5} →

y
20
5
4
10
3
2
1
x
0
3
1
2
1
x
0
4
3
2
1
1
10
2
3
4
5
6
1
2
20
3
4
5
Problem
2
Create
your
a
onto,
c
neither
If
you
them,
a
but
a
own
not
onto
change
you
relation
also
solving
the
a
A
set
friend
gives
●
From
●
Turn
●
At
●
My
of
of
inverse
ordered
the
ordered
pairs
in
a
one-to-one,
d
both
relation
and
R
reversing
by
.
every
relation
has
an
not
onto
relation
pairs,
but
inverse
onto
one-to-one.
Since
which
is
State
you
right
the
4th
house
on
instructions
go
the
5th
a
●
Start
●
Add
●
Divide
●
Subtract
●
The
worked
function
a
real
turn
one
get
route
to
on
get
left
the
on
to
her
B21
house:
for
3
km.
back
home
out.
f
models
that
domain
number
Georgia
Avenue.
right.
them
possible
with
on
to
Drive.
light,
directions
largest
north
Kanuga
trac
you
down
its
these
home,
is
the
how
Write
1
your
down
Explain
a
the
b
relation.
Write
2
are:
one-to-one
elements
obtain
is
that
one-to-one
nor
Exploration
1
functions
and
the
from
steps
your
in
the
friend’s
house.
algorithm
below.
Continued
next
range.
x.
4.
the
5
nal
result
from
by
this
answer
is
3.
result.
y
on
page
E2.2 Doing and undoing
41
b
c
Write
down
the
Write
down
a
Look
from
algebraically.
f
to
a
g
g,
i
your
and
whether
one-to-one
state
f
and
its
g
that
that
models
functions
interchanging
function,
State
at
algorithm
g
algorithm.
Rewrite
d
inverse
function
the
x
f
and
largest
starts
the
and
y,
so
possible
with
steps
g.
in
y
Explain
that
g : x
domain
and
this
how
→
y.
and
leads
to
x.
inverse
you
Justify
can
get
that
g
is
range.
are:
functions
ii
onto
functions.
Simplify
the
functions
e
Draw
the
graphs
of
f
( x )
and
g ( x )
on
the
same
set
of
axes.
Write
graphing
the
Step
2
of
inverse
equation
of
Exploration
of
a
function:
interchange
the
x
the
1
line
goes
y
the
them.
symmetry.
through
reverse
and
of
before
down
the
process
algebraic
by
process
making
x
the
of
nding
subject,
the
then
variables.
1
The
notation
Inverse
1
The
The
the
3
If
the
inverse
of
a
function
f
is
f
functions :
inverse
domain
2
for
graph
a
function
range
of
original
the
of
and
the
of
in
relation
is
f :
A
→
B
function
inverse
function
inverse
the
of
the
a
the
a
y
=
relation
formed
when
the
interchanged.
function
line
also
is
are
is
the
reection
of
the
graph
of
x
function,
then
it
is
the
inverse
function
of
1
the
original
Example
Find
the
function,
f
:
B
→
A.
We
say
that
the
function
is
invertible
2
inverse
of
the
function
.
Write
5y
=
6x
6x
=
5y
the
function
using
5
+
5
y instead
Make
=
5( y
+
x
the
1)
Write
can
4 2
also
f ( x ).
subject.
Interchange
You
of
interchange
the
x
and
E2 Relationships
y
variables
rst
and
then
isolate
y
in
inverse
x
and
y.
notation.
A LG E B R A
Reect
ATL
●
Are
●
Is
all
the
and
discuss
linear
relations
inverse
Example
the
of
any
3
also
linear
linear
functions?
function
also
a
Explain.
function?
Explain.
3
a
Find
inverse
b
Conrm
c
Determine
your
if
of
the
answer
the
function
to
a
inverse
,
algebraically.
graphically.
is
also
a
function.
a
( y
3)(x
2)
=
1
State
the
excluded
Interchange
x
value.
and y
y
b
6
y
=
x
5
4
1
y
=
+
x
3
−
Graph
2
the
function
and
its
inverse.
3
Add
in
the
graph
of
the
line
y = x
1
2
y
=
+
x
−
to
3
check
that
one
graph
is
a
2
reection
1
of
the
other
in
y = x.
x
1
2
is
the
line
y
=
the
is
Practice
1
Write
is
a
f
=
b
g ( x )
c
h ( x )
( x )
=
=
because
it
is
a
reection
in
a
function
because
it
passes
the
vertical
line
test.
2
down
inverse
of
x
c
ATL
inverse
the
also
{(3,
{(2,
{(0,
a
inverse
of
each
function,
and
state
whether
or
not
the
function.
1),
4),
0),
(
(
(
1,
5,
2,
1),
3),
4),
(
(
(
3,
2,
1,
1),
(1,
3),
4),
1)}
(1,
(2,
1),
2),
(
(
4,
3,
4),
3),
(
(
1,
4,
2),
4),
(3,
(1,
1)}
1)}
E2.2 Doing and undoing
4 3
Problem
2
Copy
each
Determine
a
solving
graph
if
the
and
draw
graph
of
the
the
inverse
inverse
on
of
the
each
same
set
function
of
axes.
also
Use
represents
line
function.
the
vertical
test.
y
y
a
y
b
c
3
8
2
6
1
4
4
3
2
1
2
x
0
3
2
2
3
4
5
6
x
0
x
0
2
2
2
2
3
4
4
6
8
5
4
2
2
1
1
10
3
1
2
3
4
y
y
d
y
e
f
5
1
4
4
3
x
0
1
3
1
1
2
2
1
2
1
3
x
0
x
0
4
3
2
1
1
2
3
4
4
3
2
1
1
4
1
5
2
6
3
1
2
3
4
5
3
Find
the
the
inverse
function
is
of
also
each
a
function
algebraically.
Determine
if
the
y
=
- 4 x
+
1
y
b
= 4 −
− x
x
c
y
= −1 −
2
x
≠ 0
y
e
=
C
=
x
≠ −5
f
y
=
x
h
Reversible
●
How
●
Is
the
do
you
know
inverse
Exploration
and
of
a
y
=
x
+ 2,
x
≥
−2
irreversible
when
two
function
x
+
1
+ 5
3
3
y
,
x
x
g
3
1
1
y
− 5
=
2
d
of
function.
3
a
inverse
functions
always
a
i
y
x
+ 3
x
2
=
,
x
≠ 2
changes
are
inverses
of
each
other?
function?
2
2
1
Draw
the
graph
of
the
quadratic
function
y
=
x
.
State
the
largest
possible
2
domain
2
Find
and
the
the
inverse
corresponding
of
the
range
function
of
the
function
y
=
x
algebraically.
Continued
4 4
E2 Relationships
on
next
page
2
3
A LG E B R A
3
Draw
that
the
your
inverse
graph
on
the
shows
4
Determine
if
the
5
Determine
how
same
the
inverse
graph.
inverse
of
the
of
Draw
the
function
the
line
y
=
x
and
conrm
function.
is
also
a
function.
y
=
2
also
6
a
State
the
of
Reect
Is
can
restrict
the
domain
of
x
so
that
its
inverse
is
function.
range
●
you
domain
its
and
there
and
inverse
range
discuss
more
of
your
new
function
and
the
domain
and
function.
than
one
4
way
of
restricting
the
domain
of
the
function
2
y
=
x
so
that
its
inverse
is
also
a
function?
2
●
Is
●
When
the
function
y
=
x
one-to-one
and
onto?
2
you
function,
If
the
does
inverse
domain
of
restrict
of
the
a
it
the
domain
become
function
function
so
of
y
=
one-to-one
is
not
that
a
its
x
so
and
its
inverse
is
also
a
onto?
function,
inverse
that
is
you
a
may
be
able
to
restrict
the
function.
The
proof
of
this
Theorem
The
inverse
and
a
of
a
function
one-to-one
Example
f
is
also
a
function,
if
and
only
if
f
is
both
an
onto
theorem
is
the
of
level
beyond
this
course.
function.
4
2
a
Find
b
If
the
the
its
inverse
inverse
inverse
is
is
of
not
also
a
y
a
=
x
1
algebraically,
function,
function,
restrict
and
the
conrm
and
conrm
domain
of
graphically.
the
function
so
that
graphically.
2
a
y
=
x
1
Find
the
inverse.
y
2
f
(x)
=
x
−
1
1
f
(x)
=
√
x
+1
2
Graph
both
functions
2
is
the
reection
of
y
=
x
1
and
in
the
line
y
=
conrm
that
one
is
x
x
a
y
=
x
reection
in
f
(x)
=
−√
x
the
line
of
the
other
y = x.
+1
3
Continued
on
next
page
E2.2 Doing and undoing
4 5
b
6
Tr y
5
dividing
function
the
into
original
pieces
that
are
4
y
=
x
both
one -to-one
and
onto.
3
2
1
f
(x)
=
√
x
+
1
2
x
1
2
2
3
4
5
Dene the restricted
6
1
domain for each section.
2
f
(x)
=
x
−
1, x
≥
0
1
2
Restricted
domain
x
≥
0
y
6
y
=
x
5
4
2
f
(x)
=
x
−
1, x
≤
0
1
3
Write
2
the
function
inverse
using
notation.
1
x
0
3
2
1
1
2
3
4
5
6
1
f
(x)
=
−
√
x
+
1
2
2
3
Restricted
Therefore,
For
the
for
domain
the
domain
Example
x
x
≤
0
domain
≤
0
of
x
≥
0
of
f ,
.
f ,
.
5
2
a
Find
the
inverse
b
Restrict
the
domain
and
of
f ( x )
domain
of
=
f
x
so
6x
that
+
its
2
algebraically,
inverse
is
also
and
a
conrm
function,
graphically.
and
state
its
range.
1
c
State
the
d
Conrm
domain
and
graphically
range
that
of
the
f
inverse
function
is
indeed
a
function.
2
a
y
=
6x
x
+
2
You
cannot
immediately
isolate
x
2
x
6x
=
y
2
=
y
2
=
y
2
x
6x
+
9
+
9
Complete
( You
the
could
square
also
use
to
the
solve
for
x
quadratic
2
(x
3)
+
7
formula
to
solve
for
x.)
Interchange
Continued
4 6
E2 Relationships
on
next
page
x
and
y
A LG E B R A
Since
they
one
are
graph
is
inverses
the
of
reection
each
of
the
other
in
the
line
y
=
x,
other.
y
8
y
=
3
±
√
x
+
7
6
4
2
Draw
and
the
the
graphs
line
y = x.
x
8
6
4
2
0
2
4
6
8
2
4
y
=
x
6
2
y
=
x
6x
+
2
8
Identify
b
The
vertex
is
(3,
7).
Therefore
and
the
inverse
domain
c
The
of
to
f
be
is
domain
a
function,
either
of
f
is
x
x
≥
≥
3
the
domain
that
would
make
the
function
onto
for
one -to-one
by
nding
the
ver tex
of
the
formula
the
parabola
the
or
x
≤
either
3.
graphically,
or
using
.
3.
1
Then
the
domain
of
is
f
x
≥
-
1
and
the
range
of
f
is
y
≥
3.
Select
y
is
a
a
domain
function,
such
and
that
graph
the
inverse
it.
8
d
y
=
3
+
√
x
+
7
6
4
2
x
8
6
4
2
0
2
4
6
8
2
Check
y
=
that
the
symmetrical
4
graphs
about
are
the
line
y = x
x
6
2
y
=
x
6x
+
2, x
≥
3
8
The
ATL
graphs
Practice
For
each
a
Find
b
If
inverse
inverse
that
its
restricted
reections
of
each
other
in
y
=
x.
3
function
its
the
so
are
is
in
questions
1
algebraically,
not
inverse
is
a
to
and
function,
also
a
8:
conrm
restrict
function
the
and
graphically.
domain
write
the
of
the
inverse
original
f
( x )
=
with
its
domain.
1
1
function
function
5x
2
2
f
(x ) = 7 −
2
x
3
f
( x )
=
x
6
f
( x )
=
(x
3
2
x
2
4
f
( x )
=
1
5
x
f
+ 1
(x ) =
2
,
x
≠ 0
2)
x
2
7
f
( x )
=
x
2
+
2x
4
8
f
( x )
=
2x
4x
+
5
E2.2 Doing and undoing
4 7
Problem
9
Below
is
the
and
of
left
set
is
of
zero.
the
solving
the
real
graph
Reecting
inverse,
of
the
numbers.
the
below
absolute
The
range
graph
of
f
value
is
( x )
the
=
function
set
|x
of
in
the
line
3
3
2
2
1
1
x
1
Explain
Write
apply
Apply
10
In
A
you
Applying
you
have
like
players
of
handicap
you
on
bowler’s
In
a
one
or
very
golf,
while
win,
180
Write
b
Find
down
the
in
to
to
175.
f
so
real-life
The
domain
gives
this
graph
a|
=
a
a|
=
-a
for
a
that
A
a
inverse
to
of
h (s)
determine
the
‘perfect’
score
of
strikes
all
ten
an
optimal
The
9
in
of
than
degrees
will
game
pins
rest
rather
5
the
and
ball
8
a
in
if
inverse
often
in
solving
have
a
score
a
170
Handicaps
if
and
are
to
reach
question
a
function.
allow
you
her
usually
a
solution
10
‘handicap’
to
example,
bowls
is
(h).
comparisons
bowl
a
150
handicap
calculated
is
and
5,
based
(s).
of
=
are
calculated
0.9(220
h (s).
and
the
row,
the
pins
the
from
ball.
using:
s)
Explain
conrm
inverse
upright.
the
So,
a
for
an
why
your
also
a
it
needs
to
be
restricted.
result
graphically.
Use
bowling,
a
strike
itself
hits
pin
will
perfect
with
just
1,
when
just
will
5
by
game
48
pins.
E2 Relationships
other
if
9
with
can
bowl
pins.
it
pins,
tilts
bowler’s
and
connect
by
down
you
four
fall
handicap
getting
knocking
down
3,
whose
achieved
by
right-handed
number
ball
is
However,
any
A
is
only
be
a
your
is
never
function.
opponent
bowling
done
knocked
and
bowler’s
pins.
ball
are
ball,
with
be
where
the
connecting
4 8
can
one
strike,
score
ten-pin
This
with
connect
left-handed
2,
average
300.
12
0.
contexts
player’s
For
its
less
Find
0.
<
domain.
handicap
c
>
a
3
In
the
graph
a
for
x
successfully
functions
levels.
handicaps
domain
of
restricted
strategies
Davinia
score
league,
your
players
added
h (s)
a
domain
with
inverse
dierent
30,
average
bowling
about
number
would
f
mathematics
learned
is
the
for
mathematical
bowling
is
restrict
function
selected
then
a
could
inverse
sports
.
numbers
2
3
D:
x
2
3
handicap
your
3
2
the
=
|x
0
3
2
what
with
2
1
how
Objective
iii.
1
1
the
y
=
real
y
0
2
( x )
right.
y
3
f
positive
pins,
just
the
just
ball
and
the
a
1,
result
of
is
15.
than
zero.
A LG E B R A
This
mapping
diagram
shows
a
function
and
its
inverse.
A
B
f
x
f
(x)
1
f
1
f
maps
x
onto
f
( x )
and
then
f
maps
f
( x )
to
x.
1
So
the
composite
function
f
1

f
(x )
=
x.
Similarly
f

f
(x )
=
x
1
If
two
functions
and
a
and
f
are
inverses
the
given
of
each
other,
then
.
Example
Conrm
f
6
algebraically
that
functions
are
inverses:
and
b
and
a
1
and
,
so
f
and
f
are
inverse
functions.
Find
and
simplify.
Find
and
simplify.
Check
that
functions
b
both
are
Find
Find
Check
Since
,
the
functions
are
mutual
that
composite
equal
to
x
and
simplify.
and
simplify.
both
composite
inverses.
functions
are
E2.2 Doing and undoing
equal
to
x
4 9
Practice
In
4
questions
functions
1
are
to
8,
determine
inverses
of
each
algebraically
whether
or
not
the
pairs
of
other.
1
1
f
( x )
=
x
3
;
g
( x )
=
x
+
3
2
f
( x )
=
4x
+
8
;
g ( x )
=
x
2
4
2
3
f
( x )
=
-
5
x
2
;
g ( x )
=
-5
-
x
5
2
2
4
f
(x )
5
f
(x ) =
=
x
−4
+ 2
+
− 2, x
≥
− 2 ;
g (x )
=
(x
4 x
− 2)
− 2
2
,
x
≥ 0 ;
2,
x
≠ 2
g (x ) = (x
+ 2)
,
x
≥ −2
2
1
6
f
5 +
(x ) =
−
x
7
f
;
1
x
1 +
x
(x ) =
,
x
Problem
≠ −1 ;
2x
g (x ) =
2
1
x
1 +
x
g (x ) =
,
x
≠ −2
2 +
x
,
≠ −1
x
solving
− x
−
2
2
8
f
( x ) = −2 x
− 2 ;
g (x ) =
2
Exploration
In
Practice
inverses
4,
question
when,
1
Graph
2
State,
3
State
the
3
in
fact,
functions
therefore,
what
whether
For
the
or
not
two
f
functions
2
students
think
that
f
and
g
are
mutual
not.
state
why
dierence
is
the
between
information
functions
this
are
case.
and
must
be
mutual
given
in
when
this
example.
determining
inverses.
f
( x )
=
3x
1
and
g ( x )
=
2x
+
3,
nd:
1
( x )
and
1
d
are
4
1
a
the
many
and
additional
Exploration
1
8,
they
g
1
( x )
b
f

e
g
1
f

Repeat
g
g ( x )
1
( x )
step
1
for
these
c
[
f

g ( x )]
1

f
( x )
functions:
2
a
3
f
( x )
Make
=
a
Reect
2x
+
5
;
g ( x )
conjecture
and
b
=
about
the
discuss
f
( x )
=
relationship
x
+
x
1
;
g ( x )
between
=
x
+
and
1
.
You
6
should
discovered
Which
is
easier,
nding
or
proof
is
any
two
functions
f
and
g,
E2 Relationships
of
this
beyond
4.The
result
the
level
.
of
5 0
in
?
Exploration
For
have
this
this
course.
A LG E B R A
D
When
the
doing
same
●
Are
●
Can
there
and
undoing
produces
result
functions
everything
be
that
are
self-inverses?
undone?
1
The
inverse
You
can
of
say
the
that
function
f
is
f
( x )
=
2
x
is
( x )
=
f
( x )
=
2
x,
so
f
is
its
own
inverse.
self-inverse.
1
A
function
f
is
self-inverse
Exploration
1
Determine
function
2
3
the
the
other
4
is
1
k
∈
,
for
=
not
which
f
not
x,
,
is
k
∈
,
whether
is
or
not
any
linear
self-inverse.
self-inverse.
Are
there
values
of
k
for
self-inverse?
x
,
is
f ( x ).
graphically
k
≠
( x )
if
( x )
and
( x )
if
Determine
f
f
,
function
than
which
form
if
Determine
f
5
algebraically
of
Determine
which
if
x
1,
is
≠
is
self-inverse.
not
1
is
Are
there
values
for
there
values
of
k
self-inverse?
self-inverse.
Are
k
for
self-inverse?
Summary
For
a
function
f
:
A
→
1
B:
The
inverse
relation
●
each
element
in
the
domain
A
must
have
in
the
range
each
element
in
function
the
domain
A
can
have
image
in
the
The
range
is
function
an
A
image
function
f
:
A
of
f
:
→
an
A
B
is
onto
element
→
B
:
A
→
B
domain
is
the
and
are
range
is
a
if
in
every
element
in
graph
of
of
the
the
inverse
graph
of
of
a
the
function
original
is
the
function
3
A
one-to-one
function
the
line
y
=
x.
B
If
the
inverse
relation
is
the
inverse
function
is
also
of
the
a
function,
original
then
1
one
element
element
The
A
in
in
the
the
horizontal
function
intersects
least
f
is
any
range
is
the
image
of
if
The
and
horizontal
only
line
if
its
drawn
if
graph
on
it
inverse
andonly
at
function
at
If
f
is
one-to-one,
intersects
exactly
one
any
if
and
horizontal
only
line
if
the
may
so
inverse
be
that
able
its
say
the
function
is
invertible
of
if
f
a
is
function
both
an
f
is
also
onto
a
and
function
a
one-to-one
two
of
to
functions
then
f
and
f
are
inverses
and
=
of
each
x.
its
drawn
on
it
For
any
two
functions
f
and
g,
point.
A
If
We
1
once.
graph
A.
function.
other,
A
→
Theorem
tests:
onto,
B
exactly
domain.
line
:
a
function
restrict
inverse
is
the
also
a
is
not
a
domain
function,
of
the
function
f
is
self-inverse
if
.
you
function
function.
E2.2 Doing and undoing
it
function,
if
f
every
of
interchanged.
B.
in
A
f
the
only
reection
one
function
when
B
2
●
a
an
the
image
of
formed
51
Mixed
1
i
practice
Determine
if
these
relations
are
functions.
3
Justify
that
ii
Find
the
inverse
iii
Determine
if
of
the
the
which
are
also
of
these
functions
have
inverses
functions.
relation.
inverse
is
a
function.
a
f
:

→

y
a
R
=
{(1,
2),
(2,
3),
(3,
4),
(4,
5)}
6
b
S
c
T
=
{(
2,
0),
(3,
1),
(
5,
1),
(6,
7),
(0,
0)}
5
=
{(0,
1),
(3,
1),
(
4,
1),
(
3,
6),
(3,
1)}
4
2
Determine
onto,
whether
neither,
or
each
function
is
one-to-one,
3
both.
2
1
a
f
:

→

x
0
y
3
2
1
1
2
3
1
5
2
4
3
3
2
b
f
: {x
:
x ≥ 1} → { y
:
y
≥ 1}
1
y
5
4
3
2
1
1
6
x
0
2
3
4
5
1
5
2
4
3
3
2
b
f
:

→

1
y
x
0
5
1
2
3
4
5
6
7
8
9
10
4
c
3
f
:

→

2
y
5
1
4
x
0
7
6
5
4
3
2
1
1
2
3
1
3
2
2
3
1
4
x
0
c
f
:

→
6

5
3
4
2
1
1
2
1
2
y
4
3
3
4
2
d
1
f
: 
→ [ y
:
y
≥
−1}
y
x
0
3
2
1
1
2
3
5
1
2
4
3
3
4
2
1
x
0
3
2
1
1
1
2
5 2
E2 Relationships
2
3
4
5
6
7
8
A LG E B R A
4
a
For
each
function
below,
nd
its
2
inverse
algebraically,
and
conrm
(x
4
d
f
(x ) = 2 +
3x
−
4,
x
≥
graphically.
−
2)
+
4
g (x ) =
;
3
3
3
4 x
+ 13
2
b
If
its
inverse
is
not
a
function,
restrict
e
the
f
(x ) =
x
+ 3x
− 1,
x
≤ −
;
g (x ) = −
−
2
domain
of
the
original
function
so
that
its
7
inverse
is
also
a
function,
and
write
down
The
weight
function
with
its
restricted
w
in
kg
of
a
certain
type
of
sh
is
the
related
inverse
3
2
to
its
length
l
in
cm,
and
domain.
can
6
by
the
function
w
=
(9.4
×
10
be
modelled
3
)
×
l
.
3
i
f
( x )
=
-3x
+
8
ii
f
(x ) =
x
By
−1
obtaining
the
inverse
of
the
model,
determine
4
the
2
iii
f
( x )
=
4
f
of
a
sh
that
weighs
0.58
kg.
2
x
iv
x
v
length
f
( x )
=
3x
2
1
(x ) =
,
2x
x
Objective
≠
1
D:
Applying
mathematics
in
real-life
2
contexts
5
Find
the
inverse
Determine
function
is
of
each
whether
also
a
or
function
not
the
function.
If
algebraically.
inverse
the
of
inverse
iii.
the
not
a
function,
restrict
the
domain
of
a
f
( x )
so
=
that
its
inverse
5x
is
b
f
also
( x )
a
=
correctly
2
f
1
(x ) =
x
6x
x
−
d
3
+
mathematical
a
strategies
solution
answer
function
is
the
best
questions,
to
use,
the
you
will
original
have
or
the
to
decide
inverse.
8
2
A
computer
=
700
company
12t
to
uses
nd
the
the
formula
depreciated
value
in
2
g (x ) =
+1
2
reach
function.
d (t)
c
selected
to
the
which
function
the
relation
To
is
apply
successfully
d
dollars
of
computers
t
months
after
they
are
5
put
on
sale.
1
2
e
h( x )
=
x
− 3
p( x ) =
f
+ 2,
x
≠
0
x
x
g
+
Calculate
b
Find
Justify
and
state
what
this
means.
,
x
≠
3
s(x )
h
=
x
+ 3 ,
x
≥
-3
t
such
that
d (t)
=
50,
and
explain
what
3
this
6
d (24)
2
q (x ) =
x
a
algebraically
that
these
pairs
of
value
means.
functions
1
c
are
inverses
of
each
Write
the
1
a
f
( x ) = −7 x
+ 4
;
g (x ) = −
(x ) =
+ 3,
x
x
for
(t),
d
≠ 1 ;
formula
means
in
this
and
explain
what
context.
+
d
7
2
f
formula
4
x
7
b
a
other.
x
5
x
3
g (x ) =
1
Find
be
,
x
after
worth
how
many
months
a
computer
will
$500.
≠ 3
1
3
−3 +
c
f
e
3
2x
(2 x
(x ) =
,
x
≥ 0 ;
Find
what
2
Review
1
in
Aryabhata,
came
up
●
The
●
is
a
6th
century
this
added
result
That
●
Then
Hindu
mathematician,
3
puzzle:
●
The
●
Find
to
is
result
8
Ask
is
a
certain
divided
is
the
with
by
Choose
by
2,
a
is
by
from
that
result.
given
34,
write
arrive
subtract
12,
the
at
Invert
the
think
of
a
following
explain
positive
integer.
operations
in
order
number:
Square
●
Multiply
the
result
●
Add
the
result.
●
Ask
the
function,
down
the
add
and
result
to
the
●
the
original
steps,
the
number.
by
2.
3
to
for
the
nal
result.
in
6,
multiply
nally
you
process
process
then
the
above
write
its
number
in
the
inverse
that
your
form
so
of
that
a
you
classmate
can
started
order,
with.
Find
the
original
number.
State
for
number.
the
divide
by
result
numbers
this
puzzle
would
have
no
result
is
solution.
2.
Make
up
your
own
puzzle
whose
the
obtain.
same
b
and
34.
number.
number,
Justify
100,
6.
4
a
=
2.
which
2
(t)
means.
classmate
the
out
to
this
perform
identify
necessary
d
number.
multiplied
subtracted
answer
a
Then
Write
Starting
that
context
on
5
such
2
with
●
t
+ 3)
g (x ) =
and
justify
the
as
the
number
you
start
with.
result.
E2.2 Doing and undoing
53
How
can
we
get
there?
E3.1
Global
context:
Objectives

Describing
Scientic
Inquiry
translations
using
vectors
and
technical
innovation
questions

What

How
is
a
vector?
F

Adding
and
subtracting
vectors
do
you
nd
the
you
perform
magnitude
of
a
vector?
CIGOL

Multiplying

Finding
the
a
vector
by
magnitude
a
of
scalar
a

vector
C

Finding

Using
the
the
dot
dot
betweentwo

Solving
ATL
product
product
of
to
two
nd
the
angle
using
vectors
Critical-thinking
Consider
5 4
ideas
from
multiple

D
problems
do
operations
on
mathematical
vectors?
vectors
vectors
geometric
How
perspectives
How
can
vectors
geometric

Can
logic
be
used
to
solve
problems?
solve
everyday
problems?
G E O M E T R Y
Y
ou

should
nd
the
two
points
already
distance
know
how
between
1

solve
linear
simultaneous
2

use
the
cosine
rule
the
points:
a
(4,
(
7)
and
3,
Solve
3
distance
of
equations
T R I G O N O M E T R Y
to:
Find
b
A N D
2)
the
3x
+
5y
2x
+
y
Find
the
(11,
and
between
=
pair
5)
(1,
1)
simultaneous
=
each
equations:
5
1
size
of
angle A.
A
8
7
cm
cm
B
11
cm
C
Introduction
F
The

What

How
map
You
can
blocks
to
shows
give
is
a
do
part
in
an
vectors
vector?
you
of
nd
Salt
directions
travel
to
in
the
Lake
this
east-west
magnitude
City,
area
of
of
USA,
Salt
direction,
a
which
Lake
and
vector?
in
has
City
the
by
a
grid
layout.
giving
the
north-south
number
of
direction.
lia
9th Ave
9th Ave
rT
A
block
m
o
d
distance
is
the
between
e
e
rF
8th Ave
two
tS
adjacent
intersections.
J
I
H
tS
tS
G
F
tS
tS
D
E
tS
C
tS
tS
B
A
tS
tS
City
8th Ave
7th Ave
7th Ave
6th Ave
6th Ave
5th Ave
5th Ave
4th Ave
4th Ave
3rd Ave
3rd Ave
2nd Ave
2nd Ave
creek
canyon
South Temple
J
E
tS
I
H
tS
tS
G
F
tS
tS
D
E
tS
C
tS
tS
B
A
tS
tS
1st Ave
1st Ave
E
South Temple
E3.1 How can we get there?
5 5
Exploration
1
Using
the
a
intersection
the
H
b
c
You
to
of
and
and
can
and
use
a
column
1st
and
2nd
Street
F
Street
and
vector
to
means
translations
in
instructions
Avenue
to
the
to
get
from:
intersection
of
Avenue
to
the
intersection
of
3rd
Avenue
to
the
intersection
of
represent
a
translation
from
one
point
The
column
the
a
intersection
the
map,
the
the
intersection
5th
each
and
of
a
of
usually
A
J
Street
A
column
vector
A
column
vector
C
units
left
column
and
and
Street
start
down
3rd
and
vector
Avenue
4th
Avenue
and
8th
4
units
for
to
to
Avenue
vectors
describes
an
at
your
the
nal
the
the
the
to
intersection
position
d
column
has
units
up.
down.
journey
from:
intersection
intersection
the
of
intersection
x
a
e
to
describe
of
after
F
a
Street
of
and
translation
of :
f
translations
on
a
coordinate
component
and
a
y
component:
x
component
tells
you
how
far
to
move
in
the
x-direction.
The
y
component
tells
you
how
far
to
move
in
the
y-direction.
Continued
E3 Logic
grid.
translation.
The
5 6
of
Avenue.
c
use
3
the
Street
below,
Write
b
will
down
4
Avenue
3rd
vector
Avenue.
and
Avenue
8th
intersection
GStreet
For
of
8th
and
right
d
means
write
and
units
c
vector
Using
3
words:
b
BStreet
You
and
down
Avenue.
vector
these
CStreet
4
D
column
a
c
Street
write
Avenue
of
6th
A
City,
Avenue
of
7th
intersection
Street
Describe
b
of
3rd
Lake
another.
The
3
Salt
intersection
Street
the
A
map
Street
the
G
2
1
on
next
page
G E O M E T R Y
AB
is
the
vector
representing
the
translation
from
A
to
A N D
T R I G O N O M E T R Y
B.
y
On
this
grid,
5
4
The
position
vector
of
a
point
is
the
vector
from
B
the
origin
O
to
the
3
point.
A
2
On
this
grid,
1
O
is
ofthe
the
zero
is
the
Find
It
is
the
position
0
vector
point
How
(4,
9)
What
the
and
B
is
the
discuss
the
point
5
(6,
4).
A
to
B
you
move
2
units
right
and
5
units
down.
1
vector
operation
coordinates
How
a
4
relate
to
the
coordinates
of
points
B ?
could
Practice
1
and
does
Aand

3
AB
Reect

2
1
From

x
1
origin.
Example
A
vector.
would
of
you
A
nd
you
and
the
perform
to
nd
the
vector
AB
if
you
knew
B ?
vector
BA
from
the
coordinates
of
A
and
B ?
1
Write
down
the
column
vectors
for
these
translations
6
on
the
grid.
B
5
i
AB
ii
AC
iii
AF
iv
GF
v
OD
vi
OC
4
G
3
b
Write
down
the
column
vectors
2
for:
A
1
i
AD
and
DA
ii
EG
and
GE
O
0
c
Conjecture
the
relationship
between
vectors
5
4
3
2
1
x
1
2
3
4
5
6
1
of
the
form
PQ
and
F
QP
C
2
d
Write
down
the
position
vector
E
of :
3
D
i
A
ii
B
iii
C
iv
O
4
5
e
Write
down
the
point
with
position
vector:
6

1

i

3

ii

1

2

iii

3


4
E3.1 How can we get there?
57
2
On
this

0
grid,
nd
two

3

a
points
so
that

the
5

b
translation

one
to
the
other
is:

d
1

4

c
 4
from
1

3

y
5
H
4
3
A
2
D
B
1
0
7
6
5
4
3
2
x
1
1
2
3
4
5
7
6
1
E
C
2
G
3
F
4
5
3
On
a
clean
grid,
coordinates

2
of
start
the
the
point
after
Write
a
M (
2,
c
R (11,
the
4)
to
2)
Problem
2

CD
.
4
Point
A
5)
Write
has
down
position
Show
points
b
Write
down
down
A,
the
column
4
the
the
three
and
b
P (4,
d
T (

AB
C
6)
5,

9
to
Q (11,
1)
to
BC
=
U (
3)
6,
3)
DC
5
of
points,


and
suitable
vector
2
=

on
position
pairs
of
6

3
5


axes.
point
where
C
one
transforms
to
the
other

translation


A
11

vector

vector
B

by


a
Write
down
solving
 2
7
Write
d
6)
S (7,
4).
translates:

6

1


that
(3,

=

3


vector
N (1,
to
coordinates
translation:
c
4

down
with
each
b
 5
5
2


a
4
at
point
journey
2
3


of
is
shown
on
this
5th Ave
section
 2
of
the
Salt
Lake
The
length
The
larger
of
city
the
map.
journey
is
4
blocks.
4th Ave
journeys
with
a
of
page,
length
vector:
a
b
4
blocks.
c,
D
labelled
next
tS
dierent
is
the
C
Each
on
tS
shows
map,
etc.
3rd Ave
5 8
E3 Logic
the
G E O M E T R Y
A N D
T R I G O N O M E T R Y
f
7th Ave
a
6th Ave
b
5th Ave
4th Ave
Tip
Single
e
are
3rd Ave
letter
printed
When
some
d
in
bold
writing
vectors
c
vectors
by
hand,
countries
convention
is
in
the
to
2nd Ave
underline
D
F
tS
C
tS
tS
B
E
tS
A
tS
tS
a,
b,
and
they
are
with
an
them,
in
written
arrow
1st Ave
b
Reect
ATL

Are
on

if
they
are
any
the
What
Vectors
are
of
is
the
does
it
equal
equal
journey
discuss
journeys
smaller
represent
not
The
and
the
starting
if
their
for
blocks
that
point
in
If
have
larger
are
to
be
map
the
equal.
exactly
dierent
the
same
as
the
one
the
same
This
same
directions,
as
each
means
that
other?
they
instructions.
even
if
the
are
Two
length
of
equal
vectors
the
same.
position
of
the
vector
south
has
vectors
are
and
2
equal
if
doesn’t
matter.
Walking
two
blocks
east
and

vector
,

Two
the
journeys
components
journeys
they
two

two
on
map?
mean
if
2
shown
no
matter
where
you
start
from.
2
their
then
components
if
are
and
equal.
only
if
and
E3.1 How can we get there?
e.g.
others
5 9
Exploration
1
A
hiker
walks
a
Sketch
b
Write
c
Find
her
2
3
due
east
and
then
4
miles
due
north.
walk.
down
the
miles
the
total
shortest
distance
distance
she
walks.
between
her
starting
position
and
endingposition.
A
2
The
diagram
Find
the
represents
shortest
the
distance
vector
from
A
AB
to
AB
B.
B
3
The
the
a
Sketch
b
Find
length
other,
written
The
the
of
is
shortest
vector,
the
of
a
theorem
magnitude
Example
(4,
the
to
represent
distance
or
the
this
from
C
(shortest)
magnitude
of
vector.
to
D
in
distance
the
vector.
terms
from
The
of
one
x
and
end
y
of
magnitude
the
of
vector
vector
v
to
is
|
v
|
components
|
v
|,
=
a
diagram
called
Pythagoras’
A
a
11)
vector
to
nd
of
represent
the
,
perpendicular
vector’s
is
given
distances,
so
you
can
use
magnitude.
by
|
v
|
=
2
and
B
=
(
|
v
| =
=
8.06
6 0
(3
s.f.)
E3 Logic
G E O M E T R Y
Vectors
a
and
2
|
a
|
=
But
and

5
b
=
are
b
A
not
7
magnitude:
|
b
|
2
equal
7
=
+
because
( −1)
they
=
50
have
two
their
vectors
vector
a

1

have
same
=
though
When
and
50

and
 5
Even
the
T R I G O N O M E T R Y
components:

=
have
2
+ 5
dierent
a
both
2
5
a
b
A N D
has
lengths
a
both
direction.
magnitude
and
and
b
are
are
the
parallel
magnitude
For
two
same,
and
vectors
they
but
point
point
direction.
to
be
in
A
equal,
in
dierent
opposite
scalar
they
directions.
directions,
quantity
must
have
does
the
a
=
−b
not
same
direction.
Numbers
quantities.
a
scalar
but
1
Find
2
the
magnitude
of
each


 24 
a
given
a
vector
as
Find
3
8


magnitude
of
is
vector

6

The
each
vector.
Give
your
answers
in
exact
3


a
1

2

b
4

also
force
you

gravity
 3
size


 4
3
Find

the
4
magnitude
of
or


13
2
Give
For
(your
weight)
vector
0
8
each
a
A (1,
c
A (
e
A (11,
pair
5)
1,
AB
AB
and
6)
4)
B (2,
and
and
and
4)
B (2,
B (4.5,

11
7)
23.8)


24
8



f
AB

AB
correct
to

7
4

=
6
nd
24
=

=
B,
AB
c

A
10
=

points
s.f.
AB

of
3

6

0.96 

to
2


e
correct
f

0.28

4


6
=
5



b

For
nd
=

AB
AB
(down).

answers
 4
each
and
c

2
AB
your
8


5
vector.


16

e

d

b
d
a
magnitude
f
each
 5
a

 12 

a
4
5
e
5

due
has
c

d
a
quantity.
direction
5

is
quantity.
form.
to

in
c
 4
the
speed
direction)
experience
2
is
vector:
b
a
Speed
velocity
Force
7

scalar
quantity,
(dened
Practice
are
3
3

s.f.
b
A (4,
6)
and
B (3,
d
A (3,
5)
and
B (5.8,
f
A (8.7,
5.3)
and
3)
11.2)
B (2.1,
14.6)
E3.1 How can we get there?
61
a
Problem
6
Four
solving
points,
A (1,
7),
B (6,
2),
C (3,
3)
and
D (2,
4)
lie
in
a
plane.
=
Show
7
Points
Find
AB
that
A
and
four
CD
B
lie
possible

8
The
vectors
u
Determine
the
2x
vectors
u
plane,

x

and
has
such
10
x
3
and

=
B
7x

of
A
coordinates
AB
that
=
(6,
2).
13 .

v =

values
2
and
for

and
y
 x
The
a
positions
=

9
in
are
equal.
are
equal.

y
7x
10

v =
2
 3x 
Determine
C
value
of
How
diagram
do
shows
you
two
x
+
2 
x
Operations

The
the

with
perform
journeys
vectors
mathematical
in
Salt
Lake
operations
on
vectors?
City:
7th Ave
6th Ave
Journey
2
5th Ave
4th Ave
Journey
1
3rd Ave
2nd Ave
tS
F
D
E3 Logic
E
tS
C
6 2
tS
tS
B
A
tS
tS
1st Ave
G E O M E T R Y
Journey

3
1

then
2



then
 4
1
2
=
Whole
2
journey

do
 6

Reect
and
you
discuss
notice
3
about:

the
relationship
between
the
x
components
of
the
LHS
and
RHS?

the
relationship
between
the
y
components
of
the
LHS
and
RHS?
A
translation
as
the
AC
To
is
add
The
sometimes
vectors
vector
=
AB
can
two
The
A
to
from
B
A
called
you
to
the
simply
addition
followed
+
also
BC
resultant
is
AB
by
+
translation
BC
=
of
AB
resultant
their
a
from
B
to
C
is
the
same
AC
and
BC
components.
law:
then
=
draw
vectors
C :
add
and
You
the
from
translation
If
AC
T R I G O N O M E T R Y
=

What
Journey
A N D
so
diagrams
that
the
the
vector
to
add
second
that
vectors
vector
joins
the
using
begins
starting
a
graphical
where
point
the
to
method.
rst
the
vector
tip
of
the
Draw
ends.
second
vector:
v
u
+
v
u
This
shows
the
result
of
performing
both
translations.
A
Using
the
vector
addition
law:
OA
+
AB
=
OB
AB
OA
You
can
rearrange
You
can
also
see
this
this
to
on
AB
the
=
OB
diagram.
OA
Another
route
O
from
A
to
B
B
is:
OB
A
or
to
AO
O,
+
then
OB
=
O
−
to
B
OA
+
OB
=
OB
OA
E3.1 How can we get there?
63
For
any
two
Reect
Explain
Hence
and
is
the
AB
why
explain
Example
A
points
A
and
discuss
BA
+
=
OB
OA
4
must
AB
why
AB
B,
be
equal
to
the
zero
vector.
− BA
=
3
point
(2,
11)
and
B
is
the
point
(
3,
6).
Find
AB
Write
Example
A
has
Find
values
of
(a,
a
3a),
and
B
has
coordinates
(b,
b
6),
and
3a
=
−2
2a
=
6
Practice
1
Find
AC
components
AB
Solve
3
for

a
3
pair

BC
1
0
BC

AB
6 4
8



BC

2
E3 Logic
4

1

9
BC



and
=
BC
10

=

 12 
=
3
AB

and

d
2
=

=

and

AB

 11 
=

6
5

and
2
b


=
vectors:
=
 6
e
of

and

AB
each
=

c
OB
b
components
–
and
.
The
b
OA
4
coordinates
the
down
5
9

=

12 
of
of
AB
the
AB
are
given
equal
in
the
simultaneous
to
the
question.
equations.
G E O M E T R Y
2
Find
the
7)
A (0,
c
A (5,
2)
and
B (
6,
e
A (
0)
and
B (
11,
2,
3

and
and
3

.
A (2a,
b
A (
c
A (2a
point
Find
a
3a
1)
+
+
1,
4,
squared
what
Draw
a
2
Explain
3
a
b
4
+
3a
b
A (
1,
d
A (3,
3)
and
B (4,
2)
9)
and
B (12,
9)
7)
is
(
12,
3).
Find
the
coordinates
of
point
B.
values
of
a
and
b
for
these
coordinate
pairs:
the
any
Explain
7)
and
B (b
5)
and
B (2b
draw
vectors
vector,
for
any
which
nd
b
4)
paper,
why
B (3b,
3,
+
1,
the
have
d,
in
that
vector
v
For
any
vectors
3)
of
these
vector
and
the
same
vector
vectors
is
.
not
v,
are
v
property.
+
v
must
parallel
parallel
to
to
any
be
parallel
each
of
the
to
v
other:
others.
.
and
multiplication
any
+
common.
has
non-zero
which
k v
For
2b
vectors
vector
why
b)
3
fourth
Hence
Scalar
A
the
and
a
Determine
For
8)
coordinates:
solving
Exploration
State
0)
of
7
a
On
pair

=

1
each
7
Problem
AB
1,
for
T R I G O N O M E T R Y

=

4
B (
BA
and
a
AB
3
AB
vectors
A N D
=
u
and
v
of
are
a
if
to
each
other.
vector :
and
v,
parallel
any
u
=
scalar
k v
value
then
u
k,
and
k v
v
=
are
parallel.
E3.1 How can we get there?
6 5
Example
Two
AB
points
is
5
have
parallel
to
coordinates
v
=
.
A (
Find
1,
the
4)
and
value
B (2b,
of
b
+
2).
b
Write
the
position
Find
Parallel
2b
+
1
=
3k
2
=
4k
b
=
−2
Write
b
Example
Four
that
system
of
A (4,
ABCD
simultaneous
6),
B (
forms
2,
a
16),
C (
5,
12)
and
D (
2,
7)
lie
in
a
and
ABCD
CD
has
Practice
vectors
trapezoid:
are
one
parallel.
pair
of
None
parallel
of
the
sides
other
so
it
is
sides
a
are
vectors
u
a
2

=

and

Determine
the
some
parts
book)
a
a
a
for
each
BC ,
side
and
CD
parallel
of
the
world
trapezoid
v =
are
 3a
+ 1
quadrilateral
of

opposite
of
DA
vectors.
is
(andin
dened
as
value
of
have
to
1
parts
A (2,
5)
and
B (b
.
3
E3 Logic
2,
1
b).
of
the
world
The
Find
the
value
of
to
1795,
mathematical
in

v =

exactly
parallel.
one
In
pair
other
this
is
called
confusion
a
dates
a
coordinates

parallel
with
sides
parallel.
+ 3
back
points
6 6
b
trapezoid.
trapezium.
is
for
parallel.
a
AB
solve
solving

Two
other.
4
Problem
2
each
and
for
AB ,
Look
this
The
the
−2
In
1
equations
of
plane.
the
AB
multiples
AB
trapezoid.
Find
=
are
vector
6
points,
Show
the
vectors
the
vectors.
the
USA
to
a
reference
dictionary
that
directly
in
reversed
b
accepted
meanings
of
a
published
the
day.
the
G E O M E T R Y
3
Two
points
7

PQ
P (2a,
a
4)
and
Q (3b
+
1,
2a
T R I G O N O M E T R Y
3b).
The position
.
5
coordinates

=

have
A N D
Find
the
position
vector
of
the
midpoint
of
the
line
segment
of
a
point
vector
4
Four
5
points
a
Use
b
Determine
Four
vectors
points
Show
6
have
that
points
OA
9u
=
to
if
Find
and
have
AB
b
Hence
c
Given
and
show
line
DC
DC
in
that
u
are
= 19 u
2
A (3,
the
lines
AB
AB
A (4,
B (2,
and
and
6),
11),
B (
CD
CD
1,
C (15,
are
are
9),
3)
D (17,
5).
1)
in
and
D (11,
ve
+ 14 v, OC
terms
is
of
a
=
u
and
,

25 u
+ 5v
and
OD
= 15 u .
v
1

v =
and
A (
1
show
that
ABCD
is
a
rhombus.

3,
1),
B (1,
3),
C (4,
3),
D (5,
1)
and
E (
1,
R
points
are
two
sets
Use
vectors
of
four
points,
each
of
which
forms
a
to
show
that
the
three
points
P (5,
4),
Q (10,
0)
and
R (25,
on
a
straight
Reect
Why
does
show
that
discuss
showing
the
On
suitable
Label
2
The
the
that
three
Exploration
1
the
points
if
the
PQ
and
parallel.
5
vectors
lie
on
a
formed
straight
by
three
points
are
parallel
also
line?
4
axes,
vectors
angle
line
line.
and
draw
OA ,
between
OA
the
OB
points
and
and
A (2,
AB
OB
is
on
4)
and
your
B (6,
1).
diagram.
∠ AOB
A
O
B
Describe
3
you
Find
nearest
4
how
Use
the
and
could
.
use
Hence
the
nd
cosine
the
rule
size
of
to
nd
∠ AOB,
the
size
correct
of
to
∠ AOB.
the
degree.
method
in
steps
1
to
3
to
nd
the
angle
Q
a
12)
are
lie
P,
trapezoid.
vectors
8
on
2),
straight
nd
point.
parallelogram.


points
the
5).
and
the
the
to
length.
Three
From
from
O
vectors:
 1
7
the
parallel.
equal
C (1,
and
origin
is
parallel.
ABCD
=
7),
segments
position

that
that
coordinates
+ 9 v, OB
a
show
the
have
AB
Four
coordinates
vector
PQ

between
the
vectors
.
Continued
on
next
page
E3.1 How can we get there?
67
QR
5
The
diagram
Write
down
shows
an
vectors
expression
u
and
for
v
v
u
in
terms
of
v
u
x
v
x
=
and
u
=
v
u
y
y
O
6
7
Write
down
Show
that
expressions
|
v
u|
for
|
u
|
and
|
v
|
=
Use the cosine rule to show that
and
hence
that
Another
For
is
any
two
called
the
vectors
dot
and
product.
It
is
,
written
u
the
the
dot
the
equation
=
Therefore
u
Example
Find
u
the
|
u
|
=
|
v
|
=
v
|
u
|
v
you
|
v
|
have
cos
formed
the
scalar
above,
you
can
see
×
is
that
a
scalar
=
=
|
u
|
|
v
|
cos
.
between
the
vectors
and
.
Find
|
u
|
and
|
v
|
=
|
v
|
cos θ
+
4
×
5
Substitute
in
the
Give
Example
Vectors
a
vectors
and
their
your
answer
the
to
denition
a
suitable
of
the
of
and
b
=
are
perpendicular.
Find
the
value
of
|
a
|
|
b
|
cos
90°
=
angle
and
b
is
90°
=
Substitute
8)
+
(14
7x)
⇒
x
=
0
=
−6
E3 Logic
between
90°,
and
0
cos
6 8
accuracy.
x
a
(8x
product.
8
=
=
dot
degree
The
b
magnitudes.
=
Use
a
result
quantity.
7
angle
|
u
|
−2
product,
the
.
=
1
is
v
=
=
for
product
because
From
name
quantity
Evaluate
the
0.
in
scalar
a
and
b
product.
G E O M E T R Y
If
two
vectors
Equally,
if
a
a
b
=
Practice
5
1
angle
Find

the
4


0
b
are
then
either
between
2
perpendicular
each
a
=
,
pair
of
b
 5
4




16

4
13

2

8
Problem
b
are
nearest
3

perpendicular.
degree.

12 



8

and
 4

and

the
2


0.
and
4
d
that
a
=
T R I G O N O M E T R Y
and


20 


Show
to
26

and
b
or
b
2

a
=
8

c
then
vectors,
and
a
2
and
A N D
are
perpendicular.
 1
solving
20


1
3
Two
speedboats
are
travelling
with
velocities
ms
15 


7

1
and
ms
respectively,
on
an
east-north
coordinate
grid.
 24 
4
a
Show
b
Find
5
the
Triangle
Use
the
Vectors

u
2
that
angle
ABC
dot
u
has
and

v
have
between
vertices
are
3m + 3
to
the
their
A (6,
nd
same
speed.
directions
3),
B (2,
1)
of
travel.
and
C (4,
5).
∠ ABC
perpendicular:

v =
Find
5
 2m

the
value
a

of
m
b


and
 a
+
a
Write
b
Explain
the
2

Vectors
c
boats
product

=

6
both
2
an
why
of
that
Problem
b
you
involving
need
aand
=
1
perpendicular.
b
equation
values
Given
are
 2
more
a
and
b
information
to
determine
b
a,
determine
the
values
of
a
and
b
solving

7
Find
three
dierent
vectors
perpendicular
to
the

8
Parallelogram
OABC
a
Find
the
position
b
Find
the
size
of
has
vertices
vector
of
O (0,
0),
A (4,
8

vector
9)
and
2
B (6,
6).
C
∠OAC
E3.1 How can we get there?
6 9
9
A
rhombus
Find
the
OABC
value
of
has
a
For
the
vectors
a
=

=
(b
+
c)
=
a
b
+
a
Consider
the
2

OC
+
Vector
D
You
(b

How

Can
can
use
=
120°
and
|
a
|
=
4.

,
verify
that
 1
a
x

b

x
c)
=


b =
and
a
y 
b

+
a
c
c
x

c =
b
c
y 

for
any
2D
y 
vectors
a,
b
and
c
geometry
can
vectors
logic
vectors
components.
5
c =
2
a =
vectors
a
∠ AOC
c

that
c

a
Prove
=

and


11
and
b =
,
 5
a
a
c
3

10
OA
solve
in
be
used
everyday
geometric
Remember
to
that
solve
problems?
problems?
problems
you
geometric
can
add
even
when
vectors
you
do
not
know
their
pictorially.
v
u
u
+
+
v
v
v
u
ATL
Exploration
1
The
diagram
u
5
shows
a
pair
of
vectors
a
and
b
b
Copy
a
the
diagram
and
draw
3a
b
the
vectors:
2b
Start
with
copy
for
a
new
each
a
diagram.
O
2
c
a
+
b
d
a
–
b
In
this
N
is
(the
resultant
e
diagram,
the
vector
M
midpoint
is
of
the
of
a
a
+
and
then
b)
b
f
midpoint
of
AB
a
b
and
B
AC .
M
a
Given
that
AM
=
a
and
AN
=
b,
explain
a
why
AB
=
2b
and
write
down
a
similar
A
b
expression
for
AC
Tip
N
b
Explain
Form
a
why
MN
similar
=
b
a
expression
for
C
BC
Remember
MN
c
Hence
show
that
MNCB
is
a
trapezoid.
Continued
7 0
E3 Logic
on
next
page
=
MA
that
+
AN
G E O M E T R Y
3
Four
a
a
points
+
b
A,
B,
C
and
respectively,
This
sketch
Copy
the
shows
sketch
D
have
where
one
and
a
position
and
b
possible
draw
are
pair
points
vectors
a
+
non-parallel
of
A,
vectors
B,
C
and
a
D
b,
a
b,
a
non-zero
and
on
b,
A N D
T R I G O N O M E T R Y
and
vectors.
b
the
diagram.
b
a
O
b
This
sketch
Copy
the
shows
sketch
another
and
possible
draw
points
pair
A,
B,
of
C
vectors
and
D
a
on
and
the
b
diagram.
O
a
b
4
c
Find
d
Hence
This
expressions
show
diagram
for
that
AB
ABCD
shows
and
is
points
a
A,
DC
parallelogram.
B,
C
and
D
A
a
AD
=
a
+
BD
=
2a
BC
=
3a
+
b
b
D
a
2b
a
B
a
Find
AB
b
Find
DC
c
Hence
3a
show
that
and
AB
DC
are
parallel.
C
The
word
‘vector’
think
of
a
place
to
another.
some
other
from
one
folklore
a
set
of
vector
a
insects
on
from
and
does
originally
also
are
to
one
use
another.
answer
parallelogram
to
In
generation
headings
to
Latin
as
the
referred
discuss
your
a
mathematics
We
appropriate
How
is
in
organism
Reect

was
word,
something
word
as
in
the
next.
an
a
‘to
carries
contexts:
because
sociology,
to
that
other
vectors
guide
meaning
And
aircraft
in
in
is
a
you
You
from
transmit
person
aeronautics,
and
disease
that
a
can
one
mosquitos
they
vector
carry’.
passes
vector
is
ight.
6
to
Exploration
regardless
of

Why
was
it
necessary
in
step

Why
was
it
necessary
to
state
3
the
to
step
direction
state
that
5,
a
that
and
b
3c
of
a
show
vectors
and
were
b
not
you
a
were
that
and
ABCD
b?
non-zero?
parallel?
E3.1 How can we get there?
71
Practice
1
Using
6
the
diagram,
express
the
vectors
below
it
in
terms
of
u
v
and
w:
u
A
B
v
C
D
w
a
BD
b
Problem
2
The
CA
c
BC
solving
diagram
shows
a
regular
hexagon
F
ABCDEF
A
a
Find
in
terms
of
a
and
b:
E
B
O
b
a
b
AB
c
OC
d
FO
e
AC
DB
D
3
Points
P,
Q
and
R
form
a
straight
line
and
have
position
C
vectors
R
Q
p,
q,
and
r
respectively.
QR
=
4
PQ
P
q
a
Express
r
in
terms
of
p
and
p
1
b
Hence
show
that
q
=
r
q
(4p
+
r).
5
O
4
Points
S,
T
and
respectively,
U
form
where
a
SU
straight
=
3
line
and
have
position
vectors
s,
t
and
u
S
TU
s
Express
s
in
terms
of
t
and
T
u
t
U
O
5
In
this
diagram,
a
Find
b
The
M
is
the
midpoint
of
AB.
OA
=
3a
and
OB
=
u
A
2b
OM
3a
point
N
has
position
vector
ON
where
ON
=
2OM
M
Show
that
AN
=
2b
O
c
Given
that
a|
=
|b
and
a
b
=
0,
fully
describe
the
quadrilateral
OANB
2b
B
72
E3 Logic
G E O M E T R Y
A N D
T R I G O N O M E T R Y
Summary
A
column
vector
describes
a
A
translation.
translation
from
A
column
vector
has
an
x
component
and
to
C
from
is
the
A
to
same
B
followed
as
the
by
a
translation
translation
from
a
A
y
B
to
C :
+
AB
BC
=
AC
component:
is
AC
The
x
component
x-direction.
move
in
AB
the
A
is
to
The
the
tells
y
you
how
component
far
tells
to
move
you
how
in
the
far
addition
If
representing
the
translation
the
resultant
of
AB
and
BC
law:
and
then
from
B.
AC
The
the
vector
called
to
y-direction.
vector
The
sometimes
position
origin
O
vector
to
the
of
a
point
is
the
vector
=
the
zero
For
any
vectors
BC
=
=
two
points
A
and
B,
AB
=
OB
OA
and
− BA
vector
Scalar
Two
+
from
point.
AB
is
AB
are
equal
if
their
components
multiplication
of
a
vector:
are
equal.
For
If
and
then
if
and
only
The
If
magnitude
of
a
vector
is
its
the
magnitude
of
,
is
v
=
and
u
=
kv
then
u
and
given
k:
v
are
parallel.
any
two
vectors
and
,
the
is
called
the
dot
product
=
A
vector
A
scalar
has
both
magnitude
and
For
two
quantity
vectors
does
to
be
not
have
equal,
a
they
v
=
=
a
direction.
must
have
two
b
=
vectors
magnitude
and
Write
a
a
if
b
a
=
or
a
down
column
translation
of
5
vectors
right
a
translation
of
c
a
translation
from
that
and
4
represent:
3
Find
AC
are
perpendicular
then
=
0
then
either
a
=
,
b
are
perpendicular.
given:

up
AB
2
left
and
3
translation
from
(4,
(9,
5)
to
14)
(7,
to
(
9)
2,
b
AB
AB
and
BA
given
3)
and
B (
2,

A (4,
2)
and
B (
3,
c
A (
7,
1)
and
B (
4,
4)
d
A (
2,
7)
and
B (11,
7)
1


7

and
8
AB

1

=
BC
9

=
 2
the

and
BC
4

=
3
2

points:
AB
6


=
and
 15 
b

=
5).
d
0)
0


A (1,
BC

=

vectors

and
4
up
c
the
2
=

a
b
practice
b
Find
and
and

a
a
the
a
d
cos
direction.
b
Mixed
|
v
|
0.
Equally,
same
|
u
|
direction.
If
2
value
by
u
1
scalar
length.
quantity
|
v|
any
=
For
|
v
|,
vector,
if
kv
and
any
BC
4

=
 0
5)
E3.1 How can we get there?
73
2

4
Given
that
AB
,
b
given
the
nd
the
values
of
Problem
a
solving
1

and

=

coordinates:
10
The
u
vectors
5a
A (a
3,
b
A (2a,
c
A (a
3a
4
+
+
6,
2)
a)
2a
and
and
+
B (
B (
10)
b,
2b,
and
2b
Determine
+
9,
b
+
11
3)
For
the
B (2b
A ( −3a, a − 2 )
d
B (11 + b , 2b
and
4

+
the
points
1,
b
possible
with
the
magnitude
your
answer
of
in
each
of
exact
5

4

a


b
12
these
3

13
of
a
A (
3,
2)
and
is
parallel
to
2

v =
Find
the
value
of
b
9
The
diagram
shows
ve
points
O,
OX
=
XB
=
a
B,
C
andX.
B
and
16 

A,

d
 15 

values
vectors,


c
 2

form.
12

2a
4):

giving
are
coordinates

Find

a + 7
8

+ 1)
AB
5

v =

parallel.
1)
b)
B (2b
2
and
a

a
+
=
A
CX
=
XA
=
b
a
X
b
a
6
M (
3,
2)
and
N (2,
5).
Show
that
OABC
is
a
b
Find
7
a
Find,
b
ON
correct
to
the
MN
nearest
c
degree,
parallelogram.
NM
the
C
O
angle
Problem
solving
between:

3

2

a


and
1

 7
1

4

b
13
The
 1


5
3


c
2
a
quadrilateral
OABC
=
OC
a
=
and OA
2a
=
b.
Lines
OB
and
AC

and

shows

AB

diagram

and
intersect
at
B
X
 4
2
a
C
a
Explain
why
OABC
is
A
X
Problem
a
solving
trapezoid.
b
2a
8
From
of
the
list
vectors
the
nine
which
Determine
of
of
are
which
vectors
parallel
vector
is
below,
to
select
each
not
pairs
b
Show
c
Find
6
to

 20 
16


 20 

12


18 
36
15

5


 8


 25


 45

9


15

4
6
9
Show
that
the
1


vectors
9
Review
in
to
each
D:
By

b)
for
some
to
explain
constant
k
=
b
+
m (2a
OA
–
+
b)
AX ,
for
show that
some
constant
m.
Hence
which

nd
X
OX
and
divides
determine
the
ratio
are

2
other.
context
Applying
mathematics
these
questions,
a
oor
cleaning
robot
has
been
in
on
appropriate
mathematical
the
oor
of
a
large
room.
It
measures
progress
solving
authentic
real-life
across
the
oor
in
10
cm
units,
so
the
in
the
strategies

situations
1

vector
represents
a
movement
of
10
cm
 0
Think
about
how
magnitude,
direction,
the
dot
x-direction.
product,
in
AC .
contexts
select
when
(a +
considering
OX
its
ii.
k
OB
30 
placed
real-life
=
on

In
Objective
OX
lies

and
 4
perpendicular
8
b
O
X
f

a +
AC
d
e

=
any
others.

OB
other.
parallel
why

that
etc.
relate
to
the
questions
being
asked.
1
Show

3
,
 4
74
E3 Logic
that
if
the
robot
travels
on
a

it
covers
a
distance
of
50
cm.
vector
of
G E O M E T R Y
2
Find,
correct
travelled
by
to
the
the
nearest
robot
as
it
centimeter,
moves
on
the
each
distance
5
The
its
vector.
robot
axes
is
are
placed
not
in
lined
A N D
T R I G O N O M E T R Y
another
up
with
room,
the
PQRS,
but
sides.
R

3

2

a


b
 7
9

4


15

d

 3

40

e

f
3

6
c
25


S
20 

y
3
In
a
rectangular
sothat
the
the
room,
x-and
the
robot
y-directions
is
are
positioned
the
edges
of
Q
O
x
room.
The

robot
25
starts
in
one
corner
at
(0,
0),
travels

P
on
until

0
0

then
it
reaches
another
corner
and

The
robot
until
it
reaches
a
third
corners
to
2
Find
the
area
of
the
room
in
be
10

m
OP
Write
down
the
vector
which
would
location
of
robot
to
its
starting
The
robot
room
Find
4
The
and
the
travels
then
total
along
distance
robot
is
put
ABCDEF.
It
starts
inside
room.
OQ
in
the
at
two
sides
back
to
which
room
is
Its
x-direction
25


OS
and
DE
and
its
y-direction
=
 15 
a
Find
PQ
and
b
Show that
c
Find
and
PS
below,
walls
PQ
and
PS
are
perpendicular.
SR
parallel
to
Determine
parallel
to
whether
SR
and
PQ
are
parallel.
AF,
determine
whether
the
room
is
AB,
rectangular.
CD


somewhere
is
is
50
the
Hence
BC
2
start.
travelled.
L-shaped
O,
of
the

=

OR =
d
the
room’s
return
point.
straight
34


 45
c
the
follows:
30 

the
as
=

b
the
corner.
 60 
a
nds

on
Justify
your
answer.
FE
6
D
The
robot
is
placed
at
O,
somewhere
in
a
E
four-sided
the
room,
room.
U,
V,
It
W
nds
and
the
four
corners
of
X
B
C
y
You
are
given

15
OW

OX

that
OA
14

=

,

OC
30 
a
Find
vector
b
Find
the
the
2
robot

and
OE
and
the
22
Hence
nd
c
Determine
d
Calculate
the
vector
distance
which
the
is


50

=

0

a
Find
UV
b
Hence
c
Show
and
is
a
XW
 25 
distance
from
A
to
E
show
that
and
that
VW
is
the
room
trapezoid.
perpendicular
to
both
XW
OF
from
longer,
perimeter


UV
position


UX
22
0
=
 15 
AE
32

determines

=
40
information:
=
25
=

process,

OV
F
A
mapping
vector

=

its
following
x
O
In
the
and
C
AC
to
or
area
F
CE
of
the
d
Find
the
area
e
Find
the
size
of
of
the
room.
∠VUX
room.
E3.1 How can we get there?
75
How
to
stand
out
from
E4.1
the
crowd
Global
context:
Objectives
●
Making
and
●
Using
Inquiry
inferences
standard
Identities
about
data,
given
the
mean
F
deviation
dierent
forms
●
of
the
standard
deviation
and
relationships
questions
How
do
you
deviation
●
What
●
How
is
of
the
calculate
a
data
the
standard
set?
normal
distribution?
formula
SPIHSNOITA LER
●
Understanding
●
Making
●
Using
the
inferences
the
normal
about
standard
distribution
normal
deviation
C
distributions
and
the
is
the
deviation’
meaning
of
represented
‘standard
in
its
dierent
formulas?
mean
●
Can
●
Do
samples
give
reliable
results?
D
●
Using
mean
ATL
unbiased
and
estimators
standard
of
the
population
we
want
to
be
like
everybody
else?
deviation
Communication
Make
inferences
and
draw
conclusions
4 3
12 3
E4 1
Statement of Inquiry:
Generalizing
can
help
7 6
to
and
clarify
representing
trends
relationships
among
individuals.
S TAT I S T I C S
Y
ou
●
should
nd
the
mean
frequency
of
already
discrete
grouped
from
1
know
a
tables
This
how
table
gives
understand
4
5
Frequency
8
29
33
35
95
table
gives
the
2
and
for
w
the
chia
the
weight
seeds.
mean
(g)
of
an
Frequency
< 75
10
75 ≤ w
<
80
22
80 ≤ w
< 85
38
85 ≤ w
< 90
10
these
grams
weight.
70 ≤ w
Dene
in
Calculate
terms:
population
a
b
sample
ultimate
●
How
●
What
do
is
Exploration
temperature
in
of
between
population
recorded
grade.
3
This
a
The
for
mean
2
Weight,
The
grades
the
1
estimate
F
exam
Grade
80packets
a
the
Calculate
and
data
dierence
P R O B A B I L I T Y
to:
200students.
b
●
A N D
La
you
the
sample
measure
calculate
normal
the
standard
dispersion
deviation
of
a
data
set?
distribution?
1
on
the
Crieux,
rst
day
France
of
and
Temperature
La
of
Crieux,
France
the
rst
ve
Betzdorf,
months
of
the
year
was
Luxembourg.
(°C)
Betzdorf,
Luxembourg
January
10
18
February
16
10
March
14
13.5
April
12
14
May
18
14.5
Continued
on
next
page
E4.1 How to stand out from the crowd
77
1
For
this
data,
tendency
When
there
two
is
called
2
sets
of
another
dierences
the
Copy
calculate
(mean,
data
between
and
in
the
range
and
an
statistical
standard
the
mode
and
median).
experiment
measure
data
sets.
that
You
the
three
What
have
you
will
measures
do
you
the
can
now
same
use
of
central
notice?
to
mean
help
explore
and
range,
discover
this
measure,
deviation.
complete
this
table
for
the
La
Crieux
data.
Calculate
the
mean
A
of
the
dierences,
and
the
mean
of
the
squares
of
the
tardigrade
tiny
animal
lives
Dierence
(°C)
recorded
mean
of
the
of
a
that
variety
environments,
from
the
tropics
to
temperature
the
10
in
temperature
dierence
and
a
between
Square
Temperature
is
dierences.
Also
16
−
4
polar
regions.
known
Water
as
Bears,
these
16
tiny
can
creatures
withstand
14
temperatures
12
ranging
273
18
Mean
of
dierences
=
Mean
of
of
dierences
Square
mean
of
3
Determine
the
mean
root
4
of
of
the
from
5
the
Decide,
Reect
●
Squaring
do
of
this
and
Why
we
of
the
the
of
the
squares
whether
original
algorithm
the
is
of
that
squares
why
the
mean
best
for
need
of
the
represents
temperature
using
and
you
root
of
of
the
square
take
the
squares
dierences
to
=
the
=
root
of
square
dierences.
dierences
the
or
the
dierence
mean
(deviation)
values.
mathematical
notation
and
vocabulary.
1
dierences
numbers
Why
of
explain
dierences
the
square
mean
Hence,
discuss
large
mean?
for
reasons,
mean
●
78
mean
with
the
units
squares.
squares
Generalize
the
the
of
squares
makes
between
them
important?
E4 Representation
even
the
data
larger.
and
How
the
does
mean?
that
impact
°C
from
to
150
°C.
S TAT I S T I C S
In
to
Exploration
represent
larger
mean
may
The
how
be
is
close
small
a
is
of
original
units
dierences
the
The
a
square
measure
of
values
are
deviation
shows
that
standard
the
the
mean
gives
of
the
the
the
are
gives
more
data
positive
weighting
further
mean
standard
dispersion
to
deviation
the
also
root
called
data
of
It
because
dispersion
is
from
mean.
important
deviation
standard
away
of
values
to
the
from
the
these
P R O B A B I L I T Y
squared
deviation.
that
gives
an
idea
of
mean.
data
the
values
same
are
close
to
the
the
units
of
the
as
data.
The
standard
The
notation
calculated
the
from
signicant.
measure
the
The
original
which
more
standard
mean.
Use
squaring
deviation
deviations,
deviations
A
1,
the
A N D
deviation
used
for
Greek
a
for
whole
letters
shows
the
=
mean
●
=
standard
of
standard
population
for
●
how
the
representative
deviation
or
for
a
populations
the
depends
sample.
Use
population
●
deviation
on
The
Roman
●
mean
is
of
the
whether
it
convention
letters
x
=
mean
s
=
standard
of
for
the
data.
is
is:
samples
sample
deviation
x
of
Standard
to
mean
x
means
this
,
In
the
sum
the
µ)
(x
formulae
sum
sum
1
also
use
the
Greek
the
symbol
sample
Σ
(upper
case
sigma)
of ’.
notation
Exploration
of
population
deviation
‘the
Using
the
of
of
the
(called
all
you
all
the
x
values.
sigma
values
found
the
notation),
divided
standard
the
dierences
the
squares
the
mean
by
the
the
mean
number
deviation
between
the
of
by
data
a
of
set
of
values
is
values.
calculating:
values,
x,
and
the
mean,
2
µ)
(x
of
the
dierences
Tip
2
Σ(x
−
µ)
of
the
squares
of
the
dierences
You
n
2
Σ( x
−
can
standard
calculate
deviation
µ)
the
square
root
of
the
mean
of
the
squares
with
your
GDC.
n
Most
for
GDCs
the
standard
instead
E4.1 How to stand out from the crowd
use
sample
deviation,
of
.
79
Example
At
a
1
summer
weight
of
weights,
a
in
18.3,
carnival,
small
goat.
a
small
The
crowd
rst
nine
of
people
people
to
are
trying
guess
give
to
guess
the
the
following
kilograms.
21.6,
22.2,
24.0,
a
Find
the
mean
b
Find
the
standard
c
Write
a
Mean
down
20.4,
17.4,
14.9,
26.7,
23.1
weight.
what
deviation
the
=
of
the
standard
=
20.96
weights.
deviation
(2
means
d.p.)
in
The
this
nine
context.
weights
b
are
Make
the
a
population,
table
to
x
18.3
20.96
2.66
7.07
21.6
20.96
0.64
0.4096
22.2
20.96
1.24
1.5376
24.0
20.96
3.04
9.2419
20.4
20.96
−
0.56
0.3136
17.4
20.96
−
3.56
12.674
14.9
20.96
−
6.06
36.724
26.7
20.96
5.74
32.948
23.1
20.96
2.14
4.5696
=
n
c
=
9
This
implies
spread
No,
of
you’re
you’re
not
exclusive
because
twisted
are
105.5024
up
not
to
8 0
seen
mean
love
at
a
a
the
the
the
dozen
goats
nuts
or
that’s
This
nuts
estimate
heavier
photo
either.
and
for
When
with
weight
kilograms
things
Morocco,
their
branches.
often
the
3.42
looking
seeing
to
of
that
is
are
E4 Representation
is
20.96
lighter
been
an
who
that
more
or
retouched
Argania
climb
grow
ripe,
goats
on
the
in
kilograms,
than
tree,
them
its
the
and
almost
do
so
thorny,
Argania
them
at
a
mean
trees
time.
but
with
weight.
a
so
organize
use
Greek
your
letters.
calculations.
S TAT I S T I C S
Practice
1
For
each
then
in
set
a
of
data,
rst
calculator.
nd
the
Explain
mean
what
3
standard
deviation
deviation
by
hand,
represents
b
2
2
4
4
4
5
6
6
8
9
13.1
20.4
17.4
21.0
14.8
12.6
16.5
d
The
15
cm
17
cm
14
cm
44.3
kg
41.5
kg
36.5
kg
12
cm
14
cm
18
cm
41.0
kg
33.6
kg
41.8
kg
14
cm
13
cm
18
cm
51.2
kg
39.2
kg
37.9
kg
duration,
in
minutes,
for
a
train
21
24
30
21
25
27
24
27
25
29
23
22
26
28
29
a
Find
b
State
Beatriz
route
ATL
the
standard
context.
c
ATL
and
the
a
2
P R O B A B I L I T Y
1
with
each
A N D
the
the
has
on
mean
and
standard
inferences
a
ve
choice
that
of
you
routes
occasions.
The
journey
is
recorded
deviation
of
can
from
to
draw
school.
times,
the
She
to
15
16
20
28
21
Country
19
21
20
22
18
a
Calculate
b
State
4
Find
5
The
the
the
the
route
mean
heights
mean
(in
you
and
and
standard
would
standard
meters),
of
a
deviation
recommend
deviation
squad
of
of
12
and
the
1.75
1.80
1.75
1.72
1.81
1.72
1.79
1.63
1.69
1.75
Find
b
A
the
new
and
mean
player
standard
and
standard
whose
height
deviation
of
deviation
is
the
1.98
m
heights
of
the
joins
of
of
train
journeys
days:
journey.
along
minute,
each
were:
route.
with
integers
players
heights
the
the
consecutive
this
nearest
each
basketball
1.62
for
her
explain
set
15
data.
the
for
1.65
a
the
timed
recorded
Motorway
roads
times
on
of
squad.
reasons.
1,
2,
3,
4,
…,
15.
are:
the
Find
players.
the
new
mean
squad.
E4.1 How to stand out from the crowd
81
Problem
6
a
solving
Calculate
3
Each
the
6
7
number
b
Predict
c
Calculate
d
Comment
One
of
now:
is
6,
known
e
to
table
the
the
9,
.
standard
deviation
of
this
set
of
data:
is
then
to
the
and
increased
mean
standard
and
by
4.
standard
deviation.
deviation.
answers.
in
The
the
original
new
mean,
data
y,
is
set
has
changed,
unknown
but
the
so
the
data
standard
set
is
deviation
34
the
two
notation,
is
mean
your
x.
set
happen
new
on
the
10
data
will
be
Calculate
Using
in
numbers
7,
and
9
what
the
3,
mean
This
possible
the
is
mean
the
sum
values
of
a
set
of
all
of
of
the
the
unknown
discrete
x f
data
values
values
values
divided
by
x
in
and
a
the
y.
frequency
total
frequency.
A
in
formula
a
to
Reect
How
is
Example
scores
the
in
the
standard
for
similar
How
is
deviation
for
deviation
of
discrete
data
presented
is:
discuss
formula
table
population?
given
table
and
the
frequency
The
calculate
frequency
it
the
to
2
standard
the
formula
for
the
discrete
standard
data
in
deviation
a
of
2
for
this
standard
25
players
frequency
in
a
table.
golf
Find
tournament
the
deviation.
mean
are
x
f
66
2
67
3
68
4
69
5
70
4
71
4
72
3
and
Continued
8 2
a
dierent?
E4 Representation
on
next
page
S TAT I S T I C S
x
f
xf
66
2
132
67
3
201
68
4
272
69
5
345
70
4
280
71
4
284
72
3
216
Use
x
f
66
2
3.2
10.24
20.48
67
3
2.2
4.84
14.53
68
4
1.2
1.44
5.76
69
5
−
0.2
0.04
0.2
70
4
0.8
0.64
2.56
71
4
1.8
3.24
123.96
72
3
2.8
7.84
23.52
=
A N D
to
P R O B A B I L I T Y
nd
the
mean.
80
To nd the standard deviation, use the formula.
The
mean
score
Example
The
table
is
69.2
and
the
standard
deviation
1.79
the
of
total
penalty
penalty
a
season
the
for
team.
standard
all
20
Find
players
the
Number
of
minutes
minutes
hockey
s.f.)
frequency
Total
in
(3
3
gives
distribution
is
mean
on
(t)
players
a
and
0
≤
t
<
10
3
deviation.
10
≤
t
<
20
9
20
≤
t
<
30
6
30
≤
t
<
40
2
Continued
on
next
page
E4.1 How to stand out from the crowd
83
Total
penalty
Mid-value
minutes
0
≤
t
<
(t)
f
tf
5
3
15
(t)
10
10
≤
t
<
20
15
9
135
20
≤
t
<
30
25
6
150
30
≤
t
<
40
35
2
70
Use
an
to
estimate
calculate
for
the
mean.
Total
Midpenalty
f
value
minutes
0
≤
t <
(t)
(t)
10
5
3
13.5
182.25
546.75
10
≤
t
<
20
15
9
3.5
12.25
110.25
20
≤
t
<
30
25
6
6.5
42.25
253.5
30
≤
t
<
40
35
2
16.5
272.25
544.5
Use
The
mean
Practice
1
Find
the
number
of
penalties
is
18.5
and
the
standard
deviation
is
8.53
2
mean
and
the
standard
deviation
for
each
set
of
data.
a
b
x
1
2
3
4
5
f
3
4
4
5
2
Interval
f
1
4
5
2
8
0
9
12
13
3
16
17
3
20
4
21
24
25
2
28
1
c
Interval
Frequency
8 4
0 ≤
x
< 5
10
5 ≤
x
< 10
12
E4 Representation
10 ≤
x
< 15
15
15 ≤
x
< 20
12
(3
the
standard
s.f.)
deviation
formula.
S TAT I S T I C S
2
The
weights
of
90
cyclists
in
a
Number
Weight
bicycle
race
were
A N D
P R O B A B I L I T Y
recorded:
of
(kg)
cyclists
a
40
≤
w
<
50
8
50
≤
w
<
60
27
60
≤
w
<
70
30
70
≤
w
<
80
25
Calculate
i
b
the
A
an
mean
cyclist
or
3
Copy
and
ii
wear
more
without
Problem
for
weight
must
deviation
compete
estimate
a
weight
below
a
the
weight
belt
if
mean.
his
the
or
Find
standard
her
the
complete
this
1
50
to
2
6.15
43
80.34
5.186
39.56
table
standard
2.06
5.05
223
the
standard
weight
table:
197
and
is
belt.
20
Copy
weight
minimum
solving
10
4
deviation.
below
and
deviation
ll
for
in
the
the
5
missing
values,
1.72
then
calculate
the
mean
data.
MidLength
(cm)
f
value
0
5
<
<
x
x
5
2.5
3
10
7.5
4
≤
≤
xf
( x )
10
<
x
≤
15
12.5
6
15
<
x
≤
20
17.5
7
20
<
x
≤
25
22.5
7.5
11.4
389.88
30
−
6.4
163.84
75
1.4
122.5
90.72
8.6
369.8
2
Σf
=
Σ fx
= 347.5
Σf ( x
− µ)
=
E4.1 How to stand out from the crowd
8 5
Exploration
32
students’
Test
test
scores
score
Frequency
1
The
data
thebars.
is
2
were
recorded:
1
2
3
4
5
6
7
8
9
1
2
3
6
9
6
3
2
1
shown
Describe
in
this
the
histogram.
shape
of
the
The
curve
joins
the
midpoints
of
curve.
10
9
8
7
ycneuqerF
6
5
4
3
2
1
0
1
2
3
4
5
6
Test
2
Calculate
3
Determine
4
From
a
the
State
the
b
the
mean
where
original
how
and
the
standard
mean
lies
7
9
x
10
score
deviation
on
8
the
for
the
data,
to
1
d.p.
curve.
data:
many
data
items
are
more
than
many
data
items
are
within
1
standard
deviation
from
mean.
State
how
1
standard
deviation
of
the
mean.
c
Find
the
percentage
deviation
The
data
The
in
to
graph
is
8 6
the
a
the
2
has
distribution
mean
of
the
data
items
that
are
within
1
standard
mean.
Exploration
normal
close
of
and
bell-shaped
is
a
a
tailing
normal
distribution.
symmetric
o
evenly
curve.
E4 Representation
distribution,
in
either
with
direction.
most
Its
values
frequency
S TAT I S T I C S
Data
on
normal
natural
attributes,
distribution.
The
99.7%
are
such
as
normal
of
the
data
95%
are
height,
are
within
within
2
68%
1
If
you
you
know
can
that
infer
●
68%
of
●
95%
lie
●
99.7%
This
is
the
of
a
within
€8000.
values
±2
within
and
at
for
Salsa
for
of
the
looks
follow
like
a
this:
mean
within
+ σ
μ
μ
follows
a
+
normal
+
μ
2σ
3σ
distribution
then
of
is
the
you
±1
standard
of
deviations
the
of
inferences
deviation
of
the
mean.
mean.
the
mean.
about
sets
of
data.
3
have
€5000.
starting
company
dierent
Sometimes
deviations
generally
curve
deviation
deviations
Erizo
a
mean
starting
Employees
salary
would
of
you
at
€41 000
rather
salary
rival
with
work
representations
dierent
How
within
making
del
mean
Which
●
age
deviations
μ
standard
deviation
a
lie
discuss
La
have
±3
Dierent
C
are
standard
distribution
standard
useful
standard
Zorro
frequency
data
extremely
Employees
del
the
standard
standard
σ
μ
3
and
distribution
P R O B A B I L I T Y
that:
lie
Reect
with
2σ
μ
3σ
μ
weight,
frequency
A N D
a
of
€45 000
company
standard
La
Salsa
deviation
for?
of
a
formula
purposes
meaning
of
‘standard
deviation’
represented
in
its
formulas?
may
not
be
Tip
given
the
whole
set
of
data,
but
just
some
summary
Most
statistics
taken
from
the
raw
data.
Usually
this
includes
Σ x
(the
sum
of
all
GDCs
have
summary
statistics
2
data
values)
and
Σ x
(the
sum
of
the
squares
of
all
the
data
a
the
values).
page,
which
gives
2
x
and
you
input
x
a
when
set
of
data.
E4.1 How to stand out from the crowd
87
Exploration
3
Here
values
are
the
data
Population
2,
4,
4,
,
5,
1
Find
the
2
Find
3
Find
4
Find
5
Generalize
the
,
the
If
6
When
ATL
a
1
of
you
ndings
of
the
ndings
of
with
analyze
14
of
Population
1.
of
Population
2.
sigma
notation:
combined
populations
2).
using
sigma
notation:
the
2
Expand
3
Distribute
the
sigma
4
Distribute
the
n:
5
Substitute
6
Collect
7
Substitute
8
Write
like
your
Reect
formula
mathematical
standard
notation
deviation
(called
formula
sigma
and
represent
for
the
population
standard
deviation:
numerator:
notation:
=
=
and
.
terms.
again.
new
and
does
the
4
Start
the
this
ways.
1
your
formula:
discuss
new
information
8 8
11,
________
manipulate
can
Exploration
What
10,
2.
values
using
Population
your
can
you
variety
What
squared
9,
1.
values
Population
the
mean
+
squared
6,
2
________
notation),
in
of
Population
the
=
the
Generalize
of
mean
then
,
=
of
your
(Population
7
Population
4,
mean
Find
populations:
7
mean
,
two
1
mean
the
for
do
=
4
formula
you
________
for
need
to
standard
use
E4 Representation
to
deviation
calculate
mean?
standard
deviation?
it
S TAT I S T I C S
A
formula
to
calculate
standard
deviation
from
the
summary
A N D
P R O B A B I L I T Y
statistics
2
x
ATL
and
x
Example
The
length
follows
a
statistics
is:
4
of
time
normal
t,
in
are:
a
Find
the
b
Make
a
a
Mean
=
minutes,
distribution
for
and
a
is
and
mean
and
to
travel
on
15
down
runs.
a
mountain
The
summary
.
standard
generalization
skier
recorded
deviation
about
the
times
of
the
for
times
the
ski
taken
for
this
ski
run.
run.
min
Standard
deviation
=
2.18
min
Calculate
b
8.6
–
2.18
=
6.42
8.6
+
2.18
=
10.78
standard
side
data
68%
ATL
of
the
Example
The
runs
summary
statistics
between
6.42
min
and
10.78
the
mean.
values
lie
1
on
68%
either
of
between
the
these.
min.
for
a
class
of
15
students’
test
marks
are
.
a
Calculate
the
b
A
class
second
12.5
mean
marks
c
Calculate
d
Compare
(Assume
a
take
values
5
and
of
will
of
the
deviation
the
the
test
of
for
and
25
the
mean
standard
students
same
mark
performance
marks
follow
had
test.
of
all
of
deviation
mean
of
the
students
two
marks
and
for
in
classes
normal
the
63.5
Calculate
the
the
a
of
the
on
the
this
standard
second
two
this
for
class.
deviation
class.
classes.
test.
distribution.)
marks
The
Continued
on
next
mean
is
page
E4.1 How to stand out from the crowd
8 9
Substitute
known
values
into
the
b
formula
for
and
solve
for
c
so
Therefore,
d
The
of
mean
15
and
than
for
Assuming
marks
class,
ATL
in
these
1
The
the
deviation
of
marks
class
s.f.)
were
are
normally
would
students’
be
swan
since
fully
the
assume
is
the
1926.
grown
mean
generalization
2
The
mean
players
slightly
distributed,
between
marks
would
that
all
national
In
a
mute
the
bird
breeding
swans
data
of
52.05
be
lower
68%
and
of
for
the
71.95.
between
farm
was
and
are
follow
51
normal
Denmark
the
class
students’
For
and
the
second
76.
in
and
distributions.
has
Copenhagen,
been
data
a
on
protected
the
weights
collected:
kg
Find
both
25.
3
mute
50
of
class
test
rst
questions,
species
of
standard
the
the
the
68%
Practice
In
(3
kg
and
standard
about
the
standard
1.80
m
and
Σ x
and
deviation
weights
of
deviation
0.0432
m
for
mute
of
the
the
data
collected
and
make
a
swans.
heights
of
a
squad
of
15
football
respectively.
2
a
Find
b
A
new
and
c
In
player
standard
Provided
this
3
Σ x
whose
height
deviation
this
data
is
of
is
the
1.98
m
heights
normally
joins
of
the
the
distributed,
squad.
Find
the
new
mean
squad.
make
generalizations
from
data.
Houston,
deviation
of
Texas,
155
the
mean
annual
rainfall
is
1264
mm
with
standard
Exceptional
mm.
are
a
One
year
Houston
was
b
The
this
9 0
an
the
rainfall
follows
a
next
year
an
saw
1100
normal
exceptional
was
was
mm.
Assuming
distribution,
that
determine
the
rainfall
whether
or
a
total
of
1400
year.
E4 Representation
mm
of
rain.
Determine
than
1
in
not
standard
deviation
from
mean.
this
year.
exceptional
more
results
whether
or
not
the
S TAT I S T I C S
4
In
a
year
spent
on
a
Find
b
Three
42
group,
their
the
mean
new
State
Problem
5
Twenty
money
of
extra
the
and
mean
recorded
night.
The
minutes
joined
minutes
new
each
one
number
25
the
what
students
students
minutes,
Calculate
c
45
homework
spent
year
30
you
Σ x
of
=
minutes,
x,
P R O B A B I L I T Y
they
2230.
homework.
and
reported
respectively
these
would
number
was
on
group
minutes
including
information
the
total
A N D
three
need
that
on
they
their
spent
homework.
students.
to
nd
the
standard
deviation.
solving
ve
students
they
were
received.
The
asked
about
summary
the
amount
values
were
of
monthly
calculated
to
pocket
be
2
Σ x
=
additional
Find
the
mean
●
Can
●
Do
purpose
from
sample
a
and
we
of
x
$228 361
give
to
statistics
of
the
the
like
to
for
all
pocket
this
money:
$85
and
$125.
data.
results?
everybody
make
else?
inferences
population.
sample
monthly
population
reliable
be
is
their
deviation
versus
want
and
added
standard
samples
sample
mean
=
students
Sample
key
data
Σ x
and
Two
D
A
$2 237
For
standard
a
about
sample,
deviation
a
population
you
can
by
using
calculate
the
s
x
If,
for
example,
Elstar
all
apples
Elstar
whole
larger
The
you
you
apples.
But
population
than
the
weights
want
can
use
to
you
sample
(in
to
make
need
and,
grams)
make
the
of
inferences
mean
as
predictions
to
take
into
therefore,
a
an
sample
of
about
estimator
about
the
account
there
25
is
whole
the
population
average
standard
that
more
Elstar
a
for
the
apples
deviation
population
likely
to
be
a
selected
at
132
122
132
125
134
129
130
131
133
129
126
132
133
133
131
133
138
135
135
134
142
140
136
132
135
of
weight
is
of
of
the
much
larger
spread.
randomare:
Σ x
The
sample
mean
is
x
=
= 132.48 g
n
2
Σ
The
sample
standard
deviation
is
s
( x
−
x
)
=
= 4.28
g
n
n
In
statistics,
you
good
estimates
these
values
an
unbiased
value
is
an
want
of
are
the
called
estimator
unbiased
the
sample
same
values
‘unbiased
of
the
values
for
entire
estimators’.
population
estimator
(mean,
the
for
the
For
mean.
mean.
standard
deviation,
population.
example,
The
For
When
the
notation
the
apples,
etc.)
they
sample
shows
µ
to
be
are,
mean
that
= 132.48
is
the
g
E4.1 How to stand out from the crowd
91
When
you
squared
When
a
deviations
you
sample,
number
of
the
a
an
in
an
of
sample
As
a
result,
the
higher
than
better,
and
into
if
mean
sum
by
of
of
sample
standard
the
the
number
of
population
squared
less
deviation,
one,
1.
divide
items
in
standard
deviations
n
you
This
from
gives
the
the
sum
population
deviation
the
an
of
mean
from
by
unbiased
n
the
estimator
deviation.
items:
estimate
of
the
mean
for
of
the
standard
the
whole
population
is
,
estimate
deviation
for
the
whole
is:
estimate
you
used
unbiased,
account
standard
mean
unbiased
population
the
estimate
the
the
n
unbiased
population
an
divide
items
sample
the
●
you
of
a
from
calculate
population’s
For
●
calculate
that
the
of
the
the
standard
population
estimate
sample
of
the
may
deviation
standard
population’s
not
exactly
for
the
population
deviation
formula.
standard
represent
the
is
deviation
whole
slightly
This
by
range
gives
a
taking
of
the
population.
Dividing
population
Reect
●
Would
●
Are
and
you
1
machine
cm.
A
times
1.0
produced
deviation
when
from
an
unbiased
summary
estimate
statistics
of
the
fact
you
are
working
a
is:
made
sampling
from
would
a
be
sample?
Explain.
absolutely
necessary?
examples.
produces
of
8
1.1
the
by
1000
ball
The
1.0
standard
the
ball
bearings
bearings
results
in
0.8
that
each
taken
day
from
centimeters
1.4
deviation
machine
is
of
the
1.3
with
the
a
mean
production
diameter
line
and
of
the
are:
0.9
diameters
of
1.1
the
ball
bearings
day.
Continued
9 2
whole
5
inferences
measured.
Determine
calculating
6
sample
diameters
for
discuss
trust
specic
Example
A
standard
there
Give
formula
E4 Representation
on
next
page
1
the
rather
alternative
n
for
with
The
by
compensates
that
only
sample
than
with
the
population.
S TAT I S T I C S
A N D
P R O B A B I L I T Y
x
Calculate
1.0
(1.0
1.075)
0.005625
1.1
(1.1
1.075)
0.000625
1.0
(1.0
1.075)
0.005625
0.8
(0.8
1.075)
0.075625
1.4
(1.4
1.075)
0.105625
1.3
(1.3
1.075)
0.050625
0.9
(0.9
1.075)
0.030625
1.1
(1.1
1.075)
0.000625
To
=
estimate
deviation
0.275
of
production)
0.1982
Objective
iii.
apply
You
must
estimate
D:
the
decide
for
the
appropriate
ATL
Practice
1
Each
had
for
a
10
an
in
you
days,
The
12
arrive
On
whether
population
need
the
standard
in
the
in
population
from
a
sample
(day’s
use:
d.p.)
real-life
strategies
standard
deviation,
contexts
successfully
deviation
and
for
baby
the
recorded
minutes
0
whole
25
Auberon
times
5
estimate
farm,
then
of
the
to
reach
data
correctly
set
apply
a
or
solution
an
the
are
14
how
shown
2
standard
many
late
of
8
how
9
many
minutes
rabbits
are
born
one
week.
Their
weights
in
grams
480
501
462
475
460
470
430
485
435
425
465
456
475
435
466
482
455
462
435
462
478
455
information,
of
the
late
buses
town.
452
this
bus
6
453
deviations
his
here :
5
deviation
minutes
450
From
standard
4
arrived.
Find
mathematics
mathematical
(4
the
.
formula.
day
10
2
Applying
selected
cm
the
and
make
weights
for
a
prediction
the
whole
about
the
mean
and
are :
standard
year.
E4.1 How to stand out from the crowd
9 3
3
The
summary
statistics
for
a
sample
n
of
the
life
given
in
mean
length
the
and
of
table.
batteries
are
Calculate
the
standard
deviation
of
25
38 750
Σ x
the
2
Σ x
batteries
in
the
Problem
4
Katie
for
solving
recorded
50
days.
60 100 000
factory.
the
The
number
results
are
of
passengers
shown
in
a
carriage
on
her
train
each
here :
1
7
6
7
7
6
7
6
7
8
8
2
10
6
10
10
5
12
5
8
6
8
10
8
9
3
7
12
9
5
8
6
7
5
9
11
12
4
9
6
7
8
9
7
9
11
7
13
14
15
a
Construct
a
b
Comment
c
Estimate
d
Make
tally
on
the
some
chart
the
to
shape
mean
and
inferences
represent
of
the
the
data.
distribution
standard
for
day
Katie
of
deviation
to
report
the
of
on
the
her
representation.
whole
train.
ndings.
Summary
The
standard
dispersion
deviation
that
gives
an
is
a
measure
idea
of
how
of
Using
close
how
A
data
values
representative
small
standard
are
the
to
the
mean
deviation
mean,
is
of
shows
and
the
values
are
close
to
the
mean.
The
values
of
all
the
units
of
the
x
deviation
original
means
Using
this
the
are
the
same
as
the
in
x f
,
formula
A
of
of
formula
a
sum
of
all
the
the
x
sum
of
all
calculate
population
9 4
discrete
frequency
table
is
,
the
sum
divided
by
the
total
frequency.
to
calculate
the
standard
deviation
for
data
presented
in
a
frequency
table
is:
mean
values.
of
a
set
of
values
the
values
divided
by
The
normal
Its
the
standard
is:
E4 Representation
distribution
with
most
is
a
symmetric
values
close
to
the
the
values.
to
of
units
mean
number
set
data.
notation,
the
a
values
distribution,
is
a
of
discrete
standard
of
the
A
data
mean
thus
data.
that
the
the
data
original
notation,
deviation
A
and
tailing
frequency
formula
to
o
graph
evenly
is
a
calculate
in
either
bell-shaped
standard
direction.
curve.
deviation
2
from
the
summary
statistics
x
and
x
is:
S TAT I S T I C S
For
a
sample
of
n
items:
The
alternative
unbiased
an
●
unbiased
whole
an
●
unbiased
deviation
Mixed
In
estimate
population
,
estimate
for
the
of
is
the
the
of
whole
mean
for
sample
the
1
calculating
population
an
standard
deviation
from
summary
3
recorded
statistics
is:
.
standard
population
is:
and
2
the
data
provided
is
Bernie
the
weight
of
food
in
grams
his
for
ate
each
day.
Here
are
his
results:
population.
Day
Find
the
each
data
set:
mean
a
3,
4,
and
the
standard
deviation
a
2,
21
3,
kg,
4,
21
5,
kg,
5,
24
6,
6,
kg,
1
2
3
4
5
6
7
8
9
55
64
45
54
60
50
59
61
49
of
Food
b
for
the
the
mean,
hamster
1
of
P R O B A B I L I T Y
practice
questions
the
formula
estimate
A N D
(g)
Find
the
mean
25
kg,
27
kg,
29
weight
of
food
b
Bernie
assumes
consumption
3
4
5
f
2
3
2
the
standard
hamster
deviation
ate
over
of
the
the
9days.
kg
c
x
and
6
Write
c
One
down
day
Should
the
is
that
his
the
be
food
distributed.
inferences
hamster
Bernie
hamster’s
normally
eats
he
25g
can
of
make.
food.
worried?
d
Interval
Frequency
4
The
summary
1 −
5
2
lambs
6 −
10
4
Σ x
11 −
15
4
a
Show
16 −
20
5
b
The
on
a
statistics
farm
for
the
weights
of
100
were:
2
=
425
for
kg
farmer
his
special
lamb
Charlotte
buys
20
strawberry
plants
for
She
sta
other
4.25
wanted
to
see
feeds
half
each
day
with
attention.
born
water
ne.
to
kg
= 1.5
and
if
produce
any
half
with
a
special
strawberry
He
weighing
Calculate
guidelines
lambs
decided
needed
between
that
2
any
how
many
and
6.5
away
from
the
mean
these
plant
valuesare.
food.
After
one
strawberries
on
month
each
Number
each
15
13
12
of
of
fed
special
14
15
Analyze
eect
of
with
17
the
fed
with
16
strawberries
17
data
using
strawberries
17
Number
records
the
amount
of
nd
19
12
on
10
each
11
12
food
14
15
there
strawberry
15
plant
plant
whether
special
on
water
strawberr y
19
to
the
plant.
plant
15
she
kg
standard
and
deviations
the
=
2031.25
her
was
garden.
µ
that
=
2
21 −
25
2
Σ x
is
15
an
plant
food.
E4.1 How to stand out from the crowd
9 5
5
For
a
particular
data
set,
the
summary
statistics
Problem
solving
2
Σ
are:
f
=
20,
Σ
f x
=
563,
and
Σ
f
x
= 16 143
7
Find
the
values
of
the
mean
and
the
Samples
of
water
were
taken
from
a
there
were
raised
river
near
a
standard
chemical
plant
to
see
if
levels
deviation.
of
6
In
a
biscuit
biscuits
factory
were
a
sample
of
10
packets
of
lead.
the
Ten
table
samples
below.
The
were
units
collected,
are
g
shown
per
in
liter.
weighed.
6.3
9.6
12.2
12.3
10.3
12.1
10.3
8.4
9.2
4.3
Mass
196
197
199
200
200
200
202
203
203
205
(g)
a
a
Calculate
the
mean
and
standard
Use
this
data
standard
b
To
make
predictions
standard
deviations
about
of
all
the
the
mean
packets
and
the
in
the
factory,
which
values
and
standard
deviation
of
would
1
Dr
to
in
measure
mean
and
amount
of
lead
in
Over
10
g
per
liter
of
you
Comment
lead
on
in
water
whether
is
the
lead
use?
should
be
investigated
further.
context
Roussianos
Saturday
the
the
the
levels
Review
for
river.
dangerous.
mean
predict
deviation
of
b
biscuits
to
deviation.
asked
their
his
cardiology
systolic
morning
for
blood
ten
patients
pressure
each
weeks.
This
for
table
shows
dierent
standard
the
values
treatments
of
the
recommended
yearly
mean
and
deviation.
Standard
Systolic
pressure
is
the
pressure
in
the
arteries
Mean
Treatment
deviation
between
muscle
the
is
beats
resting
of
the
and
heart,
relling
while
with
the
heart
blood.
low
<75
blood
pressure
<10
treatment
Here
is
the
data
for
three
patients.
75
Patient
85
no
<10
treatment
1
high
>85
95
95
87
92
100
84
87
90
87
96
pressure
any
Patient
value
to
76
75
80
Patient
81
tests
needed
mean
the
and
yearly
use
the
statistics
sample
and
data
therefore
76
80
82
80
3
65
100
76
98
58
75
76
98
66
75
9 6
the
estimate
help
75
more
>10
treatment
2
Calculate
78
blood
<10
E4 Representation
to
determine
the
treatment
for
each
patient.
S TAT I S T I C S
2
80
as
people
soon
were
as
they
asked
to
woke
up
measure
in
the
their
pulse
4
rate
In
at
morning.
one
the
given
They
after
were
a
5
asked
km
run,
to
repeat
and
just
the
town,
birth
in
summary
statistics
before
are:
the
mothers
their
rst
P R O B A B I L I T Y
were
child.
asked
The
their
results
age
are
table.
measurements
going
to
bed.
Age
The
300
of
A N D
Σ x
=
and
6000
at
birth
rst
child
of
Number
of
mothers
2
Σ x
=
From
452
this
standard
information,
deviation
inferences
3
Birth
450
that
weight
summary
can
of
be
follows
statistics
nd
each
the
data
drawn
a
for
from
normal
the
mean
set.
and
State
the
of
21–23
20
24–26
30
27–29
42
30–32
53
33–35
52
36–38
43
39–41
25
42–44
12
45–47
8
the
data.
50
15
the
distribution.
weights
18–20
The
babies
2
in
a
a
hospital
From
be
b
If
this
drawn
a
risk.
less
Σ x
data,
from
baby’s
standard
at
were
state
the
weight
deviation
Find
than
x
a
kg
= 160
the
and
Σ x
=
inferences
524 kg
that
can
data.
is
less
than
below
value
are
kg
x,
so
the
that
considered
2.5
times
mean,
babies
to
be
the
the
baby
who
at
is
weigh
risk.
a
Predict
a
for
b
a
mean
mother’s
this
The
age
and
at
standard
the
birth
of
deviation
her
rst
for
child,
town.
summary
statistics
for
a
dierent
sample
Σ x
=
7500
and
Σ x
inferences
that
can
for
of
30
300
years
ago,
women
were
2
Reect
and
How
you
have
= 190 000
be
drawn
State
from
the
the
data.
discuss
explored
the
statement
of
inquiry?
Give
specic
examples.
Statement of Inquiry:
Generalizing
among
and
representing
relationships
can
help
to
clarify
trends
individuals.
E4.1 How to stand out from the crowd
97
E5.1
Super
powers
Objectives
●
●
Evaluating
numerical
positive
or
Writing
numerical
exponents
MROF
●
Inquiry
negative
as
expressions
fractional
with
●
a
F
exponent
expressions
with
How
do
you
afractional
●
fractional
questions
radicals
How
to
do
evaluate
you
simplify
a
number
with
exponent?
use
the
rules
expressions
of
with
indices
fractional
exponents?
Using
the
rules
expressions
fractional
of
that
indices
contain
to
simplify
radicals
and/or
●
C
exponents
How
are
rulesof
●
How
the
rules
radicals
do
the
rules
simplifyradical
●
D
simpler
and
the
of
indices
help
with
decimal
simplied?
always
better?
Critical-thinking
Analyse
parts
Is
indices
expressions?
expressions
exponentsbe
●
ATL
Can
of
related?
complex
and
concepts
synthesize
them
and
to
projects
create
new
into
their
constituent
understanding
5 1
10 3
E5 1
Conceptual understanding:
Forms
can
9 8
be
changed
through
simplication.
N U M B E R
Y
ou
●
should
use
the
already
multiplicative
know
rule
how
for
1
to:
Simplify :
exponents
3
a
●
use
the
rules
of
radicals
to
2
(x
2
4
)
b
Simplify
(2
2
2
)
c
(2 x
3
)
completely :
simplifyexpressions
128
a
45
b
c
60
×
3
8
●
evaluate
expressions
with
3
Evaluate:
1
1
unitfraction
3
2
a
4
d
64
b
8
e
64
1
evaluate
expressions
with
4
negativeexponents
1
fraction
1
3
f
leaving
where
1
b


c
Fractional
F
For
How
do
you
evaluate
●
How
do
you
use
ATL
any
positive
Example
2
2


e
2

a
2

f

3


5

exponents
●
fractional
2
as
3
6
1
d

answer
necessar y :
1
1
your
( 0.001
)
2
4

32
3
Evaluate,
a
5
c
2
●
1
exponents
the
a
number
rules
of
with
indices
a
to
fractional
simplify
exponent?
expressions
with
exponents?
real
number
x,
and
integer
n,
where
n
≠
0,
.
1
Evaluate:
a
b
c
a
Using
the
multiplicative
rule
for
exponents,
b
a
(x
c
Here,
Practice
1
b
)
Work
1

out
the
value
1

1

b

4
the
:
denominator
has
been
rationalized.
of:
2
a
x
1
1

ab
=

1

4
1

1

3
c

16

1

2
d

8


1
25

1
1
1

4

e

2
f

9

27
3
4

g

2
27

3
h

25


1000

E5.1 Super powers
9 9
2
Simplify
and
leave
your
nal
as
a
power
of
3
(4 x )
( 27 x
b
6
3
2
x
)
)
1
1
1
×
x
f
Problem
3
Write
each
1
1
3
2
2
x
÷
x
g
100
×
all
x
≠
0
expressions.
1
2
2
x
for
that

1
4
2
x
2

x
d

e
4

3
(8 x
c
Assume
1
1
1
1
a
answer
x
(9 x
h
2
2
)
×
9x
solving
of
these
using
exponents
instead
3
a
of
the
radicals.
1
1
x
b
c
4
x
x
4
Solve:
1
1


1
x

2
= 2

y

3
a
b


9
= 0.3

a
Using
the
multiplicative
for
exponents
(x
b
)
ab
=
x
:
3
1
3
2
(x
)
2
=
x
=
x
4
1
4
(x
rule
3
3
)
4
2
3
3
What
ATL
do
expressions
Exploration
1
Copy
and
Using
with
fractional
exponents
like
x
or
x
mean?
1
complete
integer
this
and
table:
unit
Using
a
radical
Using
fractional
A
fraction
exponents
unit
fraction
exponents
hasnumerator
3
(x
)
(x
)
x
5
3
(x
)
2
(x
x
)
x
2
2
Simplify
(x
)
2
and
(x
2
)
.
Are
(x
)
2
and
(x
)
equivalent?
m
Generalize
3
Copy
10 0
and
your
result
complete,
for
using
(x
)
your
E5 Simplication
where
result
m
and
from
n
are
step
2:
integers
and
n
≠
0.
1.
N U M B E R
Reect
●
Did
Is
discuss
everyone
inthe
●
and
table
there
in
in
more
the
1
class
get
Exploration
than
one
the
same
results
for
the
last
two
rows
1?
way
of
writing
x
andx
in
the
other
two
columns?
For
ATL
any
positive
Example
real
number
x,
and
integers
m
and
n,
where
without
using
a
and
could
also
calculation
Which
Write
1
Simplify
a
help
these
method,
simplify
or
=
the
the
.
same
one
in
expressions
Complete
answer
as
Example
with
2?
fractional
expressions
without
using
a
2
2
2
3
3
8
b
each
27
c
fraction
and
an
integer.
why.
exponents.
25
d
64
e
2
3
125
f
343
of
x
f
(32
solving
of
these
using
4
2
x
b
Where
exponents
3
c
applicable,
x
d
leave
1
4
3
b
×
x
of
x
your
nal
answer
x
as
a
1
4
3
×
radicals.
x
2
2
c
the
5
x )
(
1
2
instead
3
x )
(
2
)
unit
calculator.
3
3
a
2.
Explain
2
(x
into
the
Example
3
Simplify.
a
you
this:
gives
3
3
3
this
this
4
Write
a
to
–
exponent
2
like
that
the
2
Problem
2
verify
easier
hint
Practice
discuss
simplify
and
was
a
0,
calculator.
Split
You
≠
2
Simplify
Reect
n
d
(3
)
power
2
2
5
e
(x
2
2
2
3
)
5
)
E5.1 Super powers
101
Example
3
ATL
Simplify
Using
the
multiplicative
a
=
exponents,(x
b
rule
for
ab
)
=
x
,
gives :
=
=
=
=
Reect
ATL
and
discuss
If
3
to
The
method
in
Example
3
dealt
with
the
negative
sign
in
the
necessary,
the
by
taking
the
reciprocal.
Conrm
that
taking
the
cube
root
rst,
rst,
all
give
the
same
nal
answer.
Which
method
is
in
Reect
or
and
squaring
add
you
exponent
wrote
rst,
hint
discuss
2,
to
easier?
help
you
simplify
expressions
with
fractional exponents.
Practice
1
Find
the
3
value
of
each
expression.
Leave
your
answer
as
a
simplied
fraction.
3
3
3
2
a
4
1

2

these
1

3

2


1
16

4

e
27
3
2
c
49
2
2
1
3
12
× 16
12
÷ 8
Problem
16
1
6
b
1
4
8
81
completely.
12
× 125
3
d
expressions
25
1
4
125
2

f

Simplify
a
36
e
64
16
3

3
d


c
16


9

4
b
25
×
343
1
3
1
2
2
÷ 27
f
45
÷ 15
solving
In
3
Write
down
an
equivalent
expression
that
contains
fractional
question
write
1
a
2
b
5
as
d
both
powers
numbers
of
the
3
same
base.
Simplify:
Forexample:
1
1
3
1
1
1
2
3
2
a
3
2
×
4
9
b
Rules
C
Writing
of
●
How
are
●
How
do
a
radical
10 2
as
4
2
2
×
2
c
indices
the
the
an
rules
rules
of
of
and
indices
indices
exponent
3
can
E5 Simplication
2
×
3
27
d
16
3
÷ 2
radicals
and
help
help
rst
2
c
2
4
4,
exponents.
the
rules
simplify
you
of
radicals
radical
simplify
some
related?
expressions?
expressions.
2
×
9
2
=
3
×
N U M B E R
Example
4
Write
as
a
power
of
2.
Write
Exploration
1
Use
the
rules
of
as
a
power
of
2,
Here
are
them
by
radicals
to
simplify:
the
expressions
writing
8
as
a
from
power
step
of
2,
1
written
and
using
using
the
exponents.
rules
of
Simplify
exponents:
Justify
why
are
the
two
methods
for
simplifying
expressions
with
square
equivalent.
Write
Use
the
rules
Generalize
5
Use
the
Use
the
of
exponents
Write
as
to
show
that
for
,
to
show
that
for
and
.
for
down
the
nth
a
rule
root
of
ab
ndings.
exponents
,
.
Why
ndings.
;
can
equal
exponents
:
Practice
1
of
your
rules
For
of
your
rules
Generalize
6
simplify.
b
roots
4
then
b
a
3
number
2
a
2
each
to
show
that
,
for
where
b
not
0?
.
.
4
a
single
power
of
2:
3
8
2
b
4
2
16
3
a
2
c
d
4
2
Problem
2
Write
as
solving
powers
of
prime
numbers:
3
5
7
×
In
4
2
8
3
a
part
b,
rst
write
8
 4
7
b
9
×
9
c
d
4
3
8
9
as
a
power
of
3.
2
3
Write
as
products
of
powers
of
prime
numbers:
3
a
6
b
50
f
c
4
162
d
100
6
4
e
4
14
72
E5.1 Super powers
10 3
Objective
iii.
In
4
move
Q4,
C:
write
Write
Communicating
between
dierent
radical
these
exponents
expressions
expressions
to
simplify.
as
as
2
b
mathematical
exponents
your
3
2 ×
of
powers
Give
3
a
forms
of
to
simplify
prime
answers
them.
numbers.
as
powers
5
4
9 ×
representation
9
the
prime
rules
of
numbers.
6
10
7
c
Use
of
×
7
d
3
6
In
n
3
4
a
n
e
f
g
Reect
Use
a
×
a
the
and
rules
of
g
a
and
is
h,
prime.
h
m
10
4
assume
m
m
5
parts
n
10
a
discuss
exponents
4
to
justify
why
these
expressions
cannot
be
simplied:
Write
a
general
rule
for
when
you
can
simplify
products
and
quotients
ofradicals.
When
with
radicals
fractional
have
the
same
exponents
to
radicand
simplify
but
dierent
them.
For
indices,
write
them
The
example:
radicand
number
square
●
ATL
is
under
root
the
the
sign.
●
Example
5
Simplify
Write
Give
When
write
radicals
them
10 4
as
have
dierent
exponents
with
radicands,
the
same
E5 Simplication
but
base
one
to
is
the
a
the
radicands
answer
power
simplify
of
them.
as
a
the
as
exponents
radical,
other,
with
with
prime
the
same
number
base.
radicand.
N U M B E R
ATL
Example
6
Write
as
powers
of
prime
a
simplied
radical,
by
expressing
the
radicands
as
numbers.
Write
the
multiples
Write
Practice
1
Simplify
these
3
2
×
of
using
as
primes.
exponents.
5
completely,
leaving
your
answer
as
a
simplied
radical.
3
a
radicands
radicands
3
6
2
×
2
b
5
2
3
4
×
×
3
3
11
12
3
c
7
d
e
3
11
5
3
7
4
5
3
3
f
g
3
2
×
2
5
h
3
2
4
3
5
× 2
3
4
5
i
3
j
3
3
2
8
5
Write
as
simplied
radicals,
with
prime
number
radicands.
3
4
3
16
4
a
b
×
7
3
3
12
3
3
c
12
×
18
d
e
3
2
2
4
4
6
5
g
125
81
h
4
3
5
Problem
Solve
these
5
i
3
3
3
4
125
f
343
j
5
49
×
343
3
5
24
solving
equations,
where
the
unknown
is
always
a
power
of
3.
3
27
1
y
a
=
3
b
9
=
c
6
=
3
3
x
3
Other
D
z
9
exponents
Think:
●
Can
●
Is
expressions
simpler
with
always
decimal
exponents
be
simplied?
has
in
better?
it,
can
A
rational
number
is
a
number
that
can
be
written
as
a
1
Convert
a
2
these
0.1
Write
b
the
as
a
b
decimals
into
Use
your
decimal
a
ratio
expressed
fraction.
0.375
fractions:
c
exponents
1.6
as
d
fractions,
0.3
and
hence
write
these
radicals:
0.3
3
and
be
‘ratio’
3
decimal
expressions
word
fraction.
a
Exploration
‘rational’
the
answers
exponent
c
to
step
can
be
2
d
to
justify
written
as
why
a
e
any
0.8
5
number
f
with
a
5
nite
radical.
E5.1 Super powers
10 5
as
π
You
cannot
evaluate
an
expression
like
3
2
or
5
without
a
calculator,
because
An
the
exponents
are
irrational
numbers.
However,
you
can
simplify
irrational
expressions
number
π
like
π
4
3
×
3
or
(7
, using
the
rules
for
exponents.
π
π
4
×
3
=
+
π
How
=
could
and
denition
why
Practice
1
Write
discuss
of
it
a
is
radical
dicult
9
d
(3
?
this:
means
evaluate
that
.
6
these
expressions
as
radicals
of
powers
1.25
b
2
get
5
is
to
0.2
a
you
7
Reect
Suggest
fraction.
approximation
for
The
a
2π
2
)
as
4
3
an
(7
be
Examples:
written
3
cannot
2
)
prime
c
100
1.5
)
e
Problem
numbers:
1.5
16
0.3
of
0.75
4
f
10 000
solving
0.2
2
Evaluate 32
3
Copy
and
.
4
your

× 4
method.
complete:
0.2
a
Explain
=
0.2
2
b
27

0.1
× 81
=
0.4
3
c
32
0.5
× 16

0.1
× 64
=
2
0.1
4
Solve
x
=
5
Simplify:
2
0.4
2
27
6
a
0.6
×
2
b
0.2
0.8
3
9
×
4
Summary
For
any
positive
real
number
x,
and
integer
n,
When
radicals
dierent
where
n
≠
0,
indices,
any
the
write
same
them
radicand
with
but
fractional
=
exponents
For
have
positive
real
number
x,
and
integers
to
simplify
them.
m
For
example:
m
andn,
where
For
n
≠
0,
=
(
)
=
:
For
When
is
The
radicand
is
the
number
a
power
square
root
the
the
dierent
other,
write
radicands,
them
as
but
one
same
base
to
simplify
exponents
them.
sign.
When
is
not
radicals
a
power
simplied
10 6
of
have
under
with
the
radicals
E5 Simplication
have
of
into
the
one
dierent
other,
radical.
radicands,
then
they
but
can
one
not
be
N U M B E R
Mixed
1
Work
practice
out
the
value
Problem
of :
1
solving
1
1
1
8
1


1

2
a


9

9
2
25
c

1

2
b

1
1

Write
27
contains


25

2
e


expression
that
exponents.
1
4
b
3
3
c
d
3
4
100


125
2

9
2
equivalent
g


16
8

2
f

an
fractional
1
a
9

down
3
d

Write
as
simplied
radicals
with
prime
number
Simplify:
radicands.
1
1
1
3
2
6
a
(49x
3
2
)
b
(64 x
4
3
)
c
(4 x
2
)
×
3
3x
a
4
4
2
b
16
24
×
36
5
c
d
3
4
2
8
(
3
Work
out
the
value
2
2
3
4
b
0 000
)
of :
3
a
10
c
Problem
3
3
1000
solving
2
27
d
9
10
Solve
these
equations:
2
4
Find
the
value
of
each
of
these
x
expressions.
1

x
Leave
your
answers
as
rational
a
numbers.
8
=
32
b
5
=
4
2
1

a
4


2
b

5
2
3

3
8

these
9

5


×

64
expressions
11
2
d


9
as
rational
of
prime
Simplify,
Leave
your
leaving
of
prime
5
125
b
0.9
81
b
as
simplied
0.1
0.4
× 5
c
9
0.2
× 81
0.4
× 3
3
9
e
0.4
125
12
36
0.2
12
d
3
3
8
4
12
× 64
4
0.3
16
3
64
answers
0.3
×
5
a
your
numbers:
0.4
25
1
1
1
16
numbers.
5
1

5
radicals
completely.
powers
64

a
answers
9
3
3
 3
c

Simplify
125

=

2

x
c
× 9
c
d
1
1
16
12
6
6
6
The
simplied
25
5
b
16
×
4
c
Simplify
these
expressions
×
343
Without
your
answers
as
has
been
fractional
encoded
using
exponents.
using
a
calculator,
completely.
simplify
Use
your
the
following
results
and
the
completely.
key
Leave
and
2
7
expressions
7
below
surds
1
2
2
2
2
×
1
3
1
2
a
quotation
3
Simplify:
simplied
to
decode
a
quote
by
Stefan
Banach.
radicals.
4
64
3
3
a
16
×
4
8
b
4
12
×
48
c
4
4
Key:
=
L,
=
I,
2
=
A,
=
O,
3
3
9
8
4
d
e
4
×
3
4
4
×
f
2
8
×
6
8
×
8
3
81
3
=
M,
8
=
D,
=
S,
1
=
N
3
4
12
9
g
h
3
4
27
12
Mathematics
1
1
1
1
1



1
32


2
1
5
15
2
×
20
2
18
 10
10
2

2
1

5

27
3
4
15

2

3

.’
1
2
1
1
1
1
1

1

1
3
64
12
3
3
×
64
3
4
×
81
2
you
have
2
144
3
×
6
3
4
4
18
2
256
2
How
64
1
8
and
4
2
3
Reect
6
4

6 ×
2
3

discuss
explored
the
statement
of
conceptual
understanding?
Give
specic
examples.
Conceptual understanding:
Forms
can
be
changed
through
simplication.
E5.1 Super powers
10 7
2
Ideal
work
for
lumberjacks
E6.1
Global
context:
Objectives
●
Evaluating
Orientation
Inquiry
logarithms
with
and
without
●
a
in
space
and
time
questions
What
does
log
b
mean?
a
F
calculator
●
Writing
an
●
exponential
logarithmic
statement
as
a
SPIHSNOITA LER
●
Solving
Writing
exponential
and
solving
C
equations
real-life
exponential
Using
nd
logarithms
using
a
do
to
logarithms
make
problems
solve?
equations
How
can
anything
in
mathematics
be
situations
natural
considered
natural?
logarithms
●
ATL
How
easier
D
●
you
statement
●
from
can
calculator?
●
●
How
Is
change
measurable
and
predictable?
Critical-thinking
Consider
ideas
Statement
of
from
multiple
perspectives
Inquiry:
E6 1
Generalizing
that
can
changes
model
in
duration,
quantity
helps
frequency
establish
and
relationships
variability.
E10 1
E12 1
10 8
N U M B E R
Y
ou
should
already
know
how
to:
2
●
use
index
laws
1
Simplify
2
x
5
×
x
Evaluate:
1
3
a
●
rewrite
surds
quantities
and
with
roots
as
3
model
a
b
real-life
anexponential
problem
2
2
c
16
Write with an exponent:
exponents
3
6
a
●
1
5
with
4
3
9
b
c
5
In 2007, the Zika virus was growing
equation
at a rate of 18% per week on Yap
Island. Two people were infected
initially. Use an exponential model
to determine the number of people
infected after 52 weeks.
Introduction
F
●
What
does
log
to
b
logarithms
mean?
a
●
How
can
you
nd
logarithms
using
a
calculator?
x
An
exponential
equation
●
a
is
the
base
●
x
is
the
exponent
Exploration
Write
each
Therst
1
Find
the
is
in
the
form
a
=
b
power
1
statement
one
or
is
done
as
for
exponent
an
exponential
equation,
and
solve
to
nd
x
you.
of
10
of
2
that
gives
the
answer:
100
1000
10
000
0.1
0.0001
2
Find
the
exponent
that
gives
the
answer:
4
8
16
128
Continued
on
next
page
E6.1 Ideal work for lumberjacks
10 9
3
Find
the
exponent
of
6
that
gives
the
answer:
36
216
4
What
patterns
Determine
do
what
Determine
you
notice?
would
what
happen
would
to
happen
the
to
answer
the
if
answer
you
if
add
you
1
to
the
subtract
1
exponent.
from
the
exponent.
In
steps
1
to
theanswer,
3
the
you
exponents
can
use
are
trial
all
and
integers.
When
you
can’t
spot
improvement.
x
5
To
solve
estimate
=
10
to
150,
nd
x
estimate
correct
to
the
2
value
decimal
of
x
and
try
it.
Improve
your
places.
x
6
Solve
2
7
Solve
5
8
You
=
45.
Give
your
answer
to
2
decimal
places.
=
77.
Give
your
answer
to
2
signicant
x
can
solve
also
use
graphing
exponential
software
equations
gures.
1.1
to
y
graphically.
x
For
example,
the
graph
y
10
of
to
the
solve
10
=
60,
exponential
draw
function
f
Read
(x)
=
60
(1.78, 60)
2
x
=
and
the
the
value
graph
of
x
at
of
y
=
the
60.
intersection
x
f
(x)
=
10
1
to
2decimal
places.
0
x
x
a
Solve
=
2
answer
to
6
2
graphically.
decimal
Give
your
places.
x
b
The
Solve
three
7
=
14
methods
●
the
●
trial
●
graphical
for
inspection
and
solving
Give
your
exponential
answer
to
equations
2
in
signicant
gures.
Exploration1are:
method
improvement
Reect
and
●
you
Would
graphically.
discuss
use
the
trial
1
and
improvement
method
or
the
graphical
x
method
●
What
110
to
are
solve
the
the
equation
advantages
and
E6 Quantity
9
=
85?
Explain
disadvantages
of
your
the
choice.
three
methods?
N U M B E R
The
trial
and
perspectives
improvement
on
solving
method
and
exponential
the
graphing
method
give
two
dierent
equations.
x
Here
is
is
another
dened
as
method.
‘the
In
the
logarithm
to
exponential
base
10
of
equation
10
=
500,
the
quantity
x
500’.
1.1
You
can
write
log
500
=
x
10
log
(500)
Tip
2.69897
10
Using
a
calculator:
log
500
=
2.69897…
x
=
2.70
10
The
(2
log
your
You
can
use
the
equivalent
button
on
d.p.)
statements
for
to
not
calculator
‘log
’.
You
is
do
10
x
10
=
500
and
log
500
=
x
nd
the
need
to
enter
10
unknown
exponent.
Rewriting
the
exponential
logarithmic
statements
statements
changes
number
as
the
unknown
Before
variable
from
an
exponent
to
a
regular
variable.
Whenever
you
see
calculators,
an
logarithms
exponential
or
logarithmic
statement
you
can
use
the
equivalence
concept
it
from
another
log
any
you
two
can
positive
numbers
a
and
b,
there
exists
a
third
number
c
so
that
write:
published
tables.
or
We
the
equivalent
say
gives
base
When
that
the
of
the
a
the
c
is
calculate
logarithms
at
b’.
new
logarithm
In
logarithm
the
and
equations
equation
to
base
logarithmic
b
is
called
involving
using
statement
the
a
of
b’,
or
‘the
exponent
statement
the
form.
,
argument
exponents
other
.
or
This
of
the
you
is
of
a
that
called
the
logarithm.
logarithms,
allows
a
it
to
can
see
be
the
helpful
.
49
When
a
not
an
exponential
with
base
10
are
often
called
‘common’
logs
and
written
log
statement.
.
x,
base.
you
most
does
as
means
Logarithms
On
to
equation
Rewrite
without
button.
1
Solve
=
a
perspective.
Example
x
of
the
statement
logarithmic
‘the
answer
solving
rewrite
from
exponential
in
Modern
calculators
touch
the
values
perspective.
were
For
of
to
dierent
consider
10.
see
‘log’
written
calculators
allow
this,
you
you
without
can
will
choose
need
to
a
base,
the
base
use
the
you
of
can
the
change
assume
the
logarithm.
of
base
If
base
your
is
10.
calculator
formula
E6.1 Ideal work for lumberjacks
111
The
Change
of
base
formula:
,
where
c
is
any
new
change
formula
in
Example
of
base
base.
is
derived
E12.1.
2
x
Solve
2
=
50
using
a
calculator
and
the
change
of
base
formula.
x
2
=
50
Rewrite
as
a
logarithmic
Use
=
5.64385619
=
5.64
Work
Example
(3
on
your
calculator.
out
x
Find
the
value
of
The
the
equivalent
Hence,
Write
500
b
using
answers
2
logarithmic
statement
x
=
to
3
down
log
your
x
10
=
150
calculator,
nd
c
64
the
=
value
60
of
d
x
in
an
equivalent
exponential
statement
=
2
b
log
4 =
c
log
0.1
=
for:
-1
8
x
=
5
e
log
5
3
Write
b
=
4
f
log
b
down
an
y
equivalent
2
3
logarithmic
statement
1
3
=
9
1 12
b
4
0
x
2
a
=
3
=
0.125
E6 Quantity
c
1000
= 10
2
each
3
log
equal,
x
10
s.f.
8
d
are
for:
x
=
45
case.
e
Give
2
your
=
an
so
for:
2
a
bases
as
1
down
10
10.
x.
=
x
a
=
base
c
s.f.)
Rewrite
Write
of
with
Write
1
change
formula
3
.
Practice
the
statement.
6
surds
as
exponents.
exponential
the
statement.
exponents
are
equal.
N U M B E R
Use
this
table
Powers
of
to
help
2
you
answer
Powers
of
questions
3
4
–
6
Powers
of
4
Powers
of
5
Tip
3
3
0.125
2
3
3
0.15625
4
3
0.008
5
Learn
results
2
2
0.25
2
2
0.04
5
solve
and
1
1
0.5
2
1
0.2
5
0
0
1
1
2
3
4
3
8
3
2
3
4
9
4
4
27
4
16
5
81
32
Without
log
625
=
a
243
x
b
log
log
nd
243
the
=
64
5
256
5
1024
5
f
x
log
Without
a
log
using
5
=
a
x
b
6
Find
value
c
of
a
log
value
17
=
of
x
to
x
b
the
log
7
Solve,
=
x
g
log
a
=
40
b
Logarithms
of
stars;
in
are
used
geology
semitones;
inside
nautilus
of
Solve
these
5
in
to
measure
Problem
8
3
the
=
value
x
of
x
d
log
1024
1 =
h
x
log
121 =
0.04
=
x
5
x
256
=
x
log
d
4
nearest
=
4
log
c
256
=
x
2
hundredth.
log
c
x
31 =
x
d
log
3
x
3
=
5
=
x
8
calculator:
x
a
nd
6
using
0.625
3
3
log
2
3125
x.
3
the
625
5
x
0.125
log
125
4
calculator,
5
25
4
2
5
5
3
3
0.00001 =
1
2
4
calculator,
5
e
5
5
3
using
16
4
5
2
5
4
3
without
calculator.
1
4
3
4
2
5
2
3
2
1
1
3
logarithmic
0
4
2
2
a
1
1
2
4
0
3
you
exponential
equations
a
2
help
1
0.25
4
3
to
2
0.625
4
3
these
x
=
100
astronomy
measure
and
shells
c
the
perhaps
whose
to
4
x
=
measure
intensity
most
of
85
the
form
apparent
2
a
in
=
90
magnitude
earthquakes;
beautifully,
chambers
d
in
nature,
logarithmic
music,
as
seen
to
in
the
spiral.
solving
exponential
equations.
2x
x
a
+ 1
2
x + 2
2 x
=
b
4
3
2 x - 2
=

x +1
c
9
2
1

In
=

2
8,
with
x + 2

1
x

d
2x
=

3

3

2
1
+
and
2

e
2x
=

3

write
them

the
use
same
the
base
laws
2
9
exponents.
E6.1 Ideal work for lumberjacks
1 13
of
Using
C
●
You
an
can
How
use
do
logarithms
logarithms
exponential
equation
logarithms
into
Example
to
statement
one
you
solve
as
can
make
an
to
solve
problems
complex
equations
easier
exponential
equivalent
logarithmic
to
solve?
equations.
statement
Rewriting
changes
the
solve.
4
,
Solve
accurate
to
2
decimal
places.
Rewrite
x
1.67195...
=
2x
=
=
0.34
0.33567
+
as
a
logarithmic
statement.
1
(2
d.p.)
Solve
Practice
Solve
each
2
equation,
giving
your
answer
2 x
x
1
2
3
5
5
4 × 2
=
x
(2
5
d.p.)
2
=
4
3
6
2 × 2
2 x
=
3
(2
s.f.)
=
x
6
(2
d.p.)
x
2x
2
6  (2
s.f.)
accuracy
stated.
5
(3
d.p.)
5
(3
s.f.)
1
=
6
(2
d.p.)
3
8
= 16
3
You
and
can
also
decay
select
real-life
use
s.f.)
an
logarithms
to
solve
equations
that
result
from
real-life
growth
problems.
D:
Applying
appropriate
mathematics
mathematical
in
real-life
strategies
contexts
when
solving
authentic
situations
Exploration
use
(2
3
Objective
In
of
2
=
7
ii.
degree
+ 1
x
=
the
+ 1
2
1
to
2,
you
appropriate
114
will
write
strategy
to
an
solve
E6 Quantity
equation
it.
to
describe
a
real-life
situation
and
then
for
x.
N U M B E R
Exploration
An
A
amoeba
certain
one
The
a
on
a
how
number
Time
single-celled
amoeba
amoeba
Determine
1
is
2
splits
long
of
of
it
will
amoebas
take
Write
the
number
3
Write
A(t)
as
4
Write
and
solve
an
and
●
How
do
●
How
could
Example
you
of
discuss
know
you
when
verify
on
amoebas
to
be
=
t.
Copy
t
=
powers
of
1000
1
=
an
cell
division.
hour.
There
on
the
complete
2
t
=
is
experiment.
amoebas
and
A(2)
of
by
every
beginning
A(1)
as
reproduces
time
t
=
amoebas
the
there
0
=
3
A(3)
the
t
=
slide.
table:
=
4
A(4)
=
2.
function.
exponential
equation
to
nd
t
when
A(t)
=
100.
2
to
that
use
logarithms
your
answer
to
solve
makes
an
equation?
sense?
5
number
of
drosophila
experiment
is
a
Find
population
b
Write
c
Find
the
an
the
for
A(0)
exponential
an
at
depends
amoebas
that
separate
slide
t =
2
Reect
two
microscope
(hours)
Number
The
into
organism
200.
The
at
exponential
number
of
(fruit
ies)
population
the
end
function
days
until
in
a
laboratory
increases
of
the
for
the
the
rst
by
8.5%
and
the
each
second
number
population
at
of
beginning
an
days.
drosophila
reached
of
day.
100
after
000
x
days.
drosophila.
a
Number of days ( x )
Number
0
of
drosophila
(D)
200
Growth
factor
b
=
1
+
r =
1
+
0.085
=
1.085
1
2
Exponential
b
a
=
200,
b
=
growth
is
modelled
by
1.085
x
y
c
=
a(1
+
r)
where
x is
Rearrange
Rewrite
x
After
77
=
days
as
a
to
the
number
isolate
the
logarithmic
of
days.
term
in
statement.
76.18
there
would
be
more
than
100 000
drosophila.
E6.1 Ideal work for lumberjacks
x
115
Practice
1
At
the
The
2
3
end
of
the
population
a
Find
the
b
Write
c
Predict
A
group
of
snow
2000
at
exponential
the
year
15
in
snow
the
increased
population
an
of
year
then
the
population
by
end
function
which
foxes
the
are
1.3%
of
2001
for
of
per
and
P (t),
the
population
introduced
a
city
was
300 000.
year.
2002.
population
should
into
a
after
exceed
nature
t
years.
350 000.
reserve.
The
number
0.4 t
where
Find
100
3
The
is
the
how
snow
can
be
number
modelled
of
years
years
after
by
since
their
the
exponential
their
N
equation
= 15 ×
3
introduction.
introduction
when
there
should
be
at
least
foxes.
and
08:00
N
many
half-life
Copy
at
t
foxes
of
caeine
complete
one
in
the
the
table
human
for
a
body
student
is
approximately
who
consumes
5
120
hours.
mg
of
caeine
morning.
Time
period
Amount
of
Time
(t hours)
caeine
(C
08:00
t
=
0
120
13:00
t
=
1
60
18:00
t
=
2
mg)
23:00
a
Find
the
common
b
Find
the
exponential
c
Calculate
d
The
the
ratio
for
the
equation
amount
of
geometric
for
caeine
the
in
sequence.
half-life
the
of
student’s
caeine
body
in
after
terms
35
of
t
hours.
Each
of
eect
hours
Coco
is
starts
(w)
which
is
negligible
there
is
only
below
a
0.02
mg.
negligible
Determine
eect
from
the
the
time
for
a
race
training,
Time
(t,
min)
and
she
to
records
can
run
1
70.56
2
69.149
3
represents
5
many
run
10
her
10
10
km
km
in
times
72
each
The
form
week.
minutes.
km
a
times
geometric
sequence.
Use
w
w
to
67.766
4
66.411
5
65.082
her
weeks
116
hours.
caeine.
=
0
and
nd
the
decrease.
Assuming
period
number
solving
training
Beforeshe
Week
caeine
after
Problem
4
of
it
time
will
continues
be
before
E6 Quantity
to
decrease
Coco
can
at
run
the
10
same
km
in
rate,
less
calculate
than
one
how
hour.
=
1
rate
of
N U M B E R
5
A
large
lake
decreasing
has
at
a
Make
a
b
Write
an
c
When
a
a
population
rate
table
to
the
show
population
how
long
Natural
●
How
●
Is
Exploration
exactly
one
6
10%
exponential
Calculate
D
of
year
the
due
number
model
of
100 000
year
for
perch
will
of
the
falls
take
for
perch.
to
perch
P
number
below
this
The
number
of
perch
is
pollution.
after
of
to
2
perch
25 000
lake
1,
be
it
at
is
a
and
P
3years.
after
at
a
t
years.
‘critical
critical
level’.
level.
logarithms
can
anything
change
of
it
of
per
in
mathematics
measurable
MYP
Standard
would
grow
and
14.1
with
considered
natural?
predictable?
explored
the
be
interest
how
$1
invested
compounded
at
in
a
bank
dierent
for
intervals:
annually
semi-annually
quar terly
monthly
weekly
daily
hourly
ever y
minute
Innity
is
without
ever y
a
oorless
walls
or
room
ceiling.
second
Author unknown
The
the
number
total
The
number
continually
In
of
times
investment
2.71828…
growing
general,
population
P
is
all
is
called
populations
nal
the
interest
is
compounded
is
n.
As
n
approaches
innity,
$2.71828…
e
and
it
is
the
base
rate
for
growth/decay
of
all
processes.
growth
the
is
the
approaches
that
grow
equation
continually
,
are
modelled
by
the
where:
population
original
r
is
the
growth
t
is
the
time
population
when
t
=
0
rate
period.
E6.1 Ideal work for lumberjacks
1 17
Reect
How
and
could
the
exponential
To
calculate
If
3
formula
be
used
for
problems
involving
decay?
the
then
The
discuss
natural
growth
the
rate
or
equivalent
the
time
you
logarithmic
logarithm
is
written
can
use
logarithms
statement
as:
with
is
,
y
base
>
e.
0.
ln y.
x
Your
calculator
The
rst
mention
appendix
until
The
to
Euler
spiral
skydiver
e,
Napier’s
galaxy
Example
a
of
have
on
spiral,
the
an
his
the
ln
on
the
and
in
of
an
logarithm,
logarithms,
work
cover
where
button,
natural
work
completed
logarithmic
A
should
button.
e
was
but
it
noted
wasn’t
in
1618
as
ocially
an
named
‘e’
1748.
this
shape
book
does
is
not
a
naturally
alter
but
occurring
the
size
increases.
6
jumps
parachute.
As
o
he
a
cli.
falls,
,
his
where
Sensibly,
speed
his
of
the
skydiver
descent
speed,
,
is
has
remembered
modelled
is
in
m/s
speed
of
40
and
by
the
time
t
is
to
wear
equation
in
seconds.
Find:
a
his
b
the
initial
time
speed
it
takes
for
him
to
reach
a
m/s.
a
Initial
=
Initial
50(1
speed
1)
is
=
0
speed
occurs
when
t
=
0.
0
m/s.
b
Substitute 40 for the speed and solve for t.
Write
=
t
=
8.05
1 18
-0.2 t
seconds
(2
d.p.)
E6 Quantity
the
equivalent
logarithmic
statement.
N U M B E R
Example
In
2000,
an
Assuming
per
year,
7
island
that
in
the
the
Caribbean
population
was
had
a
population
continually
of
2600
increasing
at
a
people.
rate
of
1.68%
calculate:
a
the
predicted
b
the
time
taken
for
c
the
time
taken
for
decrease
at
a
population
rate
in
2020
the
population
to
double
the
population
in
part
of
2%
per
b
to
return
to
2600
if
it
began
to
year.
a
Standard population equation.
=
The
3638.3
Use your calculator.
population
in
2020
will
be
3638
people.
b
Double
Rewrite
=
The
as
a
the
population
logarithmic
is
5200.
statement.
41.258…
population
will
double
in
42
years.
c
For
exponential
Rewrite
=
The
1
a
rate
is
logarithmic
negative.
statement.
will
return
to
2600
people
in
35
years.
4
Carbon-14
and
as
the
34.657……
population
Practice
decay,
dating
is
a
common
method
used
to
determine
the
age
of
fossils
bones.
0.000121t
One
formula
where
A
P
femur
the
is
used
the
bone
carbon-14
to
calculate
percentage
found
found
in
in
of
an
a
the
age
t
(in
carbon-14
Australian
normal
years)
left
in
of
the
aboriginal
femur.
an
item
item
P
(written
burial
Determine
is
the
site
=
e
as
a
,
decimal).
contains
femur’s
11%
of
age.
E6.1 Ideal work for lumberjacks
1 19
2
Radioactive
substances
decay
half-life
time
for
(the
taken
over
half
of
time
the
and
are
often
substance
to
described
by
their
decay/disappear).
b t
Theformula
used
to
describe
this
decay
is
A
=
A
e
,
where
A
is
the
amount
0
left
after
time
t,
A
is
the
original
amount
of
the
radioactive
substance
0
and
b
is
decay
Iodine-131
has
diagnosing
issues
a
Find
b
Hence
c
Find
start
3
the
A
the
a
half-life
decay
write
how
with
biologist
constant.
with
down
the
8
days.
thyroid
constant
many
3
of
the
b
for
of
is
a
common
iodine-131
formula
grams
It
substance
used
in
gland.
for
the
iodine-131
if
you
begin
exponential
will
be
with
decay
present
3
of
after
grams.
iodine-131.
28
days
if
you
grams.
monitoring
a
re
ant
infestation
notices
that
the
area
infected
by
0.7 n
the
ants
and
4
n
can
is
the
be
modelled
number
of
population
size.
the
initial
b
Find
the
population
c
Calculate
group
of
20
= 1000e
after
Find
A
A
weeks
a
how
by
long
rabbits
after
it
is
one
will
,
the
where
initial
A
is
the
area
in
hectares,
observation.
week.
take
the
introduced
population
to
a
rabbit
to
farm.
cover
After
50 000
t
years
hectares.
the
0.6 t
number
of
rabbits
a
Predict
b
Determine
Give
your
the
N
is
number
how
modelled
of
long
answers
to
rabbits
it
an
will
by
the
after
take
3
for
appropriate
exponential
equation
N
=
20e
years.
the
number
degree
of
of
rabbits
to
reach
400.
accuracy.
0.8 t
5
The
model
where
t
is
for
the
measured
a
Write
down
b
Calculate
The
in
of
it
wombats
will
take
of
a
small
Dutch
2010
2011
Population
5101
5204
0
in
that
were
for
town
Year
=
wombats
in
a
nature
reserve
is
P
=
50e
the
introduced
original
to
the
reserve.
population
to
quadruple.
the
population
is
was
modelled
obtained
by
the
for
two
standard
consecutive
equation,
and
2010.
a
Calculate
the
rate
b
Calculate
the
predicted
c
Calculate
the
number
12 0
,
years.
many
long
P
solving
population
Assume
t
how
how
Problem
6
population
of
growth,
r.
population
of
years
E6 Quantity
in
until
2020.
the
population
exceeds
10 000.
years:
that
N U M B E R
Reect
●
What
and
do
discuss
you
think
are
4
the
limitations
of
continuous
growth
and
decay
models?
●
Why
do
you
think
ln
is
called
a
‘natural’
logarithm?
Summary
x
An
exponential
equation
has
the
form
a
=
b :
Generally
all
continually
●
a
is
the
base
●
x
is
the
exponent
growth
or
any
exists
a
two
positive
third
numbers
number
c
so
equation
a
that
and
you
b,
there
can
write
is
the
is
or
the
exponential
the
equivalent
modelled
that
by
,
grow
the
population
where:
power
P
For
populations
are
nal
the
population
original
population
when
t
=
0
statement
logarithmic
r
is
the
growth
t
is
the
time
rate
statement
period.
;
We
say
or
‘the
In
the
the
that
c
is
‘the
exponent
of
logarithmic
base
of
the
logarithm
a
that
to
gives
base
the
of
b’,
answer
b’.
statement
logarithm
a
,
and
b
is
called
a
is
called
If
then
is
,
The
Change
of
Mixed
1
Write
of
the
base
natural
logarithmic
0.
logarithm
is
written
as
ln
common
logarithm
is
written
down
formula:
the
equivalent
logarithmic
4
Find
the
value
of
x
to
the
nearest
hundredth.
for:
x
7
=
23
b
10
=
a
95
log
7
=
x
b
log
12
=
x
21 =
x
4
2
x
x
8
=
6
d
4
=
c
47
log
312
=
d
x
log
9
5
x
e
2
12
=
Write
e
1200
down
statement
an
equivalent
exponential
a
log
125
=
b
3
1
log
=

=
3
d
log
9
log
m
=
Find
2401 =
3
log
of
x
without
=
b
x
c
e

1
calculator:
20
d
3 × 3
f
3 × e
=
20
x
+ 1
= 15
2 x
=
45
+ 1
=
30
the
9
=
a
x
value
of
x
without
using
a
calculator.
1
2
2 x
=
2x
=

g
x
=
2 × 3
Find

x
d
log
7
log
d.p.
x
b
8
5
+
4
3 x
- 1
= 125
3
log
log
2
5
2x
e
to
b
2 x
e
a
log
2
c
x
n
value
16
of
= x
20
4 x
a
value
ln 5
=
4
6
the
f
x

a
3
a
-2
3
log 1000
the
x

7
e
Find
=
x
5
c
5
log 650
for:

7
3
=
1 =
x
c
3
+

3
1
1

=
6


x
f
log
8
=
9

x
8
3
6
6
=
y.
aslog y
logarithm.
x
c
statement
practice
statement
a
>
equivalent
the
The
argument
y
the
x
6
E6.1 Ideal work for lumberjacks
1 21
7
The
table
price
in
below
shows
Canada
from
the
average
1990
to
movie
ticket
b
1994.
in
c
Price
of
Calculate
the
Find
the
year
average
price
of
a
movie
ticket
2016.
when
the
average
price
will
reach
$20.
a
Year
movie
1990
ticket
8
($)
A
kettle
to
cool.
of
water
The
is
heated
temperature
and
can
then
be
allowed
modelled
by
4.00
0.2 t
the
1991
4.22
T
exponential
is
the
equation T
temperature
(in
= 100e
°C)
and
t
,
is
where
the
time
in
minutes.
a
1992
4.45
1993
4.69
1994
4.95
Find
the
exponential
model
for
the
a
Find
the
initial
b
Determine
c
Elizabeth
the
can
ticket
price
as
a
function
of
t,
Review
of
years
in
Charles
Francis
Richter
and
magnitude
the
of
intensity.
1935
it
It
is
formula
earthquake
a
the
an
American
who
created
which
by
scale,
standard
the
quanties
measuring
logarithmic
been
earthquake
One
scale,
earthquakes
has
was
physicist
Richter
size
1990.
context
seismologist
1
since
their
and
measure
since
of
intensity.
for
modelling
using
the
the
Richter
magnitude
scale
of
an
is:
I
c
M
=
log
10
I
n
where
M
is
the
magnitude,
I
is
the
intensity
of
c
the
‘movement’
and
is
the
of
the
intensity
earth
of
the
from
the
earthquake,
‘movement’
of
the
n
earth
a
on
The
day
a
normal
intensity
in
as
earth
the
from
250 000
this
of
is
the
100
on
microns.
of
the
earthquake
microns.
on
basis.
movement
intensity
earthquake
12 2
the
Oklahoma
December,
the
day-to-day
Determine
the
Richter
E6 Quantity
a
normal
Last
movement
was
of
recorded
the
drink
is
40
the
leave
number
temperature
the
after
water
3
minutes.
when
the
average
temperature
movie
temperature.
size
scale.
of
it
to
cool.
°C.
Find
how
long
she
must
N U M B E R
2
To
calculate
the
pH
of
a
liquid
you
need
to
4
The
decibel
scale
measures
the
intensity
of
+
know
the
inmoles
concentration
per
liter
of
the
of
hydrogen
ions
(H
)
liquid.
sound
D
where
I
On
The
pH
is
then
calculated
using
the
pH
=
- log
(H
is
the
the
equation
intensity
decibel
scale,
ratio
the
= 10  log
of
purr
a
of
,
sound.
a
cat
measures
logarithmic
D
+
formula:
the
using
=
10
log
330
=
25.2
(3
s.f.)
decibels.
).
10
Sound
a
HCl
is
a
strong
acid
with
a
hydrogen
intensity
is
measured
on
a
trac
light
ion
system:
concentration
of
of
0.0015
Find
the
pH
the
Find
the
hydrogen
HCl
moles
per
liter.
solution.
Green:
b
liquid
which
has
a
ion
pH
concentration
of
of
no
ear
protection
needed.
1 – 75
DB
a
9.24.
Orange: ear protection recommended. 80 – 120 DB
c
A
solution
nor
base)
if
is
said
its
to
pH
be
is
7.
neutral
Find
(neither
the
acid
concentration
Red:
of
hydrogen
to
be
ions
which
considered
would
make
a
ear
The
pOH
of
an
aqueous
solution
measures
of
hydroxide
ions
(OH
)
in
a
the
the
formula:
pOH  =  - log
sound
determine
+
DB
intensity
whether
or
of
not
these
ear
noises
protection
solution,
equipment
using
120
the
and
number
necessary.
neutral.
Calculate
3
protection
solution
(OH
is
recommended
or
necessary.
)
10
a
a
Find
the
pOH
of
an
aqueous
A
chainsaw
solution
that
has
a
hydroxide
ion
concentration
b
moles
× 10
per
an
intensity
ratio
of
× 10
of
5
6.1
has
11
1.04
A
owing
river
has
an
intensity
ratio
of
liter.
3100 000 .
b
Find
the
hydroxide
ion
concentration
in
an
c
aqueous
solution
that
has
a
pOH
of
A
rocket
launching
has
an
intensity
ratio
2.3.
16
8.2
Reect
and
How
have
you
Give
specic
× 10
discuss
explored
the
statement
of
inquiry?
examples.
Statement of Inquiry:
Generalizing
relationships
and
changes
that
can
in
quantity
model
helps
duration,
establish
frequency
variability.
E6.1 Ideal work for lumberjacks
12 3
of
Slices
E7.1
of
Global
context:
Objectives
●
Drawing
cosines
●
and
and
Converting
SPIHSNOITA LER
●
Using
●
Solving
the
pi
Inquiry
using
the
tangents
angles
radian
unit
of
between
formula
problems
circle
to
nd
sines,
degrees
for
involving
What
is
a
●
the
and
of
sin θ
length
angular
and
and
time
What
is
the
●
How
radian?
cos θ,
unit
circle?
radians
of
and
an
arc
are
the
trigonometric
related
to
the
unit
Which
parameter
functions
circle?
linear
C
graphs
space
questions
●
●
Drawing
in
F
angles
displacement
●
Orientation
where θ
is
translation
of
a
aects
the
sinusoidal
horizontal
function?
in
●
What
is
the
dierence
between
phase
radians
shift
●
Describing
transformed
trigonometric
the
form
f
( x ) =
a sin ( bx
- c
) + d
D
shift?
the
order
matter
How
do
we
on
dene
when
you
sinusoidal
‘where’
perform
functions?
and
‘when’ ?
Critical-thinking
Gather
and
Statement
organize
of
Generalizing
in
Does
transformations
●
ATL
horizontal
functions
●
of
and
space
can
relevant
information
to
formulate
an
argument
Inquiry:
and
help
applying
dene
relationships
‘where’
and
between
measurements
E7 1
‘when’.
E9 2
E11 1
12 4
G E O M E T R Y
Y
ou
●
should
nd
exact
ratios
●
draw
of
●
values
30°,
the
cosine
already
of
60°
of
how
trigonometric
and
graphs
know
1
sine
a
and
2
exact
on
3
a
sin x
a
Find
functions
b
=
the
2
length
tan 60°
of
an
arc
4
a
the
the
graph
Find
of
and
graph
of
sin 2x.
graph
the
c
cos x
amplitude
Describe
the
nd
of :
cos 45°
b
frequency
y
●
value
b
for values of x between 0° and 360°
transformations
sinusoidal
the
sin 30°
Sketch the graph of these functions
functions
describe
T R I G O N O M E T R Y
to:
Find
45°
A N D
of
the
of
y
=
transformations
y
3
=
cos x
cos x
+
to
get
on
the
1.
circumference
of
this
circle.
b
Find
the
length
of
arc
AB
A
B
60°
12
cm
O
Radian
F
Until
measure
●
What
is
a
●
What
is
the
●
How
now,
you
are
have
in
units
measured
in
dierent
Another
unit
of
as
the
unit
circle
radian?
unit
the
circle?
trigonometric
measured
measured
and
varied
units.
as
angles
functions
in
degrees.
kilometers,
Exploration
measurement
for
1
miles
related
Like
and
angle
is
the
the
unit
distances,
light
investigates
to
years,
radian
gradian,
circle?
which
angles
measure
also
known
can
can
for
be
also
be
angles.
as
You
do
know
the
grad.
One
grad
is
equal
to
of
one
revolution.
The
French
the
name
replace
of
the
Objective
In
the
describe
B:
a
temperature
general
in
of
scale.
honour
of
a
grad,
Hence,
Swedish
but
the
the
term
term
was
Celsius
astronomer
also
was
Anders
used
gradians.
as
adopted
Celsius
Investigating
1,
as
to
who
look
rule
to
patter ns
general
for
r ules
patterns
describe
the
in
consistent
the
central
with
angle
ndings
and
length
of
the
arc,
and
pattern.
E7.1 Slices of pi
to
work
scale.
patter ns
Exploration
write
one-hundredth
Centigrade,
proposed
ii.
means
need
to
term
with
centigrade
not
how
12 5
Exploration
ATL
1
A
unit
circle
2
circle
has
interms
Copy
and
circle
with
Central
1
a
of
radius
complete
dierent
angle
of
1
unit.
Find
the
circumference
of
a
unit
π
this
table
sized
with
angles,
the
in
lengths
terms
of
in
of
the
Length
Fraction
of
arcs
of
the
unit
π
of
circle
degrees
corresponding
arc
360°
1
2π
(circumference)
180°
90°
60°
45°
30°
The
lengths
of
rstcolumn.
a
3
Suggest
4
Consider
a
this
with
table
the
So
formula
Central
arcs
360°
circle
the
radian
,
convert
with
the
angle
to
are
=
radius
lengths
in
of
an
r,
angle
as
the
Central
measures
and
so
of
a°
to
shown
to
the
arc
s
of
this
of
the
angles
in
the
on.
radians.
right.
circle
angle
in
Copy
terms
Length
and
of
r
complete
and
of
:
s
θ
r
degrees
360°
in
radians
corresponding
arc
s
(circumference)
180°
90°
60°
45°
30°
5
Suggest
a
formula
theangle
6
By
8
radius
Suggest
12 6
a
the
of
for
your
arc
length
s
of
a
circle
with
radius
r
when
formula
in
step
5,
explain
why
radians
are
a
units.
central
the
the
radians.
without
Determine
the
in
considering
measure
7
is
angle,
in
radians,
of
an
arc
equal
circle.
denition
for
the
radian
E7 Measurement
measure
of
an
angle.
in
length
to
G E O M E T R Y
A
1
radian
radian
is
is
a
unit
the
for
angle
measuring
at
the
A N D
T R I G O N O M E T R Y
angles.
center
of
an
arc
with
s
length
equal
to
the
radius
of
the
circle.
θ
and
r
360°
1°
=
2π
=
radians
radians
Practice
1
Convert
a
these
angles
10°
f
2
1
b
72°

Convert
g
these
radians.
15°
c
126°
angles
π
to
18°
h
to
3π
5π



40°

j
300°
3π
7π
e
5
4π
8
5π
h
5π
i
12
Problem
e
d
4
g
9
225°
c
3
8π
i
150°
2π
6
f
36°
degrees.
b
a
d
j
3
3
4
solving
2
πr
3
The
area
of
a
sector
of
a
circle
is
given
by
A
θ
=
where
the
angle
is
360°
measured
You
of
use
rotations.
and
An
angular
one
radians
the
of
Angular
formula
turn
of
s
360°
revolutions
second
tip
work
displacement
Example
The
to
Rewrite
=
displacement
full
complete
degrees.
The
angular
than
5
can
in
the
out
rθ
the
links
(angle
greater
=
2π .
formula
distance
the
of
linear
a
2π
an
wheel
for
means
example,
angle
measured
has
moved
displacement
rotation)
than
For
for
an
a
wheel
that
the
angle
s
in
in
a
number
(distance
of
radius
wheel
of 10π
is
radians.
travelled)
r.
rotates
more
equivalent
to
(turns).
1
hand
the
on
second
a
clock
hand
displacement
for
is
(the
1
20
cm
long.
distance
second
is:
the
Find
the
second
linear
hand
Find
displacement
travels)
the
in
angular
Find
6
of
seconds.
displacement
the
Continued
angular
on
next
each
second
displacement
(in
for
6
radians).
seconds.
page
E7.1 Slices of pi
12 7
r
=
20
cm
cm
cm
Practice
1
The
(3
s.f.)
2
hour
hand
is
15
cm
long.
the
linear
displacement
of
the
tip
of
the
hour
hand
in
4
hours,
to
3
s.f.
b
Find
the
linear
displacement
of
the
tip
of
the
hour
hand
in
9
hours,
to
3
s.f.
The
A
a
20
b
how
Give
The
Tour
diameter
70
cm.
wheel
A
wind
tip
A
a
of
de
Find
hand
is
15π  cm .
Find
how
many
meters
your
in
bike
the
accurate
a
the
the
to
3500
wheels
minimum
entire
distance
diameter
odometer
answer
kilometers
is
has
to
3
as
the
wheels
cm.
measures
accurate
odometer
3
travelled
62
when
the
wheel
makes
s.f.
measures
after
150
rotations.
s.f.
km
used
bike
in
number
race.
the
of
completed
race
The
rotations
race,
rules
must
to
a
the
be
specify
between
Tour
de
nearest
that
55
the
cm
France
and
bike
hundred.
solving
blade
Giveyour
the
blade
rotates
travels
horse-drawn
Find
linear
road
bike
the
makes
Determine
b
the
a
France
the
turbine
the
measures
answer
of
Problem
hour
Give
many
your
the
of
many
rotations.
Find
c
wheel
how
of
passed.
odometer
The
Find
displacement
have
bicycle
rotate.
solving
linear
hours
4
clock
Find
c
3
a
a
Problem
2
on
18.85
carriage
the
linear
in
m.
has
angular
answer
through
a
Find
wheel
90°
the
of
in
3
seconds.
length
of
diameter
displacement
of
the
the
3
During
blade,
that
accurate
time,
to
2
the
d.p.
m.
wheel
in
15.25
rotations.
radians.
displacement
of
the
wheel,
to
the
nearest
tenth
of
a
meter.
90°
The
unit
with
its
circle
center
The
axes
The
section
part
of
is
at
divide
at
a
circle
the
the
the
of
origin
circle
top
of
radius
of
into
page
the
four
129
1
unit
graph.
2nd
1st
quadrant
quadrant
quadrants
shows
the
θ
180°
is
a
the
radius
unit
circle
ofthe
in
circle,
the
and
rst
quadrant.
thus
has
0°
OP
length
1.
3rd
4th
quadrant
quadrant
270°
12 8
E7 Measurement
G E O M E T R Y
A N D
T R I G O N O M E T R Y
y
In
the
right-angled
triangle
OPQ :
1
PQ
sin θ
PQ
=
=
=
OP
OQ
cos θ
so
PQ
=
sin
P
= OQ , so
=
OQ
=
cos
1
PQ
=
(cos θ, sin θ )
OQ
=
OP
tan θ
PQ ,
1
y
=
OQ
x
θ
Since
of
P
PQ
is
has
y
coordinates
and
the
(x,
length
y),
of
the
OQ
O
length
is
x
Q
1
x
Therefore:
y
sin
=
y
cos
=
tan θ
x
=
x
The
coordinates
ofthe
unit
circle
Reect
This
of
the
are
and
diagram
point
P (x
therefore
discuss
shows
the
full
y)
in
the
1st
quadrant
(cos θ , sinθ )
1
unit
90°
circle.
y
Point
P
is
in
the
1st
quadrant.
1
P
P
Point
in
P
the
2nd
quadrant
2
is
(cos θ, sin θ )
1
2
the
reection
of
P
in
the
y-axis.
1
θ
Point
P
in
the
3rd
quadrant
is
0°
180°
3
x
the
reection
of
P
in
the
x-axis.
2
Point
P
in
the
4th
quadrant
is
4
P
P
3
the
reection
of
P
in
the
4
x-axis.
1
●
What
P
2
●
P
are
the
and
3
P
coordinates
points
270°
4
What
conclusions
ratios
in
the
graph
protractor
paper,
to
can
dierent
Exploration
On
of
?
you
make
about
the
signs
of
the
sine
and
cosine
quadrants?
2
draw
draw
a
quarter
angles
of
0°,
circle
15°,
with
30°,
radius
45°,
60°,
10
cm.
75°
Use
and
your
90°
as
shown.
y
90°
75°
10
60°
8
45°
6
nis
30°
4
15°
2
0°
x
0
2
4
6
8
10
Continued
on
next
page
E7.1 Slices of pi
12 9
1
For
each
2
Copy
angle,
and
read
complete
θ in
the
this
values
table
θ in
degrees
of
of
cos
and
sin
from
the
graph.
values.
radians
0
using
π
sin θ
cos θ
0
1
0
15
30
…
90
3
4
Compare
etc.
Predict
the
105°,
5
your
sin 30°,
Repeat
4th
values
Explain
values
120°,
step
4
Reect
of
135°,
for
quadrant:
the
=
and
with
why
sine
3rd
calculator
may
and
150°,
285°,
the
they
be
cosine
165°
for
and
discuss
…,
angles
180°.
=
quadrant:
300°,
values
for
sin 0°,
sin 15°,
dierent.
in
Check
195°,
the
2nd
with
210°,
…,
quadrant:
your
270°
calculator.
and
the
360°.
2
π
●
Use
this
sin
=
unit
circle
to
explain
2
why:
sin
5π
π
6
6
30°
sin
=
π
-sin
0
7π
11π
sin
●
=
Do
you
The
your
6
results
made
trig
6
-sin
in
functions
cotangent
can
conrm
the
Exploration
secant,
also
be
conclusions
2?
cosecant
found
3π
2
Explain.
on
and
the
U
R
unit
circle.
θ
T
Pick
a
point
P
in
the
rst
quadrant
and
extend
P
the
line
enough
is
segment
for
labelled
=
1
it
to
RU
line
x
line
segment
OP
and
intersect
here.
from
up
the
Then
point
S,
to
the
line
draw
right,
y
=
the
upward
to
1
far
which
tangent
meet
the
θ
OU
and
label
this
point
T
S
O
Then
OT
13 0
represents
sec
,
OU
is
csc
E7 Measurement
and
RUis
cot
Q
.
G E O M E T R Y
Example
Write
down
these
values
without
b
=
T R I G O N O M E T R Y
2
a
a
A N D
using
a
calculator.
c
Convert
the
degrees
so
radian
measure
to
120°
you
can
draw
it
easily.
y
1
√3
2
2
1
Draw
the
Make
a
angle
in
the
correct
quadrant.
√3
right-angled
triangle
with
the
2
x-axis
and
label
the
lengths
of
the
sides.
120°
60°
x
1
2
=
cos 120°
=
or
b
=
the
x-coordinate.
90°
Convert
the
radian
measure
to
degrees.
y
(0,
1)
Draw
the
angle
on
the
unit
circle.
90°
A
point
on
the
unit
circle
has
coordinates
(cos θ,
sin θ).
x
So
=
c
π
=
sin 90°
=
the
value
of
sin θ
is
the
y-coordinate
on
the
unit
circle.
1
180°
Convert
the
radian
measure
to
degrees.
y
180°
(
1,
0)
Draw
the
angle
on
the
unit
circle.
x
On
the
unit
E7.1 Slices of pi
circle,
tan θ
13 1
=
Reect
●
When
and
you
always
●
If
you
To
nd
●
the
draw
angles
3
in
the
unit
circle,
why
is
the
hypotenuse
value
positive?
drew
theother
How
discuss
a
circle
sides
does
of
that
the
ratios
triangles
radius
triangle
compare
trigonometric
special
with
and
to
of
=
be?
the
2
for
What
value
angles
Example
in
would
four
part
the
calculated
the
2
a,
what
x-coordinate
in
Example
quadrants,
you
2
would
be?
part
can
a?
use:
SOHCAHTOA
30°
45°
√
√
2
2
3
1
60°
45°
1
1
●
similar
special
triangles
Thiscorresponds
to
the
where
unit
the
hypotenuse
is
equal
to
one
unit.
circle.
30°
√
3
1
45°
1
1
2
√
2
60°
45°
1
√
Practice
1
Write
down
these
values.
π
sin
b
tan 2π
g
2π
cos
3
f
2
2
3
π
a
1
c
d
3
5π
11π
h
sin
2
Find
two
3π
cos
e
cos
6
2
3π
cos
i
tan
6
4
Problem
5π
tan
4
4
solving
angles
that
have
the
given
trigonometric
ratio,
0
<
A
≤
2π
If
1
a
tan A
=
-1
b
cos A
the
=
c
sin A
=
given
-
2
give
cos A
=
0
e
sin A
i
Write
π

a
down
cos

π
coordinate
=


5π

d
4

2π
The
on
unit
the
13 2
circle
unit
has
equation
11π
11π
circle.
E7 Measurement

, sin
6
x


3
cos

point.
, sin
2
ii
each
3

, sin
4
for
2π
cos
b
6
5π
cos

values


, sin
6

c
the
6

2
+
y
=
the
radians.
2
3
in
is
radians,
2
3
d
domain
2
1.
Show
that
these
four
points
are
answer
in
G E O M E T R Y
Reect
●
Can
●
How
than
To
the
and
an
can
0
or
answer
unit
angle
Write
measure
you
nd
more
the
circle
Example
discuss
the
questions
you
drew
sine
than
and
?
Can
cosine
an
values
angle
of
measure
angles
that
less
than
measure
0 ?
less
?
in
in
Example
Practice
3,
3,
you
could
question
also
use
the
results
from
3
3
down
these
values
a
without
b
a
T R I G O N O M E T R Y
4
more
than
A N D
=
using
a
calculator.
c
cos 495°
Convert
the
radian
measure
to
degrees.
y
(
1,
1)
One
√
rotation
360°,
so
495°
is
one
rotation
plus
135°
more.
2
135°
1
Draw
x
1
the
=
=
cos 495°
an
angle
x-axis
sides
b
is
in
and
the
of
135°.
label
special
Make
the
a
right-angled
angles.
triangle
Label
with
the
angles
triangle
lengths
90°,
45°,
of
with
the
45°
=
sin 990°
y
Two
is
complete
270°more.
rotations
Draw
an
is
720°.
angle
of
So
990°
270°
270°
x
The
(0,
=
sin 990°
value
of
is
the
y-coordinate
on
the
unit
circle.
1)
=
-1
Continued
on
next
page
E7.1 Slices of pi
13 3
c
y
Negative
angles
start
at
zero
degrees
and
rotate
clockwise.
1
Draw
150°
clockwise
from
zero.
Make
a
right-angled
x
√
3
triangle
2
with
the
x-axis.
This
is
one
150°
=
Practice
1
Write
down

a
8π

b
3π

13π
Write
tan
(


3π
)
7π
down
a
negative
angle
(in
4
radians)
π


6

has
the
given

π


3

cos

that



i


6
sin
f
sin
h
7π



4

tan

2

tan


c

e
4
9π
cos

cos

2
values.

3

g
these
sin

d
4
trigonometric
ratio.
1
a
tan A
=
1
b
cos A
c
=
sin A
=
-1
2
3
d
cos A
=
0
e
sin A
=
2
Graphs
of
sinusoidal
functions
C
●
Which
parameter
aects
the
horizontal
translation
of
a
sinusoidal
function?
●
What
is
Exploration
1
Graph
of
2
axes
the
13 4
dierence
write
the
between
phase
shift
and
horizontal
shift?
3
functions
and
Describe
the
f
( x )
down
=
the
sin ( x )
and
relationship
transformation
on
E7 Measurement
the
on
between
graph
of
f
these
that
two
gives
the
same
set
functions.
the
graph
of
g
of
the
special
triangles.
G E O M E T R Y
π

The
graph
of
y
=
cos
T R I G O N O M E T R Y

x
is

A N D
2
a
horizontal
translation
of
the
graph
y
=
cos x

π
by
units
in
the
positive
x-direction.
2
y
y
y
y
=
2
2
2
3π
π
x
0
x
π
π
π
3π
sin x
1
0
2π
=
sin x
1
3π
2π
2π
π
3π
π
2
2
2
2
π
π
2
1
1
y
=
2π
π
y
cos x
=
x
−
2
The
graphs
of
y
=
a
cos bx
and
π

So
the
graph
of
y
= cos
x
y
=
a
has
2
have
amplitude
a
and
frequency
b.

-

sin bx
amplitude
1,
frequency
b
=
1,
and
is
translated

π
horizontally
by
units
in
the
positive
x-direction.
For
sinusoidal
functions,
2
when
The
y
=
the
four
a cos
parameter
b
parameters
( bx
- c
=
of
1,
a
) + d are
the
horizontal
sinusoidal
interpreted
translation
function
y
graphically
=
is
a sin
called
( bx
- c
the
phase
shift
) + d  or
as:
b
y
a
d
x
c
Exploration
1
Plot
the
graphs
horizontal
2
g
Plot
graphs
the
Rewrite
Using
a
in
=
your
a
y
=
a
the
( x )
form
of
the
f
y
=
a
y
=
on
cos b (x
sin
=
on
2x
the
g ( x )
( x )
=
sin (b x
+
+
g ( x )
graph
of
sin 2(x
cos
g ( x )
graph
steps
+
1
d
of
f
+
and
into
( x )
sin
that
.
gives
Write
the
down
graph
of
the
g ( x ).
).
and
the
c)
f
=
=
( x )
cos
that
.
gives
Write
the
down
graph
of
the
g ( x ).
).
2:
an
equivalent
function
of
the
form
d
cos (b x
)
and
(x
from
)
a
=
form
results
sin b (x
convert
f
translation
in
convert
y
b
g
of
translation
Rewrite
horizontal
3
4
c)
+
d
into
an
equivalent
function
of
the
form
d
E7.1 Slices of pi
13 5
The
values
sinusoidal
For
●
●
of
b
and
c
both
inuence
the
horizontal
=
functions
1,
the
y
=
a
sin (b x
When
c
>
0,
the
shift
is
in
the
positive
When
c
<
0,
the
shift
is
in
the
negative
shift
is
the
equal
to
c
is
+
b
horizontal
parameter
c)
When
called
actual
graphs
of
=
a
cos (b x
c)
+
d :
shift
to
the
to
right.
the
translation
left.
of
the
graph,
.
shift
is
in
the
positive
When
<
0,
the
shift
is
in
the
negative
window
phase
horizontal
the
the
y
x-direction,
0,
x-direction,
to
x-direction,
the
to
right.
the
left.
4
amplitude,
translation
and
x-direction,
>
Example
d
the
When
State
of
functions.
sinusoidal
The
translation
of
from
y
=
0
-
2
to
frequency,
sin (2x
period,
π)
4.
horizontal
Draw
the
shift
graph
and
of
the
vertical
function
using
a
2π
In
:
|a|
y
amplitude
is
|a|
=
is
the
amplitude
2
2
b
frequency
is
b
=
is
the
frequency
2
0
π
π
3π
2π
2
2
2
period
is
is
the
period.
4
6
y
=
−2
sin(2x
−
π)
−
4
horizontal
shift
is
8
is the horizontal shift.
vertical
translation
is
d
=
- 4
d
Example
Without
On
the
using
frequency
horizontal
vertical
=
=
a
set
GDC,
of
vertical
draw
axes,
use
the
graph
of
the
transformations
function
to
draw
y
=
the
sin x,
graph
from
translation.
0
to
2π
of
Find
the
shift
and
amplitude,
frequency,
1.5
vertical
2
shift
=
translation
=
-2
Continued
13 6
the
5
same
amplitude
is
E7 Measurement
on
next
page
translation.
horizontal
G E O M E T R Y
A N D
Rearrange
y
=
so
T R I G O N O M E T R Y
that
x
is
‘on
its
own’.
1.5
Work
Step
1:
Compare
y
=
sin x
with
outward
from
x
:
y
2
π
y
=
y
sin x
=
x
1
The
2
graph
of
is
a
horizontal
0
x
π
π
3π
translation
of
x-direction
of
units
in
the
positive
2π
1
2
2
the
original
function
y
=
sin x
2
Step
2:
Compare
y
=
sin
with
:
y
2
π
y
=
The
x
graph
of
is
a
horizontal
2
1
dilation
of
, scale
factor
.
0
x
π
π
π
2π
The
1
2
frequency
tells
you
the
number
of
2
y
=
sin(2x
−
π)
complete
sine
curve
cycles
in
the
period
2π
2
Step
3:
Compare
y
=
sin 2
with
y
=
1.5
sin 2
:
y
2
y
=
1.5
sin(2x
−
π)
1
0
Redraw
with
amplitude
1.5.
x
π
π
3π
2π
1
2
y
=
2
sin(2x
−
π)
2
Step
4:
Compare
y
=
1.5
sin 2
with
y
=
1.5
sin 2
2:
y
2
y
=
1.5
sin(2x
−
π )
1
0
Redraw, translated 2 units down.
x
π
π
3π
2π
1
2
2
2
3
y
4
ATL
=
1.5
Practice
1
State
the
vertical
a
y
c
y
sin(2x
−
π)
5
amplitude,
translation
= 0.5 sin
π
2x
π
x

+ 7
4

= - sin 3
period,
phase
or
horizontal
shift,
and
b
y
=
sin(3 x
d
y
=
6
- π ) + 2

-

y
frequency,
of :
= 3 cos 4 x

e
2
cos
(3 x
+ π
) - 3


-
+ 2
3

E7.1 Slices of pi
13 7
2
Without
using
π

a
y
= cos
x
(4 x
b
y
=
c
y
=
3  sin
d
y
=
0.5  sin
The
draw
the
graph
of
each
function:

+ 3,
2
from
0
to
2π

- 2π
) - 1,
( 0.5 x ) - 2 ,
Problem
3
GDC,
-

cos
a
(2x
from
from
0
0
to
) - 2, from
- π
π
to
6π
0
to
2π
solving
function
y
=
a sin
( bx
- c
) + d
has
amplitude
=
2,
period
π ,
=
horizontal
π
shift
=
to
the
right,
and
vertical
shift
=
-1.
Write
the
equation
of
the
function.
4
4
For
each
graph,
equivalent
of
f
( x )
=
nd
forms:
cos x.
i
as
the
a
Verify
equation
of
the
transformation
that
your
two
of
sinusoidal
f ( x )
answers
y
a
=
sin
are
functions
x,
and
ii
as
equivalent
in
a
two
transformation
using
technology.
y
b
10
4
9
8
2
7
6
5
0
5π
3π
x
π
3π
π
5π
4
2
3
2
4
1
x
0
π
c
d
y
2π
3π
4π
y
10
x
0
π
π
3π
1
2
8
3
6
5
4
7
2
9
π
0
x
π
2π
3π
4π
4
15
13 8
E7 Measurement
2
2π
G E O M E T R Y
Reect
●
In
and
Practice
functions
Some
periodic
sinusoidal
a
can
seaside
be
where
is
of
4,
5
for
sine,
which
and
phenomena,
in
graphs
which
such
as
was
in
it
easier
terms
tides,
can
of
be
to
nd
cosine?
the
Explain.
modelled
using
6
town,
the
a
Draw
b
State
c
Explain
a
natural
modelled
t
question
terms
T R I G O N O M E T R Y
functions.
Example
In
5
in
discuss
A N D
the
the
the
by
the
number
graph
values
what
depth
of
the
(in
meters)
sinusoidal
of
of
d
hours
the
the
water
over
the
course
of
a
day
,
midnight.
amplitude,
you
the
function
after
sinusoidal
values
of
function.
period,
found
in
b
horizontal
shift
and
vertical
shift.
represent.
d
d(t)
=
1.6 sin(0.5 t
+
0.47)
+
1.93
4
Draw
the
graph
for
24
hours,
3
for
values
of
t
from
0
to
24.
2
1
0
t
4
b
amplitude
period
8
=
12
16
20
24
1.6
Find
the
values
from
the
function.
=
horizontal
vertical
shift
shift
=
=
1.93
Explain
what
these
values
mean
c
in
depth
The
full
over
period
cycle
24
vertical
The
horizontal
the
,
high
The
above
context
of
the
tides.
hours.
is
of
the
shift
or
is
low
the
shift
mean
about
and
24
hours,
and
is
the
time
taken
for
a
tide.
mean
means
at
depth
over
midnight
the
the
24
hours.
water
depth
was
already
value.
E7.1 Slices of pi
13 9
Practice
1
A
buoy
6
bobs
thebuoy
in
up
π

t

b
State
the
shift.
Explain
the
The
the
(in
water
with
seconds)
can
the
be
waves.
The
modelled
of
values
the
of
model
to
i
2
seconds
ii
5
seconds.
amplitude,
these
nd
period,
represent
the
height
in
of
the
the
horizontal
context
buoy
of
shift
the
force
of
the
sun
and
moon
cause
changes
3
a
Draw
can
b
State
the
a
be
modelled
number
graph
the
values
c
Find
the
d
Explain
height
of
of
the
amplitude,
their
The
by
the
function
in
this
height
why
h
in
hours
since
ride
begins
midnight
and
h
is
the
a
Draw
b
State
c
4
diameter
of
period,
the
be
the
the
tide
of
vertical
at
Find
the
10
lasts
in
a
height
of
of
amusement
The
ride
seconds,
32
Ferris
this
the
seconds.
high
particular
by
wheel
height
and
and
car
the
is
horizontal
low
on
a
tides
giant
function
172
period,
by
the
the
car
ride
7
then
after
m
can
shifts,
5
Draw
b
Interpret
The
table
a
function
graph
the
of
this
h
the
and
of
the
and
will
occur
Ferris
is
wheel
interpret
t
so
important.
minutes
2π
h (t ) = 86 sin
horizontal
the
of
average
each
for
be
modelled
ground.
descends
car
after

(t
- 7.5)
+ 94

30
shifts,
and
interpret
using
car
along
meters
- 3π
a
a
sinusoidal
travels
track.
after
t
horizontally
The
seconds
entire
can
ride
be
+ 7

4
from
10
parameter
monthly
A

t
t
in
π
h (t ) = 3 cos
function
meaning
shows
and
minutes.
also
above
climbs
height
10

a
meters.
m.
vertical

modelled
in
function.
park
The
+ 0.67 ,

context.
starts
and
+ 3)
midnight.
when
modelled
amplitude,
values
function.
for
graph
their
Another
the
of
context.
meters
of
heights

(t
6

The
the
function.
predicting
can
in
h (t ) = 1.3 sin

the
vertical
π

is
and
problem.
after:

t
of
function
solving
which
where
h
the
function.
the
what
gravitational
tides
height
by

graph
Problem
in
t

2
Draw
Use
time
+
5
a
c
the
down
at
π
h (t ) = 1.5 sin
2
and
meters
in
to
32
the
temperatures
seconds.
function
in
in
this
Fahrenheit
context.
for
a
US
city.
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
32°
35°
44°
53°
63°
73°
77°
76°
69°
57°
47°
37°
14 0
E7 Measurement
G E O M E T R Y
a
Draw
a
graph
months,
b
Use
and
c
the
and
data
to
interpret
Write
the
●
Does
nd
their
of
the
How
Reect
●
Write
do
and
down
data,
the
Now
draw
in
function
matter
order
in
dene
b
Describe
graph
c
a
the
in
b
a
4
Summarize
f
give
the
vertical
and
horizontal
shifts,
this
problem.
perform
transformations
on
and
‘when’?
used,
in
order,
to
draw
you
the
graph
of
5.
are
used.
always
Try
get
a
the
changing
few
the
dierent
same
graph
as
in
on
the
graph
of
y
=
cos x
on
the
graph
of
y
=
sin x
the
graph
of
y
=
cos x
that
give
the
transformations
that
give
the
3.
the
transformations
on
that
give
the
.
steps
the
the
.
function
1
in
step
1
by
applying
the
two
transformations
order.
or
and
your
transformations
to
Do
transformations
whether
Repeat
represent
5
the
each
3
to
Explain.
dierent
State
you
transformations
of
Graph
models
Example
of
Describe
graph
2
of
period,
by
5?
graph
x-axis
of
Example
Describe
the
6
transformations.
a
on
y-axis.
context.
when
‘where’
ordersof
Exploration
1
that
transformations
graph
which
this
12
functions?
discuss
the
to
the
amplitude,
in
●
1
on
T R I G O N O M E T R Y
transformations
we
the
using
meaning
order
sinusoidal
●
this
temperatures
sinusoidal
Order
D
of
the
A N D
not
2
for
of
f
is
these
ndings
need
graph
there
by
to
be
( x )
=
a
dierence
your
two
graphs.
functions:
indicating
performed
a
in
sin (bx
the
on
c)
+
order
the
in
graph
which
of
y
=
sin x
d
E7.1 Slices of pi
141
Summary
On
the
P ( x,
unit
y )
on
circle,
the
the
coordinates
terminal
side
of
of
a
angle
point
For
are
sinusoidal
and
●
:
When
b
phase
A
radian
is
a
unit
for
measuring
angles.
radian
length
of
the
is
the
equal
angle
to
the
at
the
center
of
an
arc
=
1,
the
parameter
c
is
called
the
shift
When
to
1
functions
the
c
>
0,
the
shift
is
in
the
positive
x-direction,
negative
x-direction,
right.
with
radius
When
circle.
to
the
c
<
0,
the
shift
is
in
the
left.
s
●
and
The
horizontal
shift
is
the
actual
horizontal
θ
translation
of
the
graph,
equal
to
.
r
360°
2π
=
radians
c
When
>
0,
the
shift
is
in
the
positive
x-direction,
b
1°
=
radians
to
the
right.
When
to
Mixed
1
the
these
interms
of
angles
from
degrees
to
radians,
5
π
State
12°

225°
the
b
240°
c
288°
d
g
810°
h
336°
i
480°
e
80°
amplitude
shift
(vs)
or
Convert
these
angles
from
radians
to
π
2π
3π
b
j
degrees.
e
11π
using
a
y
= 2
c
y
= 0.2
sin( π x )
sin 6
y
=
π
x
-sin(6x
Without
using
30
a

2

 3

cos
b
sin
 π

c
write
a
y
e
cos
 13π
 3π



d
sin
4
f


6
a
b
y
=
3 cos
(2x ) - 3,

π

=
sin
π
x
tan ( 4 π )
5π
3π

-
2
+ 8
2

1
draw
the
graphs
of :
2
from
0
2π
to

6

- 4 ,
2
from
0
to
2π

 7π

y
= 2 cos
π
0.5 x



+
- 3,
4
from
0
to
2π

cos

4
solving

sin
j

6
7

Find
the
equation
that
denes
each
function.
y
a
Problem
π
x


-
Problem

i

GDC,

h


= -3 cos
)
an

sin
g
tan

6
 3π

y

-1
3


d
+ 2π
down:
c
 7π
=  5 cos
60


y

+
12)

 π
(4 x
b
j
4
calculator,
(p),
vertical
17π
i
20
Without
23π
h
5
3
π
g
period
12
6
7π
f
(f ),
and
of :
a
e
15
2
(hs),
5π
d
5
12
shift
700°
8π
c
frequency
125°

a
(a),
horizontal

2
x-direction,
left.
phase
shift
f
0, the shift is in the negative
practice
Convert
a
<
solving
4
4
One
car
on
a
Ferris
wheel
rotates
through
an
3
angleof
120°
in
11
seconds.
During
that
time,
2
thecar
travels
a
distance
of
10.47
m.
Find
the
7π
π
radius
of
the
Ferris
wheel
to
the
nearest
cm.
1
0
31π
0
23π
0
0
12
12
12
12
x
1
2
14 2
E7 Measurement
G E O M E T R Y
b
e
y
Find
in
the
exact
2147.3
A N D
distance
revolutions
T R I G O N O M E T R Y
travelled,
by
in
meters,
a:
1
i
14″
tire
ii
15″
tire
iii
20″
tire
x
0
π
π
3π
π
1
2
2
4
f
3π
2
Explain
how
number
of
the
radian
revolutions
measure
helps
of
you
the
nd
the
3
4
distance
travelled.
5π
π
3
3
4
4
Problem
solving
5
9
8
The
a
diameter
Explain
of
the
of
why
tire
to
a
car
you
tire
is
need
calculate
given
to
in
know
the
inches.
the
distance
The
and
seal
population
decreases
on
an
annually.
be
modelled
using
In
early
January,
In
early
July,
a
iceberg
The
increases
population
trigonometric
can
function.
diameter
a
car
the
population
is
3000
seals.
wheel
after
all
the
baby
seals
are
born,
travels.
there
b
Find
in
the
1050
distance
travelled,
revolutions
by
in
whole
meters,
3000
16″
tire
ii
17″
tire
iii
20″
tire
Draw
inches
to
cm
to
1
inch
a
graph
the
the
Find
of
the
the
radian
The
the
population
following
reduces
to
January.
measure
of
the
trigonometric
population
of
seals
function
at
any
to
time
year.
=
2.54
Write
down
of
0.6
the
trigonometric
equation
that
cm.
represents
c
by
obtain
b
metricdistances.
seals.
again
represent
of
Convert
5000
a:
a
i
are
seals
this
function.
c
Find
the
seal
population
at
the
end
of
May.
d
Find
the
seal
population
at
the
beginning
revolutions
tire.
of
November.
d
Hence,
in
i
0.6
15″
For
of
is
is
around
class,
Find
the
when
A
used
the
travelled,
in
meters,
a:
18″
tire
a
to
sundial
be
iii
to
edge
measure
of
the
angle
surveyor’s
with
made
surveyor’s
central
the
by
22″
e
Find
2
on
wheel
the
circle
(in
a
a
diameter
with
length
of
the
radians)
wheel
rotations
3.5
a
i
Write
The
ii
State
what
the
variable
State
what
the
parameter
5
d
were
the
rst
down
of
iii
arcs
Suggest
sundial.
of
the
its
arc
means
of
State
increasing
the
could
eectively
trajectory
modelled
where
c
=
launches
(until
by
0.
seals.
the
it
a
6.25
to
catapults
distance
stone
lands
on
t
function
( x )
t
x
represents.
the
a
catapult
represents.
could
inuence
into
the
from
it
parameter
b
determines
represents.
the
stone’s
point
The
stone
starts
travelling
when
it
is
as
and
stops
as
soon
as
it
hits
the
which
State
the
domain
and
range
of
enemies.
the
air
ground)
( x ) =
the
how
rotations
use
neutralize
function:
simplied
impact.
t( x
Acatapult
the
how
what
Explain
rotations
ground.
they
4500
value.
launched
a
are
makes:
b
rotations
Romans
there
radius
v
2
month(s)
school
of
c
which
tire
iv
a
in
context
going
playground.
20cm
distance
ii
in
science
12m
the
tire
Review
1
nd
revolutions
a
sin
)
in
terms
of
a
and
b
whose
can
( bx
be
- c
),
b
State
what
Suggest
the
what
parameter
this
says
c
represents.
about
the
catapult’s
position.
E7.1 Slices of pi
14 3
c
Keeping
in
explained
range
of
t
mind
in
the
part
( x ) in
a
trajectory
iv,
terms
state
of
a,
of
the
b
the
stone
domain
and
as
i
and
Suggest
why
the
vertical
displacement
of
a
sine
trajectory
curve
if
the
accurately
starting
model
point
of
A
is
10
m
above
the
of
a,
equation
of
second
rst
stone
is
launched,
State
and
its
trajectory
in
this
iii
GDC
b
and
this
stone’s
f
or
c
otherwise,
and
stone’s
write
iv
coordinates
State
stone
the
lands
State
at
the
sensitivity
very
the
of
large
distances.
Square
to
consist
over
and
a
operating
hundreds
scale
than
any
of
the
it
Hence,
stone’s
state
revolutionize
and
How
you
of
Generalizing
can
help
or
by
spread
(SKA)
radio
early
and
building
combining
over
long
is
in
access
the
2020s.
a
of
will
Africa
this
size,
radiotelescope
magnitude
allowing
understanding
It
spread
southern
to
next-
astronomy,
telescopes
instrument
of
6π
better
them
of
the
to
universe.
discuss
explored
the
statement
of
inquiry?
Give
specic
examples.
Inquiry:
and
dene
14 4
the
facility,
Reect
for
radio
orders
our
either
actual
interferometer.
Array
an
have
is
the
down
the
modelled
by
the
of
the
starting
of
it
the
is
position
launched.
position
where
ground.
reaches
ever-greater
kilometers
current
Statement
an
in
of
With
will
that
have
of
5π
resolution
by
telescopes
called
hundreds
astronomers
on
array
Kilometre
Australia.
both
telescopes
interferometer
start
of
an
is
4π
telescopes
smaller
This
generation
due
from
of
3π
increase
individual
signals
numbers
The
2π
new
nd
)
moment
coordinates
on
the
stone’s
can
is
- π
the
maximum
this
applying
‘where’
relationships
and
‘when’.
E7 Measurement
between
measurements
in
space
height,
and
height.
domain
4
Astronomers
stone’s
trajectory.
trajectory
( x ) = 5 sin ( 0.5 x
the
the
where
π
the
graph:
8
0
of
is
it
shown
range
the
ii
A
and
ground?
of
e
your
function
i
stone
Using
values
f
the
domain
trajectory.
the catapult’s position is ignored in this question.
Does
the
c
ii
d
State
trajectory.
and
range
of
this
Making
it
all
add
up
E8.1
Global context: Scientic and technical innovation
Objectives
●
Finding
the
Inquiry
sum
of
an
arithmetic
series
questions
●
What
●
How
is
a
series?
F
●
●
Finding
Finding
where
the
the
sum
sum
of
of
a
nite
an
geometric
innite
series
geometric
can
rewriting
easier
to
What
are
a
series
make
it
sum?
series,
appropriate
●
between
similarities
real-life
and
arithmetic
●
Can
you
and
patterns
dierences
geometric
evaluate
a
series
that
require
series?
that
goes
on
D
forever?
●
What
are
the
risks
of
making
generalizations?
ATL
Communication
Understand
and
use
mathematical
notation
8 1
12 1
12 2
E8 1
Statement of Inquiry:
Using
can
dierent
help
forms
improve
to
generalize
products,
and
processes
justify
and
patterns
solutions.
14 5
MROF
the
C
Y
ou
●
should
work
with
already
arithmetic
know
how
sequences
1
to:
Find
the
20th
nth term
term
of
and
each
the
arithmetic
sequence.
a
5,
b
c
●
work
with
geometric
sequences
2
9,
25,
13,
42.7,
Find
the
17,
22.5,
43.9,
the
…
20,
17.5,
45.1,
next
nth term
two
of
…
…
terms
these
and
geometric
sequences.
a
15,
b
c
●
generate
the
triangular
numbers
3
30,
100,
162,
Write
a
120,
6.25,
54,
18,
rst
the
5
generate
explicit
a
sequence
from
an
4
formula
…
2,
…
triangular
general
triangular
●
240,
…
down:
the
b
60,
25,
Generate
formula
numbers
for
the
numbers.
the
rst
each
sequence:
a
=
ve
terms
of
+
u
3n
n
1,
∈
,
n ≥ 1
,
n ≥ 1
n
n
b
u
=
3
×
2
+
,
n
∈
n
Series
F
A
●
What
●
How
sequence
sequence.
is
an
The
is
a
series?
can
rewriting
ordered
sequence
list
and
a
of
series
make
numbers.
the
series
A
it
easier
series
below
have
is
a
to
the
sum?
sum
nite
of
the
number
terms
of
of
a
terms.
In
MYP
Mathematics
1,
7,
4,
11,
2,
4
4
11
is
a
sequence,
even
though
there
is
no
pattern
in
it.
Standard
1
+
7
+
+
sequence.
In
+
2
this
+
-4
case,
is
a
the
series,
sum
is
because
it
is
the
sum
of
the
terms
of
saw
a
do
21.
that
not
follow
An
A
arithmetic
geometric
series
series
is
Exploration
1
1
of
Find
the
value
is
the
each
a
1
+
3
+
5
+
7
+
9
b
7
+
7
+
7
+
7
+
7
c
1
+
4
+
9
+
16
d
7
+
14
28
+
+
the
sum
sum
of
of
the
the
terms
terms
of
of
a
an
arithmetic
geometric
sequence.
sequence.
series.
+
11
+
25
+
56
+
36
112
+
224
Continued
14 6
E8 Patterns
on
next
page
8.1
have
a
you
sequences
to
pattern.
A LG E B R A
2
Look
at
the
a
Which
b
One
is
patterns
are
The
neither
terms
of
the
arithmetic
description
3
in
series?
geometric
of
a
series
nor
in
step
Which
1
are
geometric
arithmetic.
Suggest
series?
a
suitable
it.
series
are
given
by
the
formula
u
=
4n
=
3
2.
n
Find
the
value
of
the
series
u
+
u
1
+
u
2
+
u
3
+
u
4
5
n
4
The
terms
of
a
series
are
given
by
the
formula
u
×
4
n
Find
the
value
of
the
series
u
+
u
1
S
means
the
sum
of
the
rst
+
u
2
n
+
u
3
terms
4
of
a
series.
n
For
example,
S
=
u
4
+
u
1
+
u
2
+
u
3
4
2
ATL
5
Given
that
u
=
n
2n,
nd
the
value
of
S
n
of
Some
In
the
types
Practice
triangular
,
the
sum
of
the
rst
6
terms
6
series.
of
6
series
of
have
MYP
numbers
special
Standard
formulae
8.1
you
that
saw
help
that
you
the
nd
general
the
sum
formula
easily.
for
the
is:
n
S
( n + 1)
=
n
2
The
nth
pattern
triangular
with
n
number
is
the
T
=
1
T
1
Since
each
number
so
the
of
nth
number
of
dots
needed
to
make
a
triangular
rows:
triangle
dots
in
is
made
the
triangular
=
3
T
2
nth
of
rows
triangle
number
is
=
6
T
3
which
can
the
be
same
get
longer
found
as
=
10
4
the
by
by
one
adding
sum
of
1
the
every
+
2
rst
+
n
time,
3
+
4
…
+
n,
integers.
5 ( 5 + 1)
Using the formula, the sum of the rst ve positive integers is S
the
+
5 ×
6
=
=
= 15,
5
2
which
you
can
Example
ATL
Find
47
+
the
48
easily
by
adding
1
+
2
+
3
+
4
+
49
+
of
the
…
+
series
47
+
48
+
49
+
…
+
82.
82
Sum
=
S
82
2
5.
1
total
+
check
S
sum
46
the
of
series
the
S
1
+
series
1
2
+
+
2
…
+
+
…
82,
+
and
from
it
subtract
the
46:
=
1 + 2 + 3 + 4 + …. + 46 + 47 + 48 + 49 + … + 82
=
1 + 2 + 3 + 4 + …. +
82
=
=
2322
S
46
46
S
82
S
=
47
+
48
+
49
+
…
+
46
E8.1 Making it all add up
14 7
82
The
sum
S
of
the
rst
n
consecutive
positive
integers
is
.
ATL
You
can
use
to
n
You
can
also
S
use
nd
the
total
of
any
series
of
consecutive
integers.
( n + 1)
=
to
nd
the
sum
of
any
series
of
integers
with
a
n
2
common
factor,
Example
Find
the
5
+
=
5(1
+
+
15
2
shown
in
Example
2.
2
value
10
as
+
of
+
3
the
…
+
series
+
…
5
+
10
+
15
+
…
+
725.
725
+
The
common
factor
is
5.
145)
=
=
52
925
Practice
1
1
Find
the
total
of
the
series
2
Find
the
value
of
25
3
Find
the
value
of
4
4
Find
the
value
of
180
+
26
1
+
+
2
27
+
+
3
+
…
4
+
+
…
+
30.
53.
In
+
8
+
12
+
16
+
182
+
184
…
+
a
Problem
5
6
7
Find
b
Hence
value
a
Find
the
value
of
b
Find
the
value
of
c
Hence
nd
nd
terms
+
186
+
188
+
190
+
…
+
324.
solving
a
The
the
+
of
a
the
the
of
the
value
1
2
of
series
+
2
3
+
value
series
series
+
5
-4
of
are
the
+
+
7
4
+
-6
1
given
+
2
6
1
9
+
+
by
+
+
-8
3
-
the
+
+
8
+
…
+
5
…
+
101.
…
+
+
+
+
3
+
4
7
60.
+
+
…
+
59.
-100.
…
100
formula
9
=
u
+
3n
101.
+
7.
n
Find
the
value
of
S
5
n
2
8
The
sum
of
the
rst
n
squared
numbers
1
+
4
+
9
+
…
+
n
( n + 1) ( 2n + 1)
=
6
a
Use
this
formula
Exploration
1
to
step
conrm
your
answer
Find
the
value
of
1
+
4
+
9
+
16
+
…
c
Find
the
value
of
1
+
4
+
9
+
…
+
400.
d
Hence
14 8
the
to
1
+
4
+
9
1c
b
nd
value
of
121
E8 Patterns
+
144
+
3
and
4,
take
out
128.
+
100.
169
+
…
+
400.
+
16
+
25
+
36
in
common
factor.
A LG E B R A
Problem
9
Find
the
solving
value
of
Exploration
1
a
Copy
and
1
–
4
+
9
16
+
…
+
361.
2
complete
40
–
+
44
the
+
48
48
S
=
40
+
44
+
S
=
72
+
68
+
calculations
+
52
+
52
+
56
+
56
+
60
+
60
to
nd
+
64
+
64
the
+
68
+
68
value
+
of:
72
+
72
+
40
The
+
+
+
+
+
+
sum
2S
=
112
+
+
+
+
+
+
+
line
+
terms
in
are
reverse
b
Hence
nd
c
Deduce
the
value
of
2
Use
a
similar
a
57
+
b
1.7
c
30
+
+
Check
2.9
21
Do
●
Without
This
you
S
have
himself
+
3
+
the
+
-6
+
to
+
out
series
Gauss
it,
sense
is
of
and
for
his
+
+
rst
line.
93
7.7
+
+
+
of:
99
8.9
+
+
105
+
111
10.1
-24
the
seven
to
the
numbers
terms,
called
terms:
have
5
task
if
how
9
+
when
from
probably
the
+
the
1
to
you
had
+
as
his
to
twice
could
to
nd
nd
2S
2S ?
and
29
it
is
the
teacher
100.
already
he
series
…
method,
used
numbers
set
in
explain
Gauss’s
teacher
to
value
calculator.
all
said
all
him
a
the
with
sometimes
sum
the
1
out
all
87
-15
using
write
the
6.5
discuss
this
is
nd
81
5.3
order
asked
Gauss
knew
work
it
method
was
–
out
it
that
him
not
the
wouldn’t
the
total
slowly.
Practice
Find
+
the
the
S
to
+
answers
discover
made
75
4.1
12
Karl
calculate
+
writing
method
rstto
+
need
for
8-year-old
to
69
+
of
method
and
●
then
+
your
Reect
1
63
value
just
of
the
second
2S
in
the
in
the
the
2
value
of
these
arithmetic
series:
a
23
+
30
+
37
+
44
+
51
+
58
+
65
+
72
+
79
b
30
+
28
+
26
+
24
+
22
+
20
+
18
+
16
+
14
c
24
+
28.4
+
32.8
+
37.2
+
41.6
+
46
+
50.4
+
54.8
+
59.2
+
63.6
E8.1 Making it all add up
14 9
2
3
An
arithmetic
a
Find
the
b
Hence
Consider
40th
nd
the
a
Write
b
Hence
c
Find
Find
5
An
the
the
the
sum
Use
sum
of
+
=
S
the
of
begins
from
has
40
terms.
=15
rst
terms
+
19
+
term
in
…
and
the
+
199.
the
common
dierence.
series.
+
25
+
+
21
series
48
+
+
…
+
104.
…
could
2
to
nd
u
+
…
+
have.
the
sum
u
2
+
of
u
u
)
+
(u
n
+
+
the
arithmetic
+
u
1
n
u
+
u
2
u
)
n
why
…
1
2
clearly
+
n
n
1
Explain
+
the
Exploration
+
+
29
that
41
u
n
(u
series
n
u
n
a
of
1
=
and
series.
series
value
…
u
n
2S
…
series.
2
=
S
+
3
+
n
+
19
arithmetic
series
u
1
S
the
the
method
u
+
the
number
maximum
the
series
of
value
of
Exploration
1
6
solving
arithmetic
Find
ATL
the
sum
arithmetic
nd
the
begins
term.
the
down
Problem
4
series
+
…
+
(u
1
+
(u
+
n
u
1
)
and
(u
n
1
u
1
+
)
+
(u
2
u
2
)
n
+
u
n
are
)
1
equal
in
value.
1
You
b
Hence
explain
why
2S
=
n(u
n
c
Write
down
a
formula
+
u
1
for
S
in
useful
terms
of
n,
u
n
and
Use
the
formula
for
the
nth
term
u
=
a
+
(n
1)d
to
general
u
1
nd
use
formula
to
write
down
a
the
nth
term:
formula
n
u
=
n
for
S
for
an
arithmetic
series
with
n
terms,
rst
term
a
and
common
n
dierence
d.
Give
Reect
and
How
you
have
Exploration
For
an
2
formula
discuss
used
to
your
the
concept
series
terms
of
a,
d
and
n
only.
2
Exploration
arithmetic
in
with
of
generalization
when
moving
from
3?
rst
term
u
=
a
and
common
dierence
d,
1
the
formula
for
the
sum
where
of
u
the
is
rst
the
n
nth
terms
term,
is
given
by:
3
4
or
n
Learn
both
useful
for
15 0
of
these
dierent
two
types
formulas.
of
Examples
problems.
E8 Patterns
it
the
n
for
2
might
)
n
and
show
they
are
both
a
+
(n
1)d
A LG E B R A
Example
Find
a
=
the
3
value
15,
d
=
2
of
the
and
n
arithmetic
=
series
with
11
terms
that
begins
15
u
=
the
22,
u
=
of
the
78,
n
=
arithmetic
series
with
9
terms:
=
Find
the
when
you
know
a, d
and
n
rst
term
22,
last
term
terms,
and
n
78.
9
this
formula
when
total
5
+
8
+
11
b
1.4
+
2
+
c
15
+
29
d
57
+
52
Find
the
of
+
these
14
+
arithmetic
know
the
rst
and
last
17
+
2.6
+
3.2
+
43
+
…
+
141
+
47
+
…
+
-3
sum
+
20
3.8
+
+
23
4.4
series.
+
+
26
5
+
+
29
5.6
+
+
32
6.2
+
6.8
+
7.4
+
an
arithmetic
series
with
15
terms,
rst
term
20
b
an
arithmetic
series
with
18
terms,
rst
term
6
and
last
c
an
arithmetic
series
with
11
terms,
rst
term
9
and
second
d
an
arithmetic
series
with
32
terms,
second
e
an
arithmetic
Find
Find
value
the
the
which
series
with
common
and
term
dierence
12,
16
common
and
rst
term
5
43
term
fourth
term
dierence
39
15
term
and
22
last
291.
the
and
8
of:
a
18
you
3
a
term
4
formula
450
Practice
3
this
9
Use
2
...
4
value
1
1
+
275
Example
Find
17
11
Use
=
+
last
value
has
11
of
the
term
of
arithmetic
is
the
series
with
series
that
12
terms,
where
the
rst
term
is
102.
arithmetic
begins
17,
19.2,
21.4,
…
and
terms.
E8.1 Making it all add up
1 51
Problem
5
Find
42
6
the
and
An
solving
sum
the
of
the
arithmetic
eleventh
arithmetic
term
sequence
is
series
with
has
x
terms
+
x
1
Find
the
value
of
x
15
terms,
where
the
eighth
term
is
57.
+
x
2
+ ... +
x
3
=
726.
55
.
28
Exploration
You
it
have
seen
backwards
approach
1
Try
S
=
to
to
1
+
3
3
Copy
S
=
3S
=
1
+
=
S
=
1
2S
=
-1
Hence
5
Explain
6
2
Use
the
+
+
27
+
+
27
+
b
2.5
+
c
1
1.3
d
625
9
+
27
+
+
9
+
27
+
6
b
2500
c
160
was
32
+
the
+
the
+
try
to
rst
use
writing
the
same
+
value
2187
+
of
the
series:
6561
method.
of
243
you
by
sequence?
the
729
values
+
3S:
729
+
+
2187
+
+
6561
+
+
19 683
subtraction:
+
81
of
+
+
243
2S,
useful
by,
+
+
pattern
and
to
and
128
always
+
+
+
+
from
126
90
+
+
1.69
+
value
48
15 2
+
nd
series
+
729
+
+
2187
+
+
19 683
6561
...
series
will
+
15
+
+
value
it
this
8
0
the
why
42
the
+
+
500
a
the
3
+
to
+
3
method
14
Find
complete
when
geometric
243
81
arithmetic
happens
a
with
the
9
0
+
an
deduce
multiply
check
512
+
behind
work
by
that
2048
+
the
3.
your
of
Suggest
S
what
suggestion
you
should
works:
8192
geometric
when
value
trying
to
sequences
sum
a
means
geometric
that
this
sequence.
4
a
+
81
9
approach
1
+
problems
why
Explain
27
of
method
+
+
Practice
+
sum
What
sum
+
nd
2
2S.
the
3
+
multiply
=
9
can
3
+
4
nd
complete
and
3S
you
Gauss’s
the
and
Copy
S
to
+
Describe
2
that
nd
use
4
378
540
+
these
+
3072
1500
+
900
+
+
360
1134
3240
320
384
240
+
2.197
400
of
Exploration
+
+
+
+
+
256
24 576
540
540
+
+
E8 Patterns
+
+
the
10 206
3.71293
204.8
+
324
810
nd
value
+
of
these
30 618
+
geometric
91 854
+
+
+
+
4.826809
163.84
series.
196 608
+
194.4
1215
+
+
1 572 864
+
116.64
1822.5
+
series.
275 562
19 440
2.8561
+
to
3402
geometric
+
4
6.2748517
A LG E B R A
3
A
geometric
a
Write
b
Hence
series
down
nd
an
the
begins
2
+
6
expression
value
of
S
+
for
,
18
+
the
the
54
nth
sum
+
….
term
of
the
of
the
rst
8
series.
terms
of
the
series.
8
Exploration
A
1
geometric
Write
5
series
down
has
rst
term
expressions
in
a
and
terms
common
of
a
and
r
ratio
for
u
,
r
u
1
second,
2
Let
S
third
be
the
and
nth
sum
of
,
u
2
and
u
3
(the
rst,
n
terms).
the
rst
n
terms.
n
n
Apply
the
method
in
Exploration
4
to
show
that
(r
1)S
=
a(r
1).
n
3
Hence
write
down
a
formula
for
S
in
terms
of
a,
r
and
n
n
4
Use
your
a
a
geometric
series
with
rst
b
a
geometric
series
with
nine
c
a
geometric
series
with
seven
term
5
For
the
1000
6
+
geometric
800
write
b
show
c
explain
Show
and
a
+
640
down
that
the
clearly
when
the
geometric
a
of:
term
3,
terms,
term
geometric
seventh
409.6
of
the
common
rst
terms,
term
second
ratio
12
7
and
term
and
eight
second
168
and
terms
term
18
third
sum
in
+
327.68
rst
term
+
262.144
and
the
common
ratio
gives:
negative
fraction
above
series
ratio
geometric
second
c
sums
numbers
formula
this
can
formula
have
be
come
from.
rearranged
will
have
a
to
give:
positive
numerator
denominator.
a
a
the
formula
the
formula
common
+
value
the
that
positive
512
sum
where
either
b
nd
series:
+
the
Use
a
to
504.
a
State
7
formula
of
is
rst
the
term
value
16,
of:
seven
terms
in
total,
and
a
with
ten
terms
whose
rst
term
is
20
and
whose
10
series
term
nd
0.4
series
is
with
to
with
twenty
terms
whose
fourth
term
is
7
and
56.
E8.1 Making it all add up
15 3
You
saw
in
very
dierently
geometric
MYP
Standard
depending
series,
and
you
Mathematics
on
whether
will
explore
or
12.2
not
this
in
that
1
<
the
geometric
r
<
1.
n
 1
thebiggest
dierence
is
that
it
is
easier
to
S
use
The
debatable
sequences
same
is
section.
behave
true
For
for
now,

r
= a
when
1
<
r
<
1.
n

For
a
geometric
The
rst
The
second
version
Reect
●
What
How
and
Example
10
a
+
the
5
=
+
10,
ratio
would
easier
is
rst
to
use
easier
if
use
a
when
use
r
and
>
1
when
common
or
1
r
<
<
r

ratio
r :
-1.
<
1.
3
the
formula
for
a
geometric
series
with
1?
you
ratio
you
to
term
r
is
nd
the
sum
of
a
geometric
series
for
which
the
1?
5
sum
2.5
r
with
discuss
happens
common
Find
is
version
common
●
series
1
of
+
the
rst
13
terms
of
the
geometric
series
which
begins
...
=
and
n
=
13
1
=
20.0
(3
1
Choose
most
suitable
formula
a
a
geometric
series
with
rst
b
a
geometric
series
with
ten
c
a
geometric
series
with
twenty
15 4
to
a
sensible
5
the
r
<
1
s.f.)
Round
Practice
<
E8 Patterns
to
term
nd
16,
terms,
the
value
seven
rst
terms,
of :
terms
term
20
fourth
and
and
term
7
common
second
and
ratio
term
seventh
0.4
10
term
56.
degree
of
accuracy.
A LG E B R A
2
Find
the
a
5
+
b
7
c
96
25
3
3
+
+
the
Find
+
+
6
of
3125
7
+ 7
216
value
geometric
625
+ 7
+
term
the
each
5
+ 7
144
second
of
125
4
+ 7
Find
4
total
+
+
8
15 625
9
+ 7
324
+ 7
+
series.
486
10
+ 7
+
729
the
geometric
series
with
of
the
geometric
series
that
of
the
geometric
series
with
ten
terms,
rst
term
24
and
96.
value
begins
18
+
24
+
…
and
has
eightterms.
5
Find
the
andlast
value
term
430
467
rst
term
10,
common
ratio
9
210.
Tip
6
A
geometric
series
has
rst
term
14
are
two
possible
and
third
term
56.
In
a
Show
that
there
values
for
the
common
8,
how
b
7
Hence
A
nd
geometric
two
possible
series
has
rst
sums
term
of
15
the
and
rst
seven
second
terms
term
of
10.
the
series.
the
rst
eight
you
use
Find
the
(E7.3),
they
help
might
sum
to
nd
this
A
terms
If
in
not,
solving
geometric
series
is
formed
by
summing
the
integer
powers
of
can
number
2:
using
S
of
series.
you
8
the
terms.
number
Problem
know
logarithms
you
of
if
to
ratio.
nd
of
trial
the
terms
and
error
= 1 + 2 + 4 + ...
n
and
Determine
the
least
number
of
terms
so
S
that
>
1
your
GDC.
000 000 000.
n
Series
C
●
What
that
There
are
many
arithmetic
or
appropriate
in
are
real-life
the
arithmetic
real-life
situations
geometric
to
similarities
require
use
an
series.
In
arithmetic
and
and
where
each
or
dierences
geometric
problems
worked
geometric
between
real-life
patterns
series?
can
be
example,
solved
using
consider
why
it
Arithmetic
is
and
sequence.
geometric
sequences
Example
6
life
explored
A
company
month
of
the
to
launches
subscribe.
year,
another
a
Explain
why
b
Find
c
Assuming
the
service
the
income
in
no
the
a
In
new
the
500
music
rst
customers
monthly
in
the
rst
join
income
rst
customers
streaming
month,
1200
every
forms
service.
Customers
customers
sign
up.
pay
For
$4
the
in
contexts
in
real-
are
MYP
Mathematics
per
Standard
rest
12.2.
month.
an
arithmetic
sequence.
month.
leave
the
service,
nd
the
total
income
from
the
year.
Each
month
the
income
increases
a
by
so
b
this
1200
×
forms
4
=
an
arithmetic
500
×
$4
=
$2000.
series.
$4800
Continued
on
next
page
E8.1 Making it all add up
15 5
c
a
=
4800,
d
=
$2000
and
n
=
12
Use
S
=
(2
×
4800
+
11
×
this
formula
when
the
last
term
is
unknown.
2000)
12
=
$189 600
Example
A
social
Remember
In
the
rst
a
Explain
write
the
units.
7
networking
thenumber
to
of
new
month
why
site
gains
members
of
the
2016
the
number
new
is
site
of
members
1.5
times
gains
new
the
very
quickly.
number
15 000
members
new
per
in
Each
the
month,
previous
month.
members.
month
forms
a
geometric
sequence.
b
Find
the
number
c
Find
the
total
growth
of
new
number
members
of
new
in
the
members
2nd
that
and
will
3rd
be
months
gained
in
of
2016.
2016
if
this
continues.
d
Explain
a
The
why
number
the
of
growth
new
cannot
members
continue
forms
a
in
this
way
geometric
indenitely.
sequence
The
because
b
r
=
1.5,
it
a
is
=
increasing
by
a
constant
scale
common
ratio
is
1.5.
factor.
15 000
Number
of
new
members
in
2nd
Number
of
new
members
in
3rd
month
month
=
=
15 000
22 500
×
×
1.5
1.5
=
=
22 500.
33 750.
Use
this
formula
because
r >
1.
c
S
=
12
d
If
=
3 862 390.137…
=
3 860 000
growth
(3
s.f.)
continued
in
this
way,
after
36
months
Give
much
(7
greater
than
current
world
an
example
population
billion).
Examples
6
information
and
7
about
asked
a
common
dierence.
The
example
next
when
the
you
know
15 6
you
scenario
shows
the
to
—
you
value
of
nd
the
how
a
E8 Patterns
the
value
rst
to
series.
term
form
an
of
a
and
series,
the
given
common
equation
and
some
ratio
work
or
backwards
to
justify
your
explanation.
A LG E B R A
Example
Your
aunt
second
8
gives
you
10
birthday
and
so
a
Find
b
On
the
total
which
AUD
on,
amount
birthday
on
given
will
your
increasing
you
for
rst
by
your
have
birthday,
the
rst
been
same
13
given
a
20
AUD
amount
on
every
your
year.
birthdays.
1530
AUD
in
total?
The series increases by a common dierence, so it is arithmetic.
AUD
b
a
=
10,
d
=
10,
S
=
1530
n
Rearrange
using
⇒
the
and
solve,
quadratic
by
formula
Interpret
n
is
a
number
of
years,
n
=
-18
does
not
make
your
There
are
Dollar
–
17th
over
Australian
birthday,
20
AUD,
Practice
GDC.
the
ocially
just
written
Pound
in
total
amount
recognized
using
the
given
Dollar
dollar
is
1530
($)
in
context
The
of
the
problem.
–
replaced
1966.
solving
scheme
has
three
Level
1
investors.
Each
person
recruits
investors.
Each
Level
2
investor
recruits
four
Level
3
pyramid
scheme
four
is
2
in
Australian
Australia
A
pyramid
Level
solution
AUD.
currencies.
sign
your
6
Problem
A
your
sense.
the
On
1
or
or
Since
the
factorizing,
a
fraudulent
investors,
‘investment’
and
so
on.
scheme.
Some
members
a
b
Show
Write
that
there
down
a
are
48
formula
Level
for
3
the
number
of
Level
n
large
prots
d
Find
the
Show
number
that
over
of
investors
1 000 000 000
make
most
will
needed
to
investors
ll
eight
would
be
but
investors.
money.
c
may
investors.
lose
their
Pyramid
levels.
needed
to
ll
15
schemes
are
in
countries.
illegal
levels.
many
E8.1 Making it all add up
157
2
Competitors
at
various
Thenal
in
a
beanbag
distances
beanbag
from
is
race
the
placed
run
on
starting
40
m
a
straight
line:
from
20
the
track.
m,
24
starting
Beanbags
m,
28
m
are
and
so
placed
on.
In
each
problem,
consider
line.
your
whether
answer
sense
in
the
t ratS
context.
Give
answers
to
of
20
24
28
40
meters
Competitors
run
startingline.
They
tothe
3
starting
to
the
beanbag,
this
until
pick
all
the
it
up
and
beanbags
then
return
have
been
it
to
the
returned
line.
a
Find
the
number
b
Find
the
total
A
rst
repeat
ball
is
kicked
before
it
falls
of
beanbags
distance
into
and
the
hits
that
air
the
that
each
from
competitor
competitor
the
ground.
each
It
ground.
then
It
must
collect.
runs.
reaches
bounces
up
a
height
again,
of
6
several
m
times.
2
Each
bounce
the
is
height
of
the
previous
bounce.
3
)m(
thgieH
6
m
T ime
a
In
the
Write
b
4
Find
the
sixth
time.
European
bounce,
the
total
the
total
distance
standard
ball
travels
distance
it
paper
has
it
12
travelled
sizes
are
m
travels
A0,
(6
in
as
it
A1,
The
area
of
A1
paper
is
half
the
area
of
A0
The
area
of
A2
paper
is
half
the
area
of
A1
This
5
rst
down
(s)
relationship
dimensions
26
a
Find
the
area
b
Find
the
total
A
set
The
(3
of
block
inches),
Find
the
15 8
by
building
rst
and
total
continues
is
37
of
a
sheet
on,
of
A10
covered
blocks
50.8
length
the
smallest
up
returns
A2
and
6
second
and
to
the
so
m
down).
bounce.
ground
for
the
on.
paper.
paper.
ocial
size,
A10,
which
has
mm.
area
so
to
m
the
contains
mm
up
of
by
long
to
all
the
the
E8 Patterns
paper.
one
sheet
blocks
(2
of
inches),
last
block,
blocks
of
9
each
size
dierent
the
second
which
combined,
is
from
A0
to
A10.
lengths.
is
254
giving
76.2
mm
your
mm
long
long
(10
answer
inches).
in
mm.
your
an
appropriate
0
makes
given
degree
accuracy.
A LG E B R A
6
In
a
this
7
tness
by
2
a
Show
b
Find
c
Determine
every
rst
that
the
Hannah
20
How
or
day
does
that
the
5
she
he
km
in
runs.
can
and
sit-ups
rst
when
30
20
on
does
sit-ups
the
the
the
50
on
the
rst
day,
and
increases
day.
number
Her
total
sixth
sit-ups.
total
minutes.
Find
you
discuss
goal
amount
of
is
sit-ups
to
of
is
1100.
decrease
time
she
her
time
would
run
by
for
3%
in
the
tell
when
a
4
real-life
situation
generates
either
an
arithmetic
series?
Series
the
30
day
does
day.
runs.
geometric
D
Pedro
every
he
runs
time
Reect
In
program,
sit-ups
●
Can
●
What
and
you
arithmetic
fractions
evaluate
are
the
a
risks
sequence
u
series
of
=
that
making
8n
2,
goes
on
forever?
generalizations?
for
n
≥
1,
there
is
no
limit
to
the
size
n
that
else
n
could
can
take.
always
No
matter
name
a
how
larger
large
value
of
a
n,
value
and
of
n
hence
you
choose,
nd
a
someone
larger
value
for
u
n
Todescribe
(gets
larger
this
and
situation
larger,
mathematically,
without
limit),
u
you
also
say:
tends
‘as
to
n
tends
innity
to
innity
(also
gets
larger
n
and
larger,
without
Reect
discuss
●
Explain
why
●
Explain
what
4
ATL
and
+
6
+
8
+
this
10
In
all
terms
these
of
+
12
if
+
could
you
14
+
try
go
to
on
nd
forever:
the
4,
value
6,
of
8,
10,
this
12,
14,
…
series:
…
6
sequences
the
5
sequence
happens
Exploration
1
limit).’
sequence
n
≥
as
1.
n
For
tends
each,
to
determine
innity
what
(becomes
happens
innitely
to
the
Tip
large).
You
your
could
GDC
use
or
a
spreadsheet.
2
For
each
ratio,
r.
geometric
Describe
sequence
how
the
in
value
step
of
r
1,
nd
aects
the
the
value
of
behavior
the
of
common
the
sequence.
Continued
on
next
page
E8.1 Making it all add up
15 9
3
a
Show
that
the
areas
of
the
squares
1
and
rectangles
in
this
diagram
32
form
a
geometric
sequence
with
1
8
rst
term
1.
1
16
b
Explain
why
the
area
of
the
whole
1
diagram
represents
geometric
the
sum
of
2
a
series.
1
c
Use
the
sum of
4
Explain
diagram
the
series
how
diagram
to
you
below
to
1
determine
+
could
+
+
use
the
the
+
4
….
illustrate
that
1
2
2
Exploration
geometric
number
6
shows
series.
of
It
terms
that
is
an
even
you
can
nd
interesting
in
a
nite
the
sum
notion
amount
of
that
of
an
you
innite
can
add
decreasing
up
an
innite
time!
In
a
decreasing
geometric
series,
each
is
term
smaller
Example
than
previous
Find
S
=
the
10
+
value
8
Common
S
=
0.8S
=
0.2S
=
S
=
+
of
6.4
ratio
10
+
the
+
geometric
5.12
=
=
+
…
0.8
8
+
6.4
+
5.12
+
…
8
+
6.4
+
5.12
+
...
may
fraction.
by
the
10
=
have
50
used
Example
16 0
a
10
one.
series:
Multiply
You
the
9
similar
shows
method
how
E8 Patterns
to
to
do
convert
this.
a
recurring
decimal
into
a
common
ratio
and
subtract.
A LG E B R A
Example
Express
10
0.27
Let
as
a
fraction.
x
=
0.27
100x
=
27.27
Multiply
Then
by
100
willcancelout
100x
=
27.27272727…
x
=
0.27272727…
99x
=
x
=
-
2
Find
the
3
=
1000
b
S
=
10
+
9
c
S
=
20
+
12
innite
An
the
By
a
+
the
+
8.1
+
of
+
7.2
the
of
the
x
it
is

=
+
part
subtract.
125
+
4.32
+
series.
…
…
+
series
…
has
rst
term
has
second
36
and
second
term
12.
series.
series
term
25
and
third
term
5.
series.
each
as
a
b
geometric
series,
0.37
express
c
as
a
fraction:
0.207
d
0.148
+
0.4
solving
0.55
the
as
a
fraction
formula
sometimes
n
+
7.29
geometric
sum
geometric
250
0.2222…
Examining
S
500
considering
Express
why
each
geometric
Problem
5
of
+
sum
innite
Find
4
value
S
Find
repeating
you
7
a
An
the
27
Practice
1
so
when
for
the
possible
to
by
writing
sum
of
nd
it
the
the
as
x
=
terms
sum
of
0.55
of
an
a
+
0.0055
geometric
innite
+
0.000055
series
geometric
+
…
shows
series.
1
r
= a
when
r
>
1
or
r
<
-1.
n

1
r

n
If
r
>
1
or
r
<
-1
then
as
n
gets

numerator
of
the
very
n
large,
r
gets
very
large.
Therefore,
the
1
r
fraction
gets
very
large,
hence
S
gets
very
large.
n

 1
S
n
r
1


r
when
= a
1
<
r
<
1.
n

1
r

n
If
1
<
r
<
1
then
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n
r
large,
r
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the

fraction
gets
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and
closer
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1.
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S
gets
n

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r
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r
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1 61
You
can
nd
common
The
sum
ratio
to
Example
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r
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sum
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innity.
=
3
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|r|
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show
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a
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3
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1
8
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+
7
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2
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a
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c
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values
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possible
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term
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to
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series.
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a
sum
of
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geometric
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6
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its
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to
sum
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term
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+
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starting
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four
an
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fractal.
then
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formed
each
by
line
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Objective
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Make
a
Stage
the
shape.
number
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or
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as
mathematical
the
some
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all
to
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general
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consistent
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a
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eventual
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the
of
how
geometric
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number
●
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length
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total
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of
properties
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to:
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number
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of
of
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3
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investigate
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fractals,
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search
as
for
the Gosper
Island
or
online.
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E8.1 Making it all add up
4
16 3
Summary
An
arithmetic
arithmetic
A
series
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the
sum
of
the
terms
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sequence.
geometric
series
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geometric
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the
sum
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r
the
sum
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rst
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terms
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dierence
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rst
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where
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nth
term,
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practice
the
sum
of
each
9
series:
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arithmetic
The
a
1
+
b
34
3
+
+
5
31
+
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+
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16
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25
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+
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arithmetic
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eighth
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common
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fraction
equivalent
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begins
15
+
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b
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c
0.123
0.64
19
+
23
+
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arithmetic
series
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second
term
9
and
…
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and
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1024
11
2
series
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-2
a
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c
sum
term
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the
sum
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the
rst
104
value.
terms.
3
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value
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the
arithmetic
series
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12
begins
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+
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arithmetic
sum
4
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15
geometric
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series
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begins
sum
of
10
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the
8
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has
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geometric
eight
6
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terms.
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begins
the
series
8
total
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+
and
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of
dierence
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12
+
the
…
5
of
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of
series
ratio
has
1.25.
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term
the
8
sum
of
the
10
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fourth
terms
term
and
is
term
40,
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the
520.
common
dierence.
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geometric
sum
of
series
the
has
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six
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1995.
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and
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An
of
its
innite
a
sum
the
term.
geometric
series
has
third
term
128.
Find
the
sum
of
E8 Patterns
innity
common
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arithmetic
96
series
of
15.
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term
the
10,
value
of
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ratio.
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d.
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term
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a
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and
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and
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ten
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terms.
to
the
sum
term
rst
rst
16
geometric
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series
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rst
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seven
the
the
and
terms.
fourth
and
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geometric
common
A
11,
585.
the
and
8
term
is
and
sum
15
eight
rst
terms
terms.
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A
has
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dierence.
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the
value
7
rst
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term
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the
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the
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14
common
of
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5
series
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the
values
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d
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sum
630.
A LG E B R A
17
An
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seven
series
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x.
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term
sum
2x
of
+
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1
and
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emails.
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total
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that
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would
receive
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100
term
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3.
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sum
to
innity
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whole
year.
3
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the
value
of
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2
21
19
I
receive
$10
birthday,
12th
to
me
birthdays
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14
rst
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and
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total
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on
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to
week,
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third
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two
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so
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of
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repaired
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phones.
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down
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week
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A4
paper,
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b
as
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the
continues
sheet
each
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of
on
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total
an
number
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owners
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to
sell
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in
assume
follow
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total
in
assumption
nth
for
the
number
of
week.
of
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an
the
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d
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when
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are
terminals.
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would
to
number
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lead
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number
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paper.
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area.
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week.
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mouth.
number
forms
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pattern
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paper.
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cm
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of
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20th
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context
business
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my
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(inclusive).
managers
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birthday
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from
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geometric
sequence.
E8.1 Making it all add up
16 5
5
A
blog
gains
month,
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its
150 000
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followers
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and
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month.
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overall
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all
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of
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inquiry?
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as
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total
for
to
Edinburgh
one
number
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be
in
in
of
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warehouse.
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dimension
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Global
context:
Objectives
●
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three
●
the
formulae
and
using
coordinates
●
in
F
Using
distance,
in
three
Pythagoras
section
and
dimensions
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can
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the
angle
describe
space
in
three
you
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distances
and
angles
solids?
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does
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understanding
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dimensions
space
●
you
in
●
three
expression
questions
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in
trigonometry
cultural
dimensions?
●
midpoint
and
between
a
line
and
a
extend
to
3D
space?
plane
●
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a
can
given
●
Can
●
Is
a
you
divide
a
line
segment
in
ratio?
4th
dimension
exist?
D
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human
activity
random
or
calculated?
Critical-thinking
Analyse
parts
complex
and
concepts
synthesize
them
and
to
projects
create
new
into
their
constituent
understanding
7 3
9 1
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Statement of Inquiry:
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the
relationships
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and
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analysis
of
measurements
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167
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dimensions
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midpoint
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points
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Find
triangles
distance
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between
the
9)
2,
30)
midpoint
and
6)
(7,
and
the
answer
(7,
and
the
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b
●
5)
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dimensions
the
points:
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b
●
to:
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the
points:
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length
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of
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x,
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your
form.
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x
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●
use
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trigonometr y
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to
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angles
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nearest
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triangles
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cm
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world
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can
you
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example,
But
a
and
describe
cerealbox,
16 8
in
dimensions.
creating
Reect
●
space
●
has
a
they
of
discuss
cuboid,
or
and
computer
not
a
three
useful
new
dimensions?
angles
using
dimensions
for
3D
very
giving
or
in
solids?
coordinates,
are
program
building
to
useful
display
but
for
a
3D
real
something
instructions
using
the
describing
in
3D
an
object
shaped
would
like
you
a
cuboid,
give?
such
as
on
a
space,
printer.
1
measurements
E9 Space
in
dimensions
a
are
space
distances
Two
in
model
what
cm
a
G E O M E T R Y
To
fully
three
You
describe
objects
perpendicular
can
describe
such
as
cuboids
measurements:
positions
in
3D
in
width,
space
three
depth
using
x,
dimensions,
and
y
you
A N D
T R I G O N O M E T R Y
use
height.
and
z
coordinates.
Tip
The
x-axis,
y-axis
and
z-axis
are
perpendicular
to
each
other.
The
convention
is
y
to
z
draw
y-axes
the
as
if
horizontal
of
a
the
x
x-
on
the
base
cuboid,
z-axis
and
and
pointing
upward.
O
A
at
surface
y-axes
is
named
The
the
in
x
3D
y
space
plane,
is
and
called
the
a
plane.
planes
The
plane
including
the
containing
other
pairs
the
of
x-
axes
and
are
similarly.
cuboid
below
is
drawn
on
a
3D
coordinate
grid.
y
z
D
C
F
G
x
O
E
From
the
origin
O
y-direction,
and
To
point
nd
the
1
to
point
unit
in
with
A
the
is
0
units
in
z-direction.
coordinates
(3,
the
x-direction,
The
4,
1)
coordinates
start
at
the
0
units
of
A
origin
in
are
and
the
(0,
0,
1).
Tip
move:
Some
GDCs
dynamic
3
units
in
the
positive
x-direction
to
point
E
4
units
in
the
positive
y-direction
to
point
F
1
unit
software
3D
and
graphing
can
graphs.
draw
Practise
plotting some points.
in
the
Example
Two
points
Calculate
C
has
so
it
z-direction,
to
point
C
1
have
the
the
is
positive
coordinates
distance
same
x
vertically
and
C (3,
4,
7)
and
D (3,
4,
18).
CD
y
coordinates
above
as
D,
D
The
7
(
18)
=
distance
dierence
CD
=
25
move
With
a
between
C
and
their
z
D
is
the
coordinates.
units
Objective
iii.
between
25
C:
Communicating
between
some
diagram.
specically
dierent
questions
in
Sometimes
mention
forms
Practice
it
might
1
of
you
help
to
mathematical
will
need
draw
a
to
representation
translate
diagram,
written
even
if
the
coordinates
question
into
does
not
it.
E9.1 Another dimension
16 9
Practice
1
Write
1
the
(x,
y,
z)
coordinates
for
each
lettered
vertex
in
these
cuboids.
z
y
a
b
z
D
(4,
9,
1)
E
G
B
(0,
7,
–2)
H
A
x
F
x
O
C
(–5,
2
Find
the
distance
between
a
(1,
1,
4)
and
(1,
1,
11)
b
(3,
2,
0)
and
(9,
2,
0)
c
(
2,
11)
and
(
4,
The
4,
space
straight
diagonal
line
joining
of
these
11,
a
one
pairs
of
0,
O
–2)
points.
11)
cuboid
vertex
is
the
H
G
to
E
the
opposite
the
center
vertex,
passing
through
F
D
of
the
C
cuboid.
A
A
face
diagonal
one
vertex
one
of
the
to
is
1
box
faces
of
the
the
theorem
2
passing
joining
B
through
shape.
cm
triangle
to
line
AG is
the
space
BG
the
face
is
diagonal
diagonal
1
measures
Sketch
straight
another,
Exploration
This
a
nd
by
BCA.
the
3
cm
Use
length
by
6
cm.
G
Pythagoras’
of
CA.
Give
F
your
answer
in
exact
(surd)
form.
C
H
2
Sketch
the
triangle
GCA.
2
cm
B
E
Label
the
lengths
of
GC
and
CA
D
6
What
is
the
size
of
∠
cm
ACG ?
3
Explain
3
Use
4
Find
5
Another
the
box’s
A
you
Pythagoras’
Using
6
how
length
box
the
Find,
same
space
cuboid
in
is
theorem
BH.
1
to
What
cm
deep,
method
as
nd
do
4
the
you
cm
you
length
notice
wide
did
to
of
GA
about
and
nd
8
GA
cm
GA,
to
1
decimal
and
place.
BH ?
long.
nd
the
length
of
this
diagonal.
measures
terms
of
x
x
cm
and
by
y,
y
the
cm
by
length
z
cm.
of
one
of
its
face
diagonals.
Continued
17 0
cm
A
know.
E9 Space
on
next
page
G E O M E T R Y
Hence
show
that
the
space
diagonal
has
A N D
T R I G O N O M E T R Y
length
z
y
x
In
a
cuboid
with
dimensions
x,
y
and
z,
the
space
diagonal
has
length
The
is
space
also
volume
Reect
How
many
dierent
●
How
many
space
●
In
the
is
rst
the
Practice
2
Find
the
length
a
cereal
b
a
room
c
a
pencil
Find
a
the
box
box
that
of
cardboard
b
a
smartphone
c
a
storage
length
pencil
16
cm
Sets
of
known
4
If
a
3
9
cm
cuboid
cuboid
between
of
diagonal
cm
by
space
1,
G
have?
explain
and
have?
Are
why
ts
by
the
4
7
of
12
m
2
cm
50
by
cm
dimensions
13
space
an
7
by
m,
26
of
20
space
diagonal
cm
which
is
4
14
m
tall
cm.
each
by
are
inches
diagonal
exact
diagonally
2
integers
4
cm.
cuboid,
giving
your
answer
15
cm
123.8
by
of
13
a
by
22
mm
by
inches
cube
cm
58.6
by
18
whose
mm
by
7.6
mm
inches.
sides
are
5
cm.
form.
has
5
into
Find
which
three
5
length
form
2
+
integer
are
cuboid
called
12
shaped
of
the
sides
box
pencil
of
include
a
with
2,
2,
3}
and
2
+
23
{2,
correct
to
right-angled
{3,
4,
5}
and
3
s.f.
triangle
{5,
12,
are
13},
13
and
also
Pythagorean
14,
dimensions
2
=
sides
a
the
Examples
2
and
a
the
can
triples.
2
=
2
+
equal?
cuboids:
an
integer
space
quadruple.
2
2
the
all
A
these
cm
by
diagonal
measuring
in
by
numbers
include{1,
and
they
accuracy:
measuring
2
+
cuboid
thefour
of
2
Pythagorean
2
because
a
solving
cm
three
as
a
does
Exploration
space
whose
answer
just
by
the
box
box
Problem
A
in
does
measuring
is
degree
a
your
the
oor
length
the
diagonals
distance
measuring
a
Find
face
shown
of
with
suitable
Give
4
box
diagonal
2
diagonals
shortest
the
2
a
to
3
discuss
●
GA
1
and
diagonal
called
23,
27},
because
1
2
+
2
diagonal,
Examples
2
+
2
of
these
2
=
3
2
=
27
E9.1 Another dimension
171
Exploration
1
Points
D (19,
of
a
h,
C (4,
17,
a
7,
17)
cuboid
and
11)
are
of
depth
Write
2
and
two
width
d
as
down
D(19,
17,
17)
vertices
w,
height
h
shown.
the
values
d
of
w
w,h
b
2
and
Hence
Points
P
nd
(x
1
P
(x
2
,
y
2
,
depth
a
w
,
z
)
the
y
1
2
cuboid
C(4,
d
,
1
)
CD
and
(x
are
=
–
x
x
two
vertices
of
w,
height
.
Write
for
show
P
h
h,
The
distance
The
distance
Reect
two
d
that
and
the
P
(x
d
,
y
1
,
z
1
w
)
1
distance
is
.
2
formula
between
two
points
P
(x
and
discuss
dimensions,
the
,
y
1
,
1
z
)
and
does
Example
d
this
to
(x
2
,
y
2
,
2
z
)
is:
2
3
distance
between
(x
,
y
the
distance
formula
)
and
(x
1
in
,
2
three
y
)
is
given
by
2
dimensions?
2
distance
your
relate
P
1
1
the
)
2
h
1
Give
z
a
and
similar
and
1
Find
,
2
1
between
How
y
d
Hence
In
,
2
1
width
2
●
11)
2
of
expressions
b
z
distance
7,
answer
between
correct
points
to
3
(5,
7,
3)
signicant
and
(
4,
8,
2).
gures.
=
Substitute
the
values
into
the
formula.
=
Be
=
9.11
(3
17 2
s.f.)
E9 Space
careful
with
the
negative
numbers.
G E O M E T R Y
Practice
1
Find
2
the
a
(0,
c
(23,
11,
e
(
11,
a
Find
b
Use
3,
0)
and
18)
Three
Show
4
Show
In
this
plane.
CE
points
the
of
points,
11)
3,
OA,
19)
where
to
nd
coordinates
plane
an
O
is
the
the
giving
b
(5,
d
(
6,
f
(5.5,
4,
origin
distance
is
an
in
between
ABCD
is
O (0,
isosceles
coordinates
forms
ABCD
Theangle
and
(33,
forms
have
PQR
cuboid,
17,
method
have
OAB
that
pair
6)
(17,
and
each
your
8)
and
3,
2)
11,
and
from
answer
(6,
and
4)
A
is
the
8,
the
7,
form.
(
9)
4.5,
point
origin
exact
10)
(0,
and
in
to
1,
(4,
B (3,
13.5)
6,
12).
14,
18).
solving
points
Three
and
similar
that
3,
distance
Problem
3
between
(2,
16)
the
a
T R I G O N O M E T R Y
3
distance
0,
A N D
P (0,
isosceles
a
0,
0),
A (3,
6,
6)
and
B (4,
9,
1),
Q (1,
triangle.
5,
Find
9)
and
the
R (5,
area
7).
of
1,
2).
triangle
G
diagonal
labelled
PQR
F
horizontal
the
4,
triangle.
E
θ
H
B
θ
C
A
Example
3
D
Find
the
Give
angle
your
between
answer
to
the
the
diagonal
nearest
CE
and
the
plane
ABCD
in
this
cuboid.
degree.
F
G
E
H
2
cm
B
θ
C
3
A
4
cm
cm
D
2
AC
⇒
2
=
3
AC
=
2
+
5
4
=
25
Find
the
length
AC.
cm
E
2
Sketch
cm
the
triangle
ACE.
θ
C
A
5
tan θ
=
θ
=
tan
θ
≈
22°
cm
Use
the
tan
ratio.
1
=
21.801…°
E9.1 Another dimension
17 3
Practice
1
Find
Give
the
4
angle
your
between
answer
to
the
the
diagonal
nearest
FD
and
the
plane
ABCD
in
this
cuboid.
degree.
F
G
E
4
cm
H
B
C
θ
5
cm
A
12
2
a
In
cuboid
length
of
PQRSTUVW,
PR
is
149
cm
show
D
that
the
W
V
cm.
T
3
cm
U
b
Find
the
angle
between
VP
and
the
plane
S
R
PQRS
correct
to
the
nearest
degree.
7
cm
P
10
3
a
In
cuboid
the
ABCDEFGH,
diagonal
BH
and
nd
the
the
plane
angle
ABCD
Q
cm
between
to
1
H
G
d.p.
E
F
b
Hence
nd
the
angle
between
BH
and
the
plane
EFHG
8
cm
D
C
4
cm
A
4
4
In
this
triangular
right-angled
prism,
triangles.
BCE
The
and
other
AFD
faces
B
cm
A
are
B
are
2
rectangles.
E
F
a
Find
the
length
BC,
correct
b
Find
the
length
DB,
c
Find
the
angle
to
correct
to
3
s.f.
3
s.f.
10
CDEF,
d
Find
correct
angle
that
to
BCE,
DB
the
the
makes
nearest
angle
with
the
plane
degree.
between
D
planes
the
5
nearest
ABCDE
The
is
point
which
ABCD
is
a
is
the
Find
b
Hence
c
Find
EB
d
Find
the
CDEF,
correct
5
C
cm
to
degree.
square-based
E
a
and
8cm
right
directly
midpoint
of
the
pyramid,
above
the
with
point
AB
=
8
E
cm.
M,
base.
BD
nd
BM
D
C
M
angle
Giveyour
between
answer
to
EB
the
and
the
nearest
base
ABCD
degree.
A
8
1 74
E9 Space
cm
B
cm
cm
G E O M E T R Y
Problem
A N D
T R I G O N O M E T R Y
solving
3
6
The
volume
Tothe
nearest
longest
7
Micol
east
As
face
holds
and
4
soon
cuboid
decimal
the
string
due
the
lets
length
to
the
The
440
place,
cm
.
what
The
is
base
the
measures
length
of
the
5
cm
●
How
does
●
How
can
midpoint
a
of
go
of
kite
1
m
Rachel,
of
the
the
kite
above
who
kite
it
string,
is
ground
Conjecture
a
level.
holding
rises
11
m
assuming
the
He
kite
(x
2
2
Use
B
,
2
your
(19,
19,
cm.
vertically
that
it
stands
at
y
,
your
you
6
m
the
due
ground.
makes
a
straight
line
understanding
divide
a
line
of
2D
segment
space
in
a
extend
given
to
3D
space?
ratio?
3
of
the
similar
z
2
on
upward.
line
segment
joining
(x
,
y
formula
for
the
midpoint
)
and
(x
1
,
y
2
between
)
is
given
by
2
P
(x
1
P
8
space
1
and
by
cuboid’s
kite.
Dividing
Exploration
1
is
of
north
Rachel
Micol
C
ATL
a
diagonal?
m
as
Calculate
from
of
,
1
y
,
1
z
)
1
).
2
formula
to
nd
M,
the
midpoint
between
A (7,
11,
14)
and
38).
Continued
on
next
page
E9.1 Another dimension
175
3
Use
4
Explain
the
the
distance
why
knowing
midpoint
5
Find
the
6
Hence
7
Prove
formula
of
that
AM
=
that
BM
AM
is
=
not
BM
enough
to
show
that
M
is
AB
how
your
that
verify
AB
distance
explain
to
you
can
formula
be
from
certain
step
1
that
gives
M
is
the
the
midpoint
midpoint
of
of
AB
segment
P
P
1
Example
Find
M
M,
4
the
midpoint
of
A (3,
2,
9)
and
B
(1,
=
=
8,
The
3,
midpoint
given
by
Example
P,
Q
Q
is
and
the
the
mean
of
the
x,
y
and
z coordinate
values
in
A
and
B
3)
M
of
the
line
segment
joining
P
(x
1
is
3).
Find
(2,
2
R
,
1
y
,
1
z
)
and
P
1
(x
2
,
2
y
,
2
z
)
2
M
5
have
coordinates
midpoint
of
PR.
(
Find
p,
p,
3,
q
8),
and
(1.5,
q,
9),
and
(7,
2,
r)
respectively.
r
Use
the
midpoint
formula.
Therefore:
Form
equations
using
the
x,
y
and
z
⇒
coordinate
Practice
1
Find
the
5
midpoint
a
A(0,
0,
b
C (1,
c
E (
d
G (1,
e
I (
f
K (1.7,
1,
2,
and
1)
− 6,
0,
11,
17 6
0)
8)
32,
and
1)
and
12)
− 0.4,
of
each
B (4,
8,
D (25,
and
11.8)
of
points:
16)
11,
F (− 4,
H (8,
and
pair
10,
E9 Space
4,
23)
8)
J (− 4,
and
7)
7,
5)
L (2.3,
4.1,
12.1)
values.
Solve
for
p,
q
and
r
G E O M E T R Y
2
Two
M
3
points
is
the
Three
and
have
points
R (4a
+
Points
and
same.
5
A,
Q
is
P,
the
Find
a
both
The
c
b
<
h
line
2
:
ATL
and
g
<
midpoint
parts,
Two
3)
and
M (11,
coordinates
P (a
the
2,
b,
midpoint
coordinates
that
the
2c),
of
have
of
of
Q (a
PR.
+
in
:
b
a,
+
b
3,
3b)
and
c
PR.
The
P (a,
letters
is
used
of
3,
of
11),
AC
B (
and
on
the
nds
ratio
dierent
in
P
3
2,
11,
1),
midpoint
C (7,
of
BD
11,
are
7)
the
a
to
b,
i
integers
1
Q (d,
e,
represent
exactly
statements
c),
f
)
the
and
R( g,
integers
h,
i ).
from
1
to
9
once.
to
9
are
to
the
variables
a
to
i
such
that
satised:
how
the
1
:
point
1.
that
How
proportions,
would
you
divides
would
for
nd
a
you
line
nd
example
the
point
a
the
that
segment
point
ratio
1
divided
into
that
:
2
a
two
divided
or
the
line
a
ratio
segment
n ?
points
lies
0.5).
2,
Find
4
Tip
the
x-y
plane
have
coordinates
P
(x
1
P
3,
B
ABCD
coordinates
integer
A (1,
midpoint
quadrilateral
formula
m
have
assignment
Exploration
1
1,
the
e
into
ratio
is
mathematical
Generalizing,
the
D
each
thus
segment
3?
into
R
midpoint
<
Q
Show
the
and
these
<
3c).
and
19).
possible
of
a
equal
C
3,
Q
inclusive,
A (1,
Find
coordinates
3b,
Describe
Points
AB.
T R I G O N O M E T R Y
solving
B,
D (10,
of
have
3,
Problem
4
coordinates
midpoint
A N D
P
1
and
divides
it
in
the
ratio
m
:
,
y
1
)
and
P
1
(x
2
,
y
2
).
When
2
n
that
2
you
P
write
divides
P
3
in
the
P
1
ratio
m
:
2
n,
y
P
(x
2
,
y
2
)
the
2
so
n
order
matters
the
m
section
closer
to
P
and
the
1
n
section
is
closer
P
3
C
to
P
2
m
A
P
(x
1
a
Explain
why
triangles
AP
P
1
b
Find
the
scale
factor
that
,
y
1
B
)
1
,
3
BP
P
1
,
and
Hence
show
that
P
A
P
3
transforms
P
BP
1
c
CP
2
=
and
to
are
all
similar.
2
AP
2
P
1
derive
a
3
similar
expression
1
for
AP
3
d
Use
your
expressions
for
P
A
and
AP
1
P
are
given
by
to
show
that
the
coordinatesof
3
.
This
is
called
the
section
formula
3
Continued
on
next
page
E9.1 Another dimension
17 7
–
is
2
Points
3
A
and
a
Use
b
Verify
Verify
the
4
Two
have
section
that
that
matches
C
AB
when
the
points
coordinates
formula
=
nd
m
=
n
pointP
,
=
1,
the
coordinates
Suggest
which
lies
a
Use
UV
P
and
2UT
formula
the
(16,
=
ratio
30,
section
If
(x
1
P
,
y
1
has
3
3)
to
:
C
(10,
divides
16).
AC
in
the
ratio
2
:
1.
2
formula
in
two
dimensions
(x
,
y
1
,
z
1
formula
and
)
and
P
1
for
divides
(x
2
it
the
in
,
y
2
,
z
2
)
coordinates
the
ratio
in
three
2
m
:
of
the
n
2
nd
the
where
coordinates
U
and
V
respectively.
Use
the
of
have
T,
the
point
coordinates
distance
that
(1,
formula
5,
to
divides
8)
verify
that
3VT
The
P
P
1
your
in
which
section
P
suitable
on
3
5
and
formula.
1
dimensions.
B,
2)
2 BC
midpoint
have
to
A (4,
,
formula
z
1
)
and
P
1
(x
2
,
y
2
,
z
2
)
where
P
2
divides
P
3
P
1
coordinates
in
the
ratio
m
:
n,
then
2
.
3
Example
Point
and
T
U
6
divides
are
(4,
the
3,
line
8)
segment
and
(24,
SU
18,
in
21)
the
ratio
1
:
4.
respectively.
The
Find
coordinates
the
of
S
coordinates
of
T
T
=
=
Substitute
(8,
6,
Practice
1
2
3
Use
the
2.2)
6
section
formula
a
Divide
A (5,
b
Divide
C (− 4,
c
Divide
E (11,
Use
the
section
a
Divide
A (1,
b
Divide
C (3,
c
Divide
E (
Find
point
the
(4,
17 8
11)
6)
4)
and
2,
2,
and
and
7)
D (5,
F (1,
to
to
27)
0)
24)
B (25,
6)
and
D (15,
3)
the
and
of
a
E9 Space
(24,
line
the
the
the
9,
6,
4)
0)
1,
three
− 4,
1
ratio
line
in
in
2
in
fths
1
:
:
the
given
ratio.
in
the
given
ratio.
2.
:
3.
ratio
ratio
the
of
in
3.
segment
the
the
24)
42).
segment
ratio
ratio
each
F (12,
point
point
in
in
divide
and
8,
each
in
5)
coordinates
6,
divide
B (25,
formula
3,
to
1
3
:
ratio
the
:
2.
1.
4
way
:
3.
from
the
the
values
into
the
formula.
G E O M E T R Y
Problem
4
Three
P (2,
2b).
simplest
T R I G O N O M E T R Y
solving
points
21,
A N D
have
P
form
coordinates
divides
and
AB
hence
in
A (a,
the
nd
11,
ratio
the
b
m
values
+
:
1),
n.
of
B (11
Find
a
,
the
and
36,
3b
ratio
+
m
1)
:
and
n
in
its
b
D
●
Can
●
Is
a
4th
dimension
human
activity
exist?
random
or
calculated?
2
In
two
dimensions,
the
distance
formula
(
is
2
three
dimensions
it
(
is
x
-
x
2
1
x
-
x
2
1
)
2
2
)
(
+
y
-
y
2
1
in
two
dimensions
the
midpoint
+
(
z
-
z
2
1
x

dimensions
it
+
formula
What
ATL
if
in
two
you
and
of
1
the
you
form
Use
your
,
x
1
2
Use
3
,
y
1
your
a
(0,
0,
b
(2,
-
4,
c
(1,
3,
Use
,
0,
,
(w
1
your
4
x
,
y
1
Use
your
Use
your
6
Explain
What
could
In
book
his
life
in
one
we
a
it
have
(w
z
and
)
x
4,
(7,
3,
of
,
to
in
x
+
x
1
y
2
+
y
1
2

is
,
2
and
in

2
There
is
a
clear
link
between
the

to
contain
four
4,
another
letter,
w ?
dimensions,
with
coordinates
,
z
in
two
distance
and
three
between
).
distance
between
each
pair
of
points:
0)
12,
a
y
2
15)
midpoint
formula
,
z
2
nd
the
2
5,
,
formula
for
4)
the
x
distance
formula
the
formula
for
the
in
two
midpoint
and
three
between
).
2
M,
formula
your
answers
and
to
A
whose
the
to
midpoint
between
A (7,
4,
1,
3)
to
consider
Romance
they
misunderstood
we
nd
step
the
5
distances
support
AM,
the
BM
claim
and
that
AB
M
is
the
B
geometry
of
inhabitants
that
which
and
1).
distance
mean
in
2
1,
(w
y
0,
2
A
,
conjecture
and
2,
a
2
(7,
formula
of
the
nd
(1,
1
discover
dimensions
,
to
and
Flatland:
world
day
,
how
midpoint
and
to
3,
5
of
knowledge
1
andB (
and
8)
2)
dimensions
points
conjecture
formula
3,
,

2
2
formulas
2
0)
z
)
z).
1
0,
+
1
1
dimensions.
draw
y,
to
)
z
2
y
5
x,
z
1
y
2
these
knowledge
dimensions
(w
2
three
could
(w,
+
1
.
extended
Exploration
Imagine
y
2
is

formulae
x
1
2
)

three
y
2
)

Similarly,
(
+
are
the
actually
space
cannot
Many
believe
we
in
more
than
Dimensions,
they
part
live
in?
Edwin
occupy
of
a
3D
Could
three
a
2D
Abbott
space,
universe.
there
dimensions?
be
explores
but
then
Similarly,
other
could
geometric
perceive?
E9.1 Another dimension
17 9
Some
that
as
problems
we
inhabit
many
this
as
12
makes
Is
●
Explain
●
Show
that
test
square,
to
equal
Show
Do
to
maybe
nd
is
meant
you
can
you
the
time
been
three
the
goes
understood
dimensions
time
on
better
being,
–
we
scientists
imagining
sometimes
can’t
will
by
know
develop
using
whether
a
greater
it
we
4
by
the
points
apply
must
familiar
term
T (1,
6,
geometric
isosceles
5,
8),
shapes
in
4D?
triangle
U (5,
2,
4,
4)
and
V (1,
1,
0,
0)
square
is
that
A (0,
four
dimensions?
Is
it
sucient
and
0,
the
in
BD
0,
0),
points
length,
must
B (6,
A,
also
4,
B,
C
diagonals
2,
be
13),
and
AC
equal
C (2,
D
to
and
in
form
BD
a
must
length.
10,
15,
11)
and
square.
possible
can’t
for
equal
AC
a
in
sides?
be
and
form
is
a
equal
points
2)
think
dimensions
than
For
describe
identify
four
sides
13,
have
triangle.
midpoints,
6,
as
and
three
you
that
D (− 4,
more
problem.
discuss
the
four
physics
idea.
have
that
all
with
a
isosceles
could
simply
●
but
the
what
an
How
●
of
possible
form
have
space
and
●
One
theoretical
solve
sense,
Reect
●
a
to
understanding
it
in
that
we
could
be
living
in
a
world
with
see?
Summary
●
In
a
cuboid
with
dimensions
x,
y
and
z,
●
the
The
P
1
space
●
A
at
The
diagonal
surface
plane
has
in
3D
space
containing
plane,
pairs
of
and
axes
the
are
,
y
1
,
z
1
)
the
is
x-
called
and
a
1
planes
y-axes
named
including
the
is
distance
The
distance
(x
2
, y
2
Mixed
1
Find
, z
2
A
13
inch
to
50
a
between
of
cm
the
by
P
(x
,
y
1
space
15
reasonable
The
section
If
(x
P
1
,
z
1
is
joining
given
by
2
formula
,
y
1
,
z
1
P
)
and
P
1
P
(x
2
in
the
,
y
2
ratio
,
2
m
:
z
)
where
2
n,
P
3
then
2
P
has
3
)
coordinates
and
1
cm
diagonal
by
degree
22
of
of
cm.
a
shoebox
Give
your
by
storage
18
drumstick
box
inches.
would
E9 Space
measures
Determine
t
into
3
Find
the
these
midpoint
pairs
of
of,
and
distance
points:
accuracy.
solving
shaped
inches
18 0
2
)
formula
length
cuboid
20
segment
z
practice
the
by
2
,
) is
Problem
2
line
y
2
measuring
answer
2
,
the
similarly.
1
P
the
(x
other
1
The
of
P
M
plane.
divides
●
M
and
length
●
x-y
midpoint
(x
the
13
inches
whether
box.
a
a
(1,
1,
b
(5,
5,
c
(13,
3)
and
2)
2,
(3,
and
11)
2,
(6,
and
1,
(9,
9)
10)
2,
4)
between,
G E O M E T R Y
4
A
straight
path
coordinates
axes
are
runs
B (0,
3,
measured
directly
2)
in
and
from
C (15,
points
6,
8),
with
where
b
The
the
middle
of216
meters.
angle
any
a
Find
the
length
of
the
path
and
m
the
of
its
Find
the
ends
of
dierence
the
in
height
between
the
A
single
totwo
Determine
strand
from
of
a
spider
web
runs
in
a
one
opposite
shaped
Two
points
if
like
and
3
corner
corner
a
of
a
room
at
the
oor
Find
at
cuboid
meters
the
the
4
ceiling.
meters
The
wide
room
by
6
19,
ratio
Find
the
6
angle
tombs
famous
of
the
strand
of
a
Each
of
height
the
with
tag
in
One
To
one
of
C (150,
be
Give
at
base
m.
the
played
the
Two
web
and
B
The
of
the
42).
2
:
9
In
the
its
top
your
answer.
coordinates
A (13,
2,
7)
and
Find
the
point
that
divides
AB
in
solving
points
Find
the
most
Great
and
have
7).
divides
A
B (
10
the
or
B
is
1,
coordinates
its
any
divides
in
the
closer
have
z).
A
Show that
C (5,
A
UV
points
3,
and
angle
and
answer
1,
Two
the
− 4,
ratio
to
U (
the
all
B
5,
:
the
3.
11,
17)
1
:
and
2.
which
origin.
11
A (
form
ratio
Determine
A (3,
possible
are
points
3)
1
in
coordinates
Find
and
UV
an
values
units
13,
5,
2,
8)
for
and
z
apart.
4),
isosceles
B (11,
6,
triangle.
12)
Find
area.
of
accurate
it,
to
‘tag’
the
designers
to
simulate
representing
one
corner
corner
each
arenas
of
from
the
outdoors,
other
is
in
used
the
one
the
or
with
3D
rectangular
origin
is
With
was
the
prism,
at
creating
3D
space
video
to
characters.
creating
two
coordinates
while
In
one
to
17,
runs
programmers
game
to
game,
tunnels.
GH
17)
the
geometry
E (22,
tunnel
G (28,19,
games,
create
transformational
and
space,
foot.
When
use
Arizona,
a
2
a
Tunnel
16)
and
from
H (213,
objects
programmer
EF
has
F (215,
end
30,
and
move
end
28,
18),
coordinate
19).
20).
a
a
Justify
5.
giventhat
m,
pyramid
your
built
Giza.
230.4
indoors
largest
system
120,
similar.
places.
trying
plan
furthest
the
of
were
pharaohs.
145.5
sides.
of
unit
are
context
can
originOas
the
strand
Egypt
measures
base
decimal
coordinate
with
the
players
lasers.
USA.
the
pyramids
of
Giza
sloping
two
Laser
sides
of
in
and
three
measures
Review
1
queens
the
between
to
between
pyramids
the
Pyramid
pyramids
web.
of
for
are
two
meters
oor.
Square-based
as
places.
tall.
length
the
and
answer
is
V (7,
b
your
to
8
a
sides
the
pyramid
Give
decimal
these
have
Problem
long
the
base
Find
straight
the
the
of
sides.
m.
path.
B (34,
line
has
143
two
7
5
Giza
of
base
sloping
mathematically
the
at
height
midpoint.
c
b
a
T R I G O N O M E T R Y
the
accurate
coordinates
pyramid
and
between
of
A N D
arena,
a
refuge
tower
is
located
along
the
so
that
Find
two
the
dierence
between
the
lengths
of
the
tunnels.
3
is
of
the
way
line
joining
b
There
is
a
sign
at
the
midpoint
of
the
tunnel
4
O
and
the
b
C
Find
coordinates
of
the
top
entrances
of
tower.
The
company
lasers
away.
will
Find
Determine
the
travel
that
reliably
appropriate
c
the
for
of
the
the
the
targets
whether
use
longest
within
middle
makes
hit
within
distance
the
this
your
structure
arena
lasers
up
if
to
says
175
lasers
that
feet
c
the
sign.
An
access
of
one
and
H
corridor
tunnel
tunnel.
are
F
Find
to
the
Find
runs
the
the
coordinates
from
the
midpoint
length
of
this
of
midpoint
the
other
corridor.
arena.
laser
you
could
stood
in
the
oor.
E9.1 Another dimension
of
181
3
A
new
roller
straight
The
coaster
tunnel
roller
that
coaster
ride
is
passes
gains
designed
through
speed
as
to
a
it
have
natural
passes
a
long
a
hill.
tunnel
before
turning
sharply
as
it
starting
emerges.
control
center
is
situated
at
(0,0,
0),
shortest
and
distance
nishing
between
her
points.
Suggest
and
reasons
why
the
distance
she
travels
The
will
ride
the
through
b
the
Find
on
be
greater
than
this.
a
1
scale
of
A (15,
a
1
25,
Find
of
b
unit
26)
the
this
Find
If
total
tunnel
the
the
1
and
m,
the
leaves
to
the
tunnel
the
change
total
Problem
c
to
hill
in
enters
at
the
B (75,
height
hill
50,
from
c
at
14).
one
Given
that
her
the
minimum
the
slope.
average
time
speed
needed
is
for
20
her
ms
to
,
ski
nd
down
end
other.
length
of
this
tunnel
in
meters.
solving
tunnel
is
created
by
boring
a
cylindrical
2
hole
with
estimate
Explain
d
A
cross-sectional
the
why
refuge
two
a
volume
your
point
thirds
of
coordinates
is
the
of
earth
answer
to
be
way
this
of
area
is
built
from
refuge
to
an
in
A
of
9
be
m
,
excavated.
estimate.
the
to
tunnel,
B
Find
the
point.
Another
4
A
skier
travels
down
a
straight
ski
slope.
slope,
units
measured
in
meters,
her
starting
to
the
ski
lift
are
(4,
7,
2).
The
the
point
where
she
nishes
are
(
208,
355,
d
to
the
his
the
ski
Find
The
(
you
have
nish.
slope,
he
skier
370,
to
ski
the
statement
of
and
analysis
18 2
relationships
of
activities
between
for
E9 Space
ritual
inquiry?
measurements
and
play.
loses
leaves
75)
nd
his
and
ski
down
from
the
the
starting
of
the
way
control
and
drops
one
Give
Find
the
distance
f
Find
the
angle
from
the
café.
specic
enables
the
of
the
point
where
he
of
examples.
construction
his
skis
then
at
walks
a
café
back
at
up
point
to
the
pole.
e
Statement of Inquiry:
Generalizing
route
line
pole.
discuss
explored
same
Three-quarters
coordinates
the
male
240,
slope
How
the
straight
poles.
the
drops
and
a
78).
of
Reect
in
coordinates
down
of
travels
heading
coordinates
point
relative
skier
With
from
the
elevation
café
of
to
his
his
ski
ski
pole
pole.
Mapping
the
world
E9.2
Global
Objectives
●
Finding
the
context:
Orientation
Inquiry
area
of
a
non-right-angled
●
triangle
F
●
Using
the
sine
●
Using
the
cosine
How
Using
area
sine
can
you
nd
the
area
of
a
non-right-
triangle?
How
does
the
area
formula
help
you
nd
rule
of
a
segment
of
a
and
cosine
rules
to
area
of
other
shapes?
circle
●
the
time
solve
problems
How
is
the
sine
rule
related
to
the
area
of
C
involving
a
bearings
●
triangle?
How
is
the
cosine
Pythagorean
●
D
Apply
is
How
do
related
to
the
theorem?
more
thecosine
●
ATL
Which
rule
useful,
the
sine
rule
or
rule?
we
dene
‘where’
and
‘when’ ?
Transfer
skills
and
knowledge
in
unfamiliar
situations
E7 1
E9 2
E11 1
Statement of Inquiry:
Generalizing
in
space
can
and
help
applying
dene
relationships
‘where’
and
between
measurements
‘when’.
18 3
S P I H S N O I TA L E R
●
the
and
rule
the
Finding
space
questions
angled
●
●
in
Y
ou
●
should
nd
the
given
already
area
two
of
a
know
triangle
1
how
Find
perpendicular
the
to:
area
trigonometr y
right-angled
triangle.
b
3
cm
7
use
each
a
measurements
●
of
in
2
Find
triangles
x in
cm
each
6
cm
14
cm
triangle.
a
b
7
cm
x
8
cm
5
cm
62°
x
●
nd
of
a
the
area
of
a
sector
3
Find
the
area
of
this
sector.
circle
115°
6
●
use
radians
4
Find
x in
each
cm
sector.
a
2
Area
=
x
b
cm
14
1.7
cm
x
rad
8
5
Area
F
of
a
rad
cm
cm
triangle
A
●
How
can
you
nd
the
area
of
a
non-right-angled
convention
practice
●
When
label
with
How
working
vertices
a
with
and
capital
does
area
triangles,
the
letter,
the
interior
and
the
formula
the
you
convention
angle
side
help
at
each
opposite
is
nd
the
area
of
other
shapes?
with
Practice
the
same
letter
in
lower
case,
to
things
vertex
Here
are
easier
everybody
this:
it.
For
follows
example,
in
algebra,
the
convention
1
triangles
remaining
ABC,
DEF,
LMN
and
UVW.
Copy
each
vertices
and
triangle
sides.
F
c
d
A
E
b
L
u
M
v
18 4
is
to
a
and
variables
E9 Space
in
label
alphabetical
the
makes
if
c
write
1
usually
essential,
A
that
like
a
which,
although
not
b
vertex
is
triangle?
order.
G E O M E T R Y
Reect
Here
of
is
this
and
Paco’s
discuss
calculation
A N D
T R I G O N O M E T R Y
1
of
the
area
triangle:
5
1
cm
1
2
Area
=
×
base
×
height
=
×
2
5
×
10
=
25
cm
2
●
What
●
Will
●
What
mistake
has
he
10
made?
cm
2
It
would
the
true
other
be
answer
be
greater
information
better
if
Paco
or
might
had
less
help
than
you
remembered
25
to
the
cm
?
How
calculate
area
do
the
formula
you
true
know?
area?
as
1
Area
=
×
base
×
perpendicular
height.
2
In
mathematics,
‘height’
Exploration
1
A
triangle
a
Use
means
sides
13
trigonometry
of
the
cm
height’.
1
has
length
‘perpendicular
the
cm
to
and
work
dashed
line,
15
out
cm.
the
which
meets
13
b
15
Hence
side
nd
the
at
right
area
of
angles.
the
triangle.
47°
2
Triangle
AC
=
22
ABC
cm
a
Sketch
b
Hence
has
and
sides
BC
∠
ACB
triangle
=
=
7
cm,
42°
15
ABC
2
3
A
nd
triangle
By
the
has
area
sides
considering
the
a,
of
b
the
and
length
of
triangle
in
cm
c :
A
the
line
through
A
c
b
perpendicular
to
BC,
of
in
terms
the
and
triangle
the
size
of
angle
nd
of
a
formula
the
for
lengths
a
the
and
area
b
C
a
4
Use
your
when
BC
formula
=
10
cm,
from
AC
step
=
5
3
cm
to
show
and
that
∠
ACB
=
30°,
the
area
of
triangle
2
ABCis
12.5
cm
Tip
In
Exploration
nding
its
formula
1
length
to
solve
you
and
found
then
problems
the
area
deriving
than
to
of
a
a
triangle
formula.
draw
in
It
extra
by
is
drawing
usually
lines.
an
altitude
easier
to
use
and
the
An
to
a
Area
of
a
triangle
altitude
line
one
side
triangle
passes
three
E9.2 Mapping the world
of
that
the
vertex.
height
triangle
its
a
through
opposite
The
is
perpendicular
is
of
a
one
of
altitudes.
18 5
Reect
●
What
●
Where
●
How
Use
is
the
would
the
discuss
measures
Practice
1
and
do
you
angle
you
in
2
need
to
relation
change
the
use
to
this
the
formula
new
area
formula?
sides?
for
a
triangle
that
is
not
labelled
ABC ?
2
formula
to
nd
the
area
of
B
a
each
triangle.
Give
your
answers
to
b
3
s.f.
K
c
36
m
34°
8
cm
18
58.5
m
km
E
60°
72°
C
A
12
L
cm
J
34.8
2
Find
the
area
of
each
triangle
to
the
nearest
A
a
square
km
centimeter.
E
b
126
cm
D
140°
84
56
cm
124°
B
In
triangle
Sketch
4
F
C
48
3
cm
Here
the
is
a
cm
PQR,
PQ
triangle
=
34
and
triangular
cm,
PR
hence
=
nd
67
its
cm
and
area
∠RPQ
correct
to
=
3
120°.
s.f.
tile.
20
cm
120°
15
a
Find
the
area
of
the
cm
tile.
2
b
A
to
tiler
the
divides
nearest
i
Explain
what
ii
Explain
why
Problem
5
In
triangle
Sketch
6
A
10
large
the
The
of
paint
18 6
this
it
DEF,
d
=
triangle
is
the
area
calculation
likely
side
12
and
cm,
to
8
m.
how
Maria
on
many
e
=
hence
tetrahedron
thickness
Determine
by
of
one
tile
and
rounds
up
number.
be
represents.
an
underestimate.
solving
regular
triangle
m
whole
has
is
each
tins
E9 Space
18
cm,
nd
four
going
area
faces,
to
surface
of
its
∠
E
paint
49º
be
Maria
and
correct
each
paint
will
=
all
1
of
3
which
four
mm.
to
∠
D
faces
Paint
should
buy.
=
30º.
s.f.
is
of
is
an
equilateral
the
sold
tetrahedron.
in
1.5
liter
tins.
G E O M E T R Y
Reect
and
discuss
Triangle
A N D
T R I G O N O M E T R Y
3
2
130°
50°
12
ATL
●
Find
●
Explain
●
How
the
area
why
does
of
Exploration
In
this
areas
1
A
relate
some
2
to
cm
1.
has
the
the
same
area
trigonometry
as
of
Triangle
the
unit
1.
circle ?
2
Exploration,
of
12
Triangle
Triangle
this
cm
other
parallelogram
use
the
formula
shapes.
has
Give
sides
20
Area
to
answers
cm
and
correct
45
cm.
to
Its
3
help
you
signicant
acute
angle
is
nd
the
gures.
75°
B
C
20
cm
75°
A
D
45
2
a
Copy
b
Explain
c
Find
d
Hence
An
the
a
Write
b
Hence
a.
s
a
a
and
line
segment
BCD
are
BD
identical.
parallelogram.
side
value
for
length
of
the
formula
sin
area
to
a
60°
of
nd
an
the
equilateral
area
of
an
triangle
with
equilateral
triangle
show
hexagon
that
the
can
area
be
of
divided
a
into
regular
identical
hexagon
equilateral
with
side
by
logo
is
a
sector
OAB
of
a
circle
with
A
radius
4
cm
and
side
cm.
regular
given
company
4
the
ABD
the
has
exact
your
length
ABD
of
formula
Hence
is
area
the
draw
triangle
triangle
Use
how
triangles.
length
nd
of
the
down
side
Explain
A
nd
and
triangles
area
equilateral
with
4
why
the
length
3
diagram
cm
center
O,
a
portion
of
which
B
is
shaded.
4
a
Find
the
b
Explain
c
Hence
d
Find
area
why
of
the
∆OAB
sector
is
cm
60°
OAB
equilateral.
O
nd
the
the
area
area
of
the
of
∆OAB
shaded
segment.
E9.2 Mapping the world
18 7
The
area
familiar
how
to
formula
shapes.
divide
Area
of
a
for
The
a
triangle
key
shapes
can
results
into
be
are
simple
used
given
parts
parallelogram
with
below,
than
to
Area
of
other
but
results
it
is
more
memorize
an
to
nd
the
areas
important
to
of
learn
theseformulae.
equilateral
triangle
a
θ
s
b
Area
=
ab
Area
of
a
Degrees:
sin θ
Area
=
segment
Area
=
r
θ
r
Radians:
Practice
Give
1
Area
3
answers
Find
the
=
correct
shaded
to
3
areas.
signicant
The
a
gures
central
angle
unless
is
otherwise
given
in
directed.
degrees.
b
c
In
1b,
two
there
parts
to
are
the
220°
32°
63°
2
6.2
11
2
Find
the
total
cm
shaded
areas.
The
a
central
angle
cm
is
given
in
radians.
b
7
cm
cm
c
0.2
9
cm
2π
0.9
13
cm
3
Problem
3
Find
the
solving
area
of
each
shape.
a
b
7
c
cm
28°
9
cm
35°
35°
2
4
m
cm
62°
35°
8
cm
5
11
18 8
E9 Space
cm
m
area
involved.
G E O M E T R Y
A N D
T R I G O N O M E T R Y
2
4
A
segment
of
a
circle
has
radius
7
cm
and
area
2
cm
2
7
Find,
correct
to
two
decimal
places,
the
central
cm
2
cm
angle
?
of
the
segment
The
C
in
radians.
sine
and
●
How
is
the
sine
●
How
is
the
cosine
Exploration
cosine
rule
related
rule
rules
to
related
the
to
area
the
of
a
triangle?
Pythagorean
theorem?
3
A
c
b
a
You
have
already
1
Explain
2
Hence
3
Show
4
Use
why
show
the
that
triangle’s
that
similarly
your
shown
this
area
can
has
also
be
area
found
using
.
that
formula
triangle
.
from
step
3
to
show
that
cm
in
the
triangle
below.
C
16
cm
a
38°
24°
B
5
Show
6
Hence
A
that
nd
.
the
size
of
the
acute
angle
at
B
in
this
triangle:
A
12
9
47°
The
For
sine
a
rule
triangle
ABC
A
:
c
b
and
a
E9.2 Mapping the world
18 9
Example
Find
the
correct
1
value
to
3
of
a,
giving
signicant
your
answer
gures.
a
11
cm
60°
40°
b
=
11
a
cm
Label
the
triangle.
60°
40°
A
B
To
nd
sine
⇒
a
rule
missing
with
side,
sides
use
‘on
the
top’.
cm
Example
Find
the
nearest
2
size
of
∠B
correct
to
the
degree.
7
cm
13
cm
65°
B
b
=
7
cm
a
=
13
cm
65°
Label
the
triangle.
A
B
To
=
B
0.488…
=
arcsin (0.488…)
=
29°
(to
19 0
the
nearest
degree)
E9 Space
nd
a
missing
angle,
use
the
sine
rule
with
angles
‘on
top’.
G E O M E T R Y
Practice
1
Find
the
A N D
T R I G O N O M E T R Y
4
length
of
a
the
sides
marked
with
b
A
letters,
correct
B
to
3
s.f.
c
A
A
40°
5
cm
110°
63°
c
14
42°
cm
11
cm
C
33°
b
33°
B
B
C
a
C
2
For
each
triangle,
nd
the
B
a
14
size
of
b
cm
the
20
unknown
angle,
B
cm
to
the
nearest
degree.
c
B
17
cm
A
C
80°
C
70°
10
20
15
cm
cm
cm
68°
C
A
A
3
Find
the
value
a
of
5
x
for
each
triangle.
Give
each
answer
b
cm
correct
to
1
d.p.
c
x
58
x
7
x
12
cm
cm
11
cm
81°
38°
4
In
triangle
XYZ,
Problem
5
A
yacht
After
6
x
=
16
crosses
the
2.6km,
another
b
Calculate
the
size
c
Calculate
the
total
diagram
point
moving
on
sights
X,
a
a
Sketch
b
Determine
c
Calculate
inkm,
a
A
to
ship
bearing
the
a
∠
Y
=
of
the
show
44º
∠CBA
sights
of
the
to
and
∠
Z
=
72º.
a
is
race
B
due
of
the
by
on
Find
the
length
of
side
y
L
10km/h.
by
of
yacht
on
a
time
illustrate
this
information.
3
all
three
from
angles
point
signicant
Y
in
to
a
the
the
the
a
bearing
of
335º
of
until
026º.
it
yacht.
∠CAB
correct
bearing
to
of
minutes
this
of
on
bearing
bearing
Thirty
on
sails
a
C
size
the
lighthouse
at
and
followed
nd
sailed
C
sails
north
path
and
at
and
again,
distance
to
a
020º
of
buoy
the
distance
size
accurate
a
which
lighthouse
triangle
line
rounds
buoy
Sketch
and
a
starting
it
a
From
cm,
42°
solving
sailing
reaches
cm
37°
cm
of
3
s.f.
350º.
later
it
The
is
at
ship
is
point
Y
325º.
triangle.
lighthouse.
Give
your
answer
gures.
E9.2 Mapping the world
19 1
Reect
Explain
and
why
discuss
you
cannot
4
use
the
sine
rule
to
nd
length
x
in
this
triangle.
x
10
cm
120°
8
Exploration
1
a
Copy
4
triangle
angles
at
cm
ABC,
point
with
the
altitude
through
C
meeting
AB
at
right
P.
C
A
B
P
Label
the
b
Label
length
c
Use
show
Expand
d
Use
e
Hence
For
why
B
the
cosine
a
the
b
as
and
x.
it
c
Hence
theorem
label
to
length
write
two
PB
terms
of
for
c
and
x
CP.
.
and
to
simplify
express
x
in
this
equation
terms
of
A
as
and
far
as
possible.
b
that
is
possible
to
rewrite
the
formula
from
step
1e
so
that
it
as
formula
so
that
it
uses
angle
C
A
rule
triangle
in
expressions
that
brackets
show
angle
Rewrite
The
AP
trigonometry
Explain
uses
a,
Pythagoras’
Hence
2
sides
ABC
:
c
b
a
Reect
●
How
the
●
does
discuss
the
How
would
wanted
How
19 2
is
the
required
you
to
5
information
information
you
●
and
modify
use
cosine
the
required
to
the
use
E9 Space
rule
related
to
use
the
formula:
formula
cosine
rule
the
to
to
the
if
the
nd
cosine
Area
=
triangle
the
rule
is
ab sin C
labelled
measure
Pythagorean
compare
of
side
theorem?
to
?
MNP
and
p ?
Explain.
G E O M E T R Y
Example
Find
the
3
T R I G O N O M E T R Y
3
length
of
side
a
correct
4
to
A N D
signicant
gures.
cm
80°
6
cm
a
A
c
=
4
cm
80°
b
=
6
cm
B
Label
sides
and
vertices.
a
C
Use
⇒
a
=
43.665...
=
6.61
Example
Find
the
the
cm
s.f.)
Check
that
your
answer
cosine
seems
rule.
sensible.
4
value
nearest
(3
the
of
x
correct
to
degree.
132
65
cm
cm
x
98
cm
B
a
=
132
cm
Label sides and vertices.
c
=
65
cm
x
C
A
b
=
98
cm
Use
17 424
12 740
=
9604
+
4225
cos
x
=
−3595
cos
x
=
−0.28218...
x
=
arccos
=
106°
(
(to
12 740
cos
x
the
cosine
rule.
Rearrange.
0.28218...)
the
nearest
degree)
E9.2 Mapping the world
19 3
Reect
discuss
6
●
Show
●
Which
version
do
you
think
is
easier
to
use
to
nd
an
unknown
side?
●
Which
version
do
you
think
is
easier
to
use
to
nd
an
unknown
angle?
●
Does
that
it
Practice
1
and
Use
the
the
cosine
matter
rule
which
can
version
rearranged
to
use?
5
cosine
rule
to
nd
the
lengths
A
a
you
be
of
the
marked
B
b
sides
correct
to
1
d.p.
A
c
6
75°
4
30
B
25°
c
14
12
a
C
140°
C
35
b
A
B
2
Use
the
nearest
cosine
rule
to
nd
the
size
of
the
marked
angles
correct
to
the
degree.
A
a
B
b
D
c
13
B
12
15
14
27
21
20
C
5
C
A
18
3
4
Find
these
side
a
∆PQR,
p
=
14,
b
∆DEF
d
=
6,
c
∆XYZ,
x
=
7.2,
d
∆BCD,
b
=
25,
Two
for
boats
13km,
a
q
e
leave
the
=
=
z
c
=
=
the
b
Find
the
angle
c
Find
the
distance
triangle
f
=
5.4,
31,
on
Draw
A
R
d
same
other
diagram
and
11,
12,
a
Problem
5
lengths
a
angles,
=
98°.
9.
Y
=
34.
illustrating
their
between
y
for
8
on
a
bearing
km.
positions.
courses.
the
two
boats.
solving
has
sides
of
length
4
a
Find
the
size
of
the
largest
b
Find
the
size
of
the
smallest
19 4
their
s.f.
D
travels
135º
3
r
side
angle
One
of
to
E
Find
Find
harbor.
side
angle
121°.
bearing
between
Find
Find
=
accurate
E9 Space
cm,
angle
6
cm
in
angle
and
the
in
7
cm.
triangle.
the
triangle.
of
060º
C
G E O M E T R Y
Applying
D
You
can
●
Which
●
How
use
triangles.
the
To
the
Referring
do
to
the
sine
useful,
dene
and
are
rule
trying
triangle
rules
to
to
discuss
the
and
sine
‘where’
cosine
which
you
and
more
we
sine
decide
information
Reect
is
the
rule
and
to
use,
cosine
or
the
A N D
T R I G O N O M E T R Y
rules
cosine
rule?
‘when’?
nd
look
missing
at
the
lengths
and
information
angles
you
in
have
and
nd.
7
here:
9
7
●
Which
measurements
can
you
calculate?
38°
What
rule
●
What
information
●
What
By
should
parts,
so
the
use
only
sine
each
two
use?
rule:
b
to
leads
rule
be
you
and
For
shown
2
=
a
Explain
needs
one.
are
2
Cosine
you
information
examining
you
will
your
given
to
in
use
cosine
the
sine
selection.
order
the
rule
to
be
cosine
you
to
use
the
sine
rule?
rule?
carefully,
rule,
able
you
only
ever
can
use
work
two
out
of
when
the
three
here.
2
+
c
2bc
cosA
Sine
rule:
21
17
x
17
Tip
19
50°
If
an
angle
and
the
30
x
side
opposite
known,
The
cosine
rule
involves
three
sides
The
sine
rule
involves
two
one
angle.
and
two
rule
triangle
sides
so
and
use
the
problem
one
cosine
Example
Find
the
angle
involves
(the
three
This
unknown),
sides
rule.
can
be
angles.
used.
This
the
sides
sine
and
then
are
triangle
and
problem
two
unknown),
so
angles
use
involves
(one
the
sine
is
two
use
If
the
not,
then
cosine
rule.
the
rule.
5
size
of
the
angle
at
A
A
10
5
26°
B
C
The
C
⇒
A
=
arcsin(2 sin 26°)
=
61.25
=
180°
=
92.7°
−
26°
(3
−
61.25°
problem
involves
two
sides
and
two
angles,
Rearrange
The
angles
in
a
to
so
use
make
triangle
the
C
sine
the
add
up
rule.
subject.
to
180°
s.f.)
E9.2 Mapping the world
19 5
Practice
1
In
each
used
6
triangle,
to
nd
the
determine
value
of
x.
x
a
whether
Explain
the
cosine
how
you
rule
have
4
b
or
sine
made
rule
your
should
be
decision.
c
19
x
11
7
21
52°
x
12
10
2
d
e
47°
f
12
15
140°
x
x
47°
16
x
15
45°
8
2
In
each
triangle,
nd
the
value
of
x,
giving
a
your
62
b
7
answer
to
3s.f.
cm
cm
78°
11
correct
130°
cm
x
30°
x
c
d
4
172
88
cm
cm
cm
x
x
6
240
cm
cm
40°
78
e
cm
f
x
130°
45
cm
14
35°
cm
x
11
3
Use
the
Give
a
sine
rule
answers
17
or
cosine
correct
to
3
cm
rule
to
nd
signicant
the
marked
sides
and
cm
angles.
gures.
b
b
5
c
cm
57°
4
cm
110°
12
6
a
cm
cm
c
10
42°
d
8
cm
30
e
cm
f
11
cm
47°
d
9
e
cm
19
f
cm
57°
61°
19 6
E9 Space
cm
G E O M E T R Y
Problem
4
Find
the
A N D
T R I G O N O M E T R Y
solving
area
of
this
triangle.
2x
x
1
1
120°
x
5
Circles
of
radius
10
cm
and
6
cm
overlap
as
+
1
shown:
A
10
cm
6
cm
O
B
C
The
circles
meet
Find
b
Hence
nd
∠ABC
c
Hence
nd
the
handy
length
and
measurements
Objective
iv.
In
justify
the
that
ATL
points
a
One
the
at
D:
the
of
the
in
space
and
straight
C,
line
of
you
mathematics
degree
accuracy
inaccurac y
comment
in
the
segment
trigonometry
can’t
Applying
Activity,
∠AOC
=
0.8
rad.
joining
A
to
C
area.
use
that
of
and
radians.
shaded
benecial
in
A
on
the
original
of
is
measure
a
accurac y
in
to
help
you
calculate
directly.
real-life
contexts
solution
of
the
measurements
original
would
measurements
have
on
your
and
the
e ect
calculations.
Activity
1
Telephone cables from a house are connected to a telegraph pole C at the
roadside, but C is not yet connected to the network.
Continued
on
next
page
E9.2 Mapping the world
19 7
On
the
other
side
of
the
road,
two
other
telegraph
poles
A
and
B
are
A
30meters
theodolite
tool
An
engineer
uses
a
theodolite
to
measure
∠
CAB
=
27°
and
∠
ABC
=
used
maps
Copy
and
complete
this
aerial
view
of
the
street,
adding
the
to
a
measure
68°
angles
a
is
apart.
size
when
or
surveying
of
construction
angle
B
and
the
length
making
sites.
AB
C
b
Find
the
c
Hence
d
actually
Calculate
ii
Find
iii
Suggest
From
It
2
the
this
b
Calculate
c
Find
the
boats
other
the
ship
km
distance
leave
on
a
size
a
a
c
Find
the
distance
diagram
Alex
of
and
113º.
the
I
standing
and
C
AC
27°
should
A
B
in
the
measurements.
The
angle
at
A
Justify
your
the
to
your
original
engineer
connect
B
degree
calculated
should
to
of
bring
accuracy.
value
of
BC
more
cable
C
on
R
of
on
a
angle
from
to
one
145º
between
point
Q
bearing
on
of
a
bearing
of
039º.
072º.
diagram.
P
travels
for
18
their
their
between
to
a
PQR.
illustrating
Mary
Mary
distance
at
I
am
are
on
a
bearing
of
050º
for
14
km,
km.
positions.
courses.
the
two
boats.
is
sea-kayaking.
on
between
point
markers
R.
of
km
R
solving
Find
Q
and
at
BC.
in
why
value
5.5
harbor:
angle
P,
of
error
point
bearing
the
are
BC
error
length
sails
information
Find
am
an
reasons
to
b
there
true
calculated
Draw
bearing
4
discovers
a
Peter,
of
pole
B.
percentage
the
Problem
3
a
4.1
Show
the
pole
three
the
P,
sails
a
Two
to
lengths
telegraph
7
point
then
the
C
23°
i
Practice
angle
the
engineer
than
1
why
connected
The
is
of
calculate
Explain
be
size
O,
every
20
m
a
bearing
Peter
near
100
a
m.
away
of
and
can
from
210º
is
430
and
a
m
from
distance
Alex
of
310
on
m
a
from
Alex.
Mary.
major
I
Peter
road.
see
Along
three
marker
P.
the
road
consecutive
The
bearing
markers:
of
P
from
R
my
current
position
is
position
is
340°
and
the
bearing
of
Q
from
my
current
030°
Q
a
Find
the
angle
POQ
b
Find
the
bearing
c
Find
my
distance
from
Q
d
Find
my
distance
from
R
of
Q
from
P
P
O
19 8
E9 Space
G E O M E T R Y
A N D
T R I G O N O M E T R Y
Summary
When
working
opposite
with
vertex
A
a
and
triangle
so
ABC,
side
a
The
is
sine
rule
on.
A
and
c
b
a
Area
of
Find
cosine
rule
triangle
Mixed
1
The
practice
the
shaded
areas
correct
to
3
s.f.
3
Find
the
angles
a
areas
are
of
given
these
in
segments,
radians.
Give
whose
your
central
answers
b
correct
4
to
3
s.f.
cm
9
cm
110°
50°
a
7
b
cm
15
cm
2.1
c
d
0.3
7
11
cm
cm
50°
11
cm
4
Find
the
labelled
side
correct
to
3
s.f.
120°
a
7
cm
b
20
cm
42°
5
5
15°
130°
cm
34°
cm
8
e
d
cm
f
42°
c
7
cm
50°
19
40°
6
cm
9
73°
cm
7.2
cm
11
cm
cm
e
f
42°
8
cm
108°
110°
g
15
21
5
cm
24
cm
cm
cm
7
cm
3
cm
62°
5
2
Find
the
areas
of
these
segments
correct
to
3
Find
the
labelled
angle
in
degrees,
correct
3
s.f.
s.f.
a
9.6
8.2
80°
a
to
cm
cm
b
b
14
6
20
cm
cm
47°
cm
115°
57°
9
18
m
15.4
cm
d
18.2
6
cm
m
c
11
m
25.9
E9.2 Mapping the world
cm
19 9
cm
6
Find
the
labelled
angle
in
radians,
correct
to
3
s.f.
8
Two
One
14
friends
walks
start
50
walking
meters
due
from
the
north.
same
The
spot.
other
walks
cm
35
8.2
meters
on
a
bearing
of
057°
a
11
cm
14.7
Find
1.2
19
cm
the
correct
distance
to
3
between
their
nal
positions,
s.f.
b
Problem
11
solving
14
cm
9
c
Find
the
perimeter
of
this
triangle,
correct
to
d
7
3
s.f.,
given
that
2.1
18
15
cm
15
cm
2
Area
7
Find,
in
degrees,
the
size
of
the
smallest
angle
=
180
cm
in
x
a
triangle
with
sides
4
cm,
9
cm
and
11
cm.
27
ATL
Review
Many
maps
Surveyors
in
are
underpins
Use
the
these
created
map
the
on
questions.
sensible
context
measure
that
degree
map,
the
as
next
Give
of
by
triangulation;
distances
and
illustrated
page
your
to
this
bearings
you
when
a
in
this
here
help
answers
is
to
high
answer
correct
to
a
network
degree
of
b
a
It
is
9.5
km
from
of
076°.
bearing
of
130°
a
from
Using
three
their
20 0
the
Rolle
to
Saint-Prex
Thonon-les-Bains
from
Rolle,
and
on
of
Find
the
are
and
joined
use
Hawaiian
and
label
∠
STR
and
Hence
nd
lies
on
a
Prex
a
on
a
d
bearing
the
across
them
to
island
size
of
a
form
of
landscape.
the
skeleton
Oahu.
angles
∠
RST,
∠
TRS
letters
R,
S
sketch
positions
and
the
from
distances
Rolle
to
from
Rolle
to
Saint-
Thonon-les-Bains.
When standing in Saint-Prex, there is an angle of
100° between Lausanne and Thonon-les-Bains.
Saint-Prex.
towns,
points
accuracy.
bearing
of170°
of
accuracy
triangulation
c
1
cm
Lausanne lies on a bearing of 038° from Thonon-
and
T
triangle
relative
E9 Space
to
to
denote
RST
each
to
other.
les-Bains. Add this information to your diagram.
the
show
e
Hence
nd
Saint-Prex
the
and
distance
from
Lausanne
Thonon-les-Bains.
to
G E O M E T R Y
The
map
border
lake
below
and
itself,
places
on
is
shows
known
and
the
the
some
as
key
Lake
mountains
shores
of
the
towns
Geneva
that
lake
is
on
in
the
most
surround
banks
of
Lac
Leman,
English-speaking
it,
nding
which
countries.
point-to-point
lies
A N D
on
Because
distance
the
of
T R I G O N O M E T R Y
French-Swiss
the
presence
measurements
of
the
between
dicult.
Lausanne
Saint-Prex
Vevey
Rolle
Montreux
Evian-les-Bains
Saint-Gingolph
Thonon-les-Bains
Yvoire
Coppe
Chens-sur-Leman
Geneva
Problem
2
The
3
solving
distance
from
Evian-les-Bains
to
The
distance
from
Vevey
The
distance
from
Montreux
is
measured
to
be
from
between
Viewed
of
66°
Lausanne,
Evian-les-Bains
from
there
and
Lausanne
is
an
angle
of
and
there
isan
a
9.6
km.
lies
on
a
The
is
distance
8.6
km
Find
the
angle
on
from
a
Vevey
bearing
between
Saint-Gingolph
angle
6.5
km.
of
to
Saint-
200°
as
Montreux
viewed
from
and
Vevey.
Saint-Gingolph.
b
Evian-les-Bains
is
Saint-Gingolph
54°
Saint-Gingolph.
Evian-les-Bains,
between
to
16.9kilometers.
Gingolph
Viewed
Montreux
Saintis
Gingolph
to
bearing
of
200°
Hence
nd
the
bearing
of
Montreux
from
from
Vevey.
Lausanne.
c
a
Find
the
bearing
of
Saint-Gingolph
Find
the
bearing
of
Saint-Gingolph
from
from
Montreux.
Lausanne.
4
b
Find
the
distance
from
Lausanne
to
Nyon
is
7.0
km
from
Chens-sur-Leman.
Coppet
Evianis
6.1
km
from
Chens-sur-Leman.
As
viewed
les-Bains.
from
Reect
and
How
you
have
Chens-sur-Leman,
between
Nyon
Find
distance
the
and
there
is
an
angle
of
Coppet.
from
Nyon
to
Coppet.
discuss
explored
the
statement
of
inquiry?
Give
specic
examples.
Statement of Inquiry:
Generalizing
in
space
can
and
help
applying
dene
relationships
‘where’
and
between
measurements
‘when’.
E9.2 Mapping the world
2 01
84°
Time
for
a
change
E10.1
Global
context:
Objectives
●
Drawing
Orientation
Inquiry
graphs
of
logarithmic
●
functions
in
space
and
time
questions
What
does
a
logarithmic
function
F
looklike?
●
Finding
●
Justifying
the
inverse
of
an
exponential
function
●
algebraically
and
graphically
How
are
logarithmic
toexponential
a
logarithmic
S P I H S N O I TA L E R
exponential
function
function
and
are
the
Identifying
and
transformations
applying
on
inverses
of
What
are
the
properties
of
logarithmic
C
function
graphs
related
functions?
corresponding
mutual
●
●
functions
that
functions?
logarithmic
●
How
are
the
properties
of
logarithmic
functions
functions
related
to
those
of
exponential
functions?
●
How
do
you
transform
logarithmic
D
functions?
●
ATL
Is
change
measurable
and
predictable?
Critical-thinking
Draw
reasonable
conclusions
and
generalizations
E6 1
E10 1
E12 1
Statement
of
Generalizing
relationships
and
20 2
Inquiry:
changes
that
variability.
can
in
quantity
model
helps
duration,
establish
frequency
N U M B E R
Y
ou
should
already
know
how
to:
x
●
rewrite
exponential
logarithmic
statements
statements
and
as
1
vice
Write
y
=
5
as
a
logarithmic
statement.
versa
2
Write
y
=
log
x
as
an
exponential
7
statement.
●
nd
the
inverse
of
a
function
3
Find
a
the
f (x)
inverse
=
3x
+
of
the
functions:
2
2
b
●
describe
transformations
on
4
g (x)
=
x
Describe
1
the
transformations
on
x
exponential
functions
the
graph
of
y
=
3
that
give
the
x
graph
●
justify
algebraically
graphically
are
mutual
that
and
two
5
functions
y
of
Justify
= 2
×
3
-1
algebraically
graphically
that
the
and
function
5
inverses
f ( x )
=
is
its
own
inverse.
x
F
From exponential to logarithmic functions
●
What
●
How
Objective
ii.
In
in
B:
describe
to
are
Fold
1,
solve
an
of
the
will
functions
need
to
r ules
look
related
like?
to
exponential
functions?
nd
consistent
the
correct
with
rules
ndings
for
the
patterns
you
generate
problem.
1
piece
of
paper
thickness
of
the
folded
function
patter ns
general
given
size
the
as
you
the
A4
Assume
logarithmic
logarithmic
patter ns
Exploration
1
a
Investigating
Exploration
order
does
paper
is
0.1
in
half.
paper
is
You
0.05
now
mm.
have
The
two
total
layers.
thickness
mm.
A4
of
size
the
paper
standard
thickness
=
0.05
used
Fold
the
paper
in
half
layer
again,
3
Find
the
4
total
a
thickness
function
number
Write
terms
down
of
the
of
a
f
of
for
folds
the
the
most
around
world.
the
two
mm
In
and
Canada,
size
is
the
USA
the
‘Letter’
standard,
layers
and
write
down
the
number
of
layers
folded
paper.
number
of
Repeat
layers
of
once
the
8.5
by
and
11.0
the
in
countries
measuring
2
part
mm
0. 1
one
is
international
inches.
more.
folded
paper
in
terms
of
x
function
number
of
g
for
folds
the
total
thickness
of
the
folded
paper
in
x
Continued
on
next
page
E10.1 Time for a change
20 3
5
Use
a
6
function
10
Use
a
7
10
a
8
It
●
area
Do
you
In
folds
of
Why
the
think
times?
2002,
a
the
●
her
Britney
of
folded
nd
●
For
in
13
at
a
paper
1
an
layers
10
give
a
total
m
Moon
total
after :
give :
folds
the
a
A4
to
thickness
d
is
384
thickness
sheet
fold
this
to
fold
a
a
is?
number
American
roll
paper,
to
of
of
of
paper
sheet
Think
folds
larger
400
of
of
in
A4
about
of
the
piece
This
fold
shows
the
mm
it
14
that
is
1
km.
384
half
of :
km
Determine
400
is
paper
the
km.
seven
more
times.
than
relationship
of
paper.
of
paper
more
in
1200
thickness,
world
of
record
and
the
for
Massachusetts
paper
thick,
long
than
Gallivan,
could
be
folded
,
thickness
was
paper
Britney
m
formula
its
the
the
student,
paper
the
t
nd
held
of
0.058
to
a
school
times.
length
order
Exploration
the
Gallivan
the
the
developed
formula
students
to
give
think
possible
that
She
of
to
paper
1
you
16-year-old
length
Use
is
Earth
impossible
to
30
many
folded
folds
folds
c
folding
do
30
c
how
the
Explain.
demonstrated
12times.
from
paper
it
layers
needed
for
many
m
discuss
record
times.
seven
●
distance
of
c
how
determine
mathematically
the
20
1
thickness
folds
determine
to
of
and
world
is
g
total
20
b
average
seven
the
b
number
The
to
cm
Reect
●
f
layers
10
the
nd
b
function
The
to
folds
function
Use
g
0.058
that
was
how
is
the
paper
showed
mm
13
would
a
where
number
she
until
paper
Use
of
half
L
is
folds.
used.
folding
that
thick.
folded
long
n
paper
in
the
a
can
group
be
formula
to
times.
sheet
of
paper
need
to
be
times?
the
inverse
of
an
1.1
exponential
x
function
is
a
logarithmic
function.
For
example,
theequivalentlogarithmicstatementis  log
y
=
for
y
=
2
,
y
x
y
=
x
2
x
Interchanging
gives
the
the
inverse
x
and
y
function
in
y
the
=
logarithmic
log
expression
y
=
2
x
2
x
Graphing
y
=
2
and
y
=
log
x
together
veries
that
they
are
x
2
f
(x)
=
2
f
1
mutual
inverses,
since
their
graphs
are
reections
of
each
0
in
the
line
y
=
x
x
To
nd
the
equivalent
inverse
of
an
logarithmic
exponential
statement
function
log
y
=
x ,
y
=
then
a
,
use
the
interchange
the
a
x
and
y
to
obtain
y
=
log
x .
The
domain
of
the
exponential
So
the
domain
a
+
x
function
y
=
a
is
,
and
its
range
is

.
of
+
y
=
log
x
is

,
but
the
range
is
a
20 4

{0},
since
log
x
a
E10 Change
≠
0.
(x)
2
=
log
x
2
other
its
inverse
x
N U M B E R
The
inverse
function.
Practice
1
State
the
of
The
an
exponential
inverse
function
of
is
the
corresponding
is
and
logarithmic
vice
versa.
1
inverse
function
of
f
( x ).
x
f
a
(x )
=
4
b
f
(x )
=
log
x
c
f
(x )
=
ln x
6
2
3 lnx
x
f
d
(x )
=
3 log
x
f
e
(x ) =
f
f
(x ) =
(11
)
5
3
4
Problem
2
The
solving
population
of
seagulls
P
in
a
coastal
village
can
be
modelled
by
the
x
exponential
a
Draw
equation P ( x )
the
graph
of
P
,
= 15 × 1.15
(on
your
where
GDC)
and
x
is
in
state
P
months.
,
the
initial
seagull
0
population.
b
State
c
Draw
d
Using
reach
the
inverse
the
the
graph
function.
of
graphs,
the
nd
●
What
●
How
are
are
the
the
of
Exploration
1
Graph
long
it
on
would
the
same
take
the
logarithmic
properties
properties
ofexponential
ATL
how
function
for
axes.
the
population
to
20seagulls.
Proper ties
C
inverse
of
logarithmic
logarithmic
functions?
functions
related
to
those
functions?
2
function
Experiment
of
functions
with
and
dierent
values
add
of
a
a
>
slider
for
the
parameter
a
1.
1.1
y
4
3
2
f
(x)
=
log
1
x
a
1
0
x
1
2
3
4
5
6
7
8
9
10
1
2
a
2
3
0
10
4
2
a
State
what
happens
b
State
the
c
State
any
d
State
whether
domain
to
and
asymptotes
the
the
value
range
of
the
function
of
of
the
y
as
a
increases.
function.
function.
is
concave
up
or
concave
down.
Continued
on
next
page
E10.1 Time for a change
20 5
3
e
Does
f
Determine
g
Explain
Graph
the
the
x-intercept
if
why
the
x
remain
function
cannot
be
the
has
less
function
a
same
all
y-intercept.
than
for
for
or
equal
dierent
values
of
a ?
Explain.
Explain.
to
0.
values
of
a,
where
.
1.1
y
3
a
2
0
f
(x)
=
log
3
1
x
0.3
1
x
0
4
3
2
1
1
3
2
4
1
2
f
(x)
=
log
2
x
0.5
3
4
a
State
b
Repeat
Explain
alike,
5
what
steps
how
and
happens
b–g
graphs
how
Summarize
2
they
your
to
the
for
of
value
this
of
results
as
a
increases
in
the
interval
(0,
1).
graph.
logarithmic
are
y
functions
with
dierent
bases
are
dierent.
in
a
table
like
x
this:
y
=
x
a
y
=
log
a
domain
range
equation
of
asymptote
intercept(s)
Properties
of
graphs
of
logarithmic
functions
y
y
y
=
log
x
;
when
a
>
1
y
=
x
x
A
logarithmic
●
domain
All
graphs
x-intercept
function
and
is
its
0)
and
of
the
range
logarithmic
(1,
when
0
<
form
is
<
1
functions
vertical
,
where
.
.
of
the
asymptote
x
form
=
have
0.
Continued
20 6
a
1
is
of
;
x
0
1
Its
x
x
0
●
log
E10 Change
on
next
page
N U M B E R
●
The
shape
graph
is
concave
●
The
Find
the
domain
the
graph
is
concave
down
and
up
value
Example
of
and
of
a
determined
by
increasing.
the
For
base
0
<
a.
a
<
For
1,
a
the
>
1,
the
graph
is
decreasing.
controls
how
rapidly
the
graph
is
increasing
or
decreasing.
1
inverse
and
of
range
and
of
the
function
verify
and
its
your
result
graphically.
State
the
inverse.
Interchange
Rewrite
as
a
the
x
and
logarithmic
y
variables.
statement.
1
Graph f
and f
and
conrm
that f and
Graphically:
1
f
are
reections
of
each
other
in y = x.
1
Since
in
f
the
and
line
f
y
are
=
x,
reections
they
are
of
each
mutual
other
1.1
inverses.
y
Domain
of
f
Domain
of
f
range
=
of
f
f
(x)
=
log
2
x
+
3
2
1
=
;
range
of
.
x
f
(x)
=
3
2
1
x
y
Reections
every
type
echoes
are
and
of
and
sonic
observed
light,
which
iii.
In
move
C:
with
of
can
many
are
properties
think
of.
studied
types
produces
reection
of
in
well
known
study
acoustics;
and
electromagnetic
beautiful
can
are
Geologists
also
and
mean
x
the
act
of
wave,
interesting
of
for
just
seismic
about
waves;
course,
reections
including
images.
peaceful
visible
Perhaps
not
meditation.
Communicating
between
Exploration
forms
physical
you
eects
often
coincidentally,
Objective
their
wave
=
3
dierent
you
logarithmic
will
forms
move
of
between
mathematical
graphs,
tables
representation
and
equations,
and
dierent
functions
E10.1 Time for a change
207
Exploration
1
Here
is
values.
the
3
graph
Explain
of
a
how
logarithmic
you
can
nd
function
a
using
,
the
table
of
with
its
table
of
values.
1.1
1.
0.
y
2.
0.63093
3.
1.
4.
1.26186
5.
1.46497
6.
1.63093
7.
1.77124
8.
1.89279
9.
2.
10.
2.0959
1
0
2
x
1
1
2
3
4
5
6
7
8
9
10
1
2
2
Copy
the
3
Identify
allow
4
the
graph
function.
Use
the
you
what
from
Draw
step
the
coordinates
to
nd
you
a
in
learned
1.
Explain
graph
on
of
the
its
how
you
inverse
graph
of
can
draw
the
inverse
of
inverse
that
function.
the
function
and
its
both.
in
steps
1–3
to
nd
a
in
this
graph
of
.
1.1
y
4
2.
.5.
3.
1.
4.
1.58496
5.
2.
6.
2.32193
7.
2.58496
8.
2.80735
9.
3.
2
0
x
2
1
1
3
4
5
6
7
8
9
10
2
4
5
10.
Approximate
.
the
State
value
what
of
you
a
from
think
this
the
3.16993
graph
exact
and
value
table
of
a
of
values
might
be.
for
Explain.
1.1
0.
1.
y
4
2.
0.693147
3.
1.09861
4.
1.38629
5.
1.60944
6.
1.79176
7.
1.94591
8.
2.07944
9.
2.19722
2
0
x
2
1
1
2
3
4
5
6
7
8
9
10
2
4
10.
6
Explain
function
20 8
an
easy
way
to
determine
from
its
E10 Change
graph
the
or
2.30259
unknown
table
of
base
values.
of
a
logarithmic
N U M B E R
The
graph
Practice
1
Match
a
f
of
=
through
the
point
(a,
1).
2
each
( x )
passes
graph
log
to
its
function.
x
b
f
( x )
=
log
2
x
c
The
C
y
4
3
3
3
2
2
2
1
1
1
0
x
2
5
x
1
1
2
0
5
2
5
1
1
2
2
2
3
3
3
4
4
4
shows
function
y
the
year.
=
log
height
The
in
height
meters
data
of
can
a
be
silver
maple
tree,
approximately
measured
modelled
by
at
x
1
1
each
x
B
4
1
of
log
4
table
end
=
y
0
2
( x )
0.3
A
y
2
f
5
the
the
x
a
Age
of
tree
1
2
3
4
5
6
7
8
9
10
0.1
3.5
6
7.5
9
10
10.5
11.5
12
12.5
x (years)
Height
y (meters)
a
Plot
b
Find
c
Conrm
you
3
Each
these
the
inverse
on
of
in
a
graph
this
graphically
found
graph
values
and
function,
that
the
estimate
and
the
state
function
in
b
its
is
value
of
domain
the
a
and
inverse
of
range.
the
function
a
below
represents
a
function
of
the
form
f
( x )
=
log
x.
a
For
each
graph:
i
state
the
natural
ii
state
the
base
of
domain
the
and
the
logarithmic
range
1
0
x
1
1
2
3
4
5
function
y
2
1
the
function.
y
2
of
2
3
4
5
2
1
1
0
2
y
2
1
x
0
2
x
1
1
1
1
2
2
3
3
4
4
5
5
E10.1 Time for a change
2
3
4
20 9
5
4
For
each
i
state
ii
nd
function
the
the
natural
inverse
function
iii
in
domain
of
order
the
for
conrm
graphically
original
function.
x
a
f
(x )
f
d
=
(x )
+
=
Explain
and
2 −
range
function,
the
that
the
of
and
inverse
to
the
function
state
be
function
a
any
domain
restrictions
on
the
function
you
found
is
the
inverse
of
the
2
e
Problem
5
below:
log (3 x
− 2)
b
f
(x )
=
e
f
(x )
=
3 log ( x
+ 2)
f
c
(x )
=
4 log (1 + 2 x )
ln( 2 x ) − 4
solving
how
logarithmic
the
change
function
of
of
base
the
property
form
y
=
log
of
x
logarithms
can
be
ensures
expressed
in
that
the
every
form
a
y
6
=
For
b ln x
each
description
below,
write
down
a
function
with
the
given
characteristics.
+
a
A
logarithmic
b
An
function,
passing
through
(1,
0)
and
(4,
2),
domain
=

1
exponential
function
whose
inverse
is
f
( x )
=
2 log
x
+
7.
5

c
A
logarithmic
function,
passing
through
(5,
1),
(1,
0)
Transformations
ATL
●
How
●
Is
do
change
Exploration
1
2
State
the
a
f
(
e
f
( x
you
measurable
eect
of
each
b
h )
a
logarithmic
logarithmic
and

25
functions
functions?
predictable?
4
x )
Using
transform
of

, 2

D
1
and
f
graphing
the
same
eect
the
transformed
transformation
f
f
( x )
tool,
on
the
( x )
+
c
k
verify
graph
on
af
the
graph
of
f ( x ).
( x )
d
f
(ax )
g
that
of
y
the
=
transformations
log
x.
Draw
a
in
graph
step
of
f
1
( x )
have
and
4
Summarize
3
Conrm
graph
your
your
to
results
results
demonstrate
in
with
a
what
you
have
discovered.
table.
another
function
of
the
form
y
=
log
x
a
4
For
each
aects
21 0
of
the
transformations
(domain,
range,
in
x-intercept,
E10 Change
step
1,
indicate
asymptote)
which
and
how.
property
it
N U M B E R
Transformations
of
exponential
functions
Reection:
For
the
logarithmic
●
the
graph
of
●
the
graph
of
y
function
=
:
is
the
reection
of
y
=
is
the
reection
of
y
=
in
in
the
x-axis.
the
y-axis.
Translation:
For
the
logarithmic
●
function
:
translates
by
k
units
in
the
When
k
>
0,
the
graph
moves
in
the
positive
When
k
<
0,
the
graph
moves
in
the
negative
●
translates
by
h
units
in
y-direction.
y-direction
y-direction
the
When
h
>
0,
the
graph
moves
in
the
positive
When
h
<
0,
the
graph
moves
in
the
negative
(up).
(down).
x-direction.
x-direction
(to
x-direction
the
(to
right).
the
left).
Dilation:
For
the
logarithmic
●
function
is
a
vertical
is
a
horizontal
:
dilation
of
scale
factor
a,
parallel
to
the
y-axis.
●
dilation
of
,
scale
factor
,
parallel
to
the
x-axis.
af ( x )
f (ax )
y
y
f (x)
=
4log x
f (x)
f (x)
=
=
log(3x)
log x
f (x)
x
0
vertical
dilation
scale
factor
y
log x
x
horizontal
4
=
0
dilation
scale
factor
y
f (x)
=
log x
f (x)
x
0
=
log x
x
0
1
f (x)
=
log
x
1
4
f (x)
=
log
3
vertical
dilation
scale
factor
horizontal
dilation
scale
factor
3
E10.1 Time for a change
21 1
Example
Describe
2
the
transformations
on
to
to
in
the
negative
is
a
translation
factor
Practice
1
f
(x )
f
the
=
(x )
is
transformations
log
x
;
g (x )
=  log
f
=
(x )
log
x
;
g (x )
=
log
f
(x )
=
(x
f
(x )
outward
from
the
variable.
a
vertical
dilation,
=
to
( x )
to
get
g ( x ).
- 3
;
g (x )
=  log
(6 x ) - 4
2
x
;
g (x )
=
- 4 log
(x
+ 2) + 8
5
2 log
(x
+ 1) - 5
;
g (x )
=  6 log
3
Problem
f
- 7) + 1
x
5
e
Work
8
x
log
applied
=  2 log
2
d
units
4
8
c
4
3.
4
b
of
3
Describe
a
.
x-direction.
to
scale
get
(x
- 4) + 2
3
solving
2
f
f
(x )
=
log
x
;
g (x ) =
log
(3 x
+ 12) - 10
7
7
3
2
Describe
a
g (x )
c
j (x )
the
=
=
e
m( x )
g
p( x )
transformations
ln ( x
applied
to
- 3) + 2
ln ( 2 x ) - 1
=
-3 ln ( x
+ 1)
f
(x )
=
ln x
to
give:
b
h( x )
=
0.5 ln x
- 1
d
k (x )
=
2 ln x
f
n(x )
=
ln (0.5 x ) + 2
+ 1
1
=
-2 ln(5 x ) - 3
q (x ) =
h
ln(2 x
- 6) + 4
5
3
Each
For
graph
each
function
shows
one,
in
an
original
describe
terms
of
the
function
f
( x ),
labelled
transformation(s).
Hence
A,
and
write
transformed
down
the
function,
equation
of
labelled
the
B.
transformed
f
y
y
a
y
b
c
1.5
2.5
4
B
A
1
2
3
B
1.5
0.5
A
2
1
0
4
2
2
0.5
1
x
4
6
A
0.5
0
1
0
x
2
1
x
2
2
4
4
1.5
0.5
2
B
1
2
1.5
2.5
3
4
Problem
solving

4
Draw
the
graphs
of
f
(x )
=
lnx
and
g (x ) =
ln
1

2
x
2

Describe
5
Draw
the
the
transformations
graphs
of
f
(x )
=
on
log
x
the
and
graph
h( x )
of
=
5
of
axes.
21 2
Describe
the
transformations
E10 Change
f
 
on
the
same
give
the
graph
-

that
- log
set
of
axes.


[5( x
+ 2 )]
on
the
of
g
same
set
5
on
the
graph
of
f
that
give
the
graph
of
h
4
N U M B E R
Summary
A
logarithmic
function
is
of
the
form
+
where
range

is
●
,
the
graph
of
y
=
is
the
reection
of
+
,
.

,
Its
since
domain
x
is
cannot

and
take
the
its
y
value
=
in
the
y-axis.
0.
Translation:
The
inverse
of
the
inverse
of
an
exponential
For
function
is
the
corresponding
logarithmic
the
logarithmic
●
The
inverse
of
is
translates
When
of
logarithmic
All
graphs
of
logarithmic
functions
of
the
k
>
0,
asymptote
x
The
of
x-intercept
(1,
0)
and
=
shape
and
a.
For
the
is
>
graph
up
value
of
a
graph
point
(a,
the
For
and
increasing
The
1,
increasing.
concave
The
a
is
determined
by
graph
0
<
a
<
is
1,
concave
k
the
graph
When
is
<
0,
how
rapidly
the
When
graph
of
h
Match
a
y
=
>
0,
h
<
passes
through
the
(to
0,
x-direction
moves
in
the
negative
by
h
units
in
the
the
(to
graph
in
the
positive
in
the
negative
right).
graph
the
moves
moves
left).
Dilation:
the
of
logarithmic
the
logarithmic
is
factor
function
of
in
is
●
:
the
the
function
:
functions
reection
b,
a
vertical
parallel
is
scale
1
graph
1).
graph
Mixed
the
(down).
x-direction
decreasing.
logarithmic
=
positive
x-direction.
●
y
the
down
Reection:
the
in
translates
decreasing.
controls
or
Transformations
●
moves
the
For
the
graph
0.
the
base
the
vertical
●
For
in
(up).
y-direction
●
units
form
When
have
●
k
functions
y-direction
●
by
y-direction.
versa.
Graphs
●
:
and
the
vice
function
function.
factor
a
to
dilation
the
horizontal
parallel
of
scale
y-axis.
to
dilation
the
of
,
x-axis.
of
x-axis.
practice
the
log
function
with
its
graph:
x
b
5
y
=
log
(− x )
c
5
0
=
log
x
0.5
A
B
C
y
y
y
1
y
x
0
x
1
E10.1 Time for a change
0
x
1
213
2
While
training
average
speed
for
for
a
a
race,
10
Joseph
km
run.
tracks
The
his
4
data,
Describe
change
the
the
transformation(s)
graph
of
f
(x )
=
necessary
log
x
into
the
to
graph
a
shown
in
the
table
modelled
by
where
speed
the
below,
function
can
s
=
be
log
approximately
of
w
the
+ 14,
the
given
same.
function,
Where
assuming
more
than
the
one
base
remains
transformation
a
s
=
and
w
=
week
of
training.
is
needed,
correct
Week
Speed
make
sure
you
write
them
a
order.
y
=
log
x
+ 2
b
y
=
log
5
14
2
14.88
c
y
=

e
y
=
g
15.76
y
=
- 2) - 3
x
d
y
=
ln
(- x ) + 2
f
y
=
ln
(2 -
h
y
=
2 log
16.04
6
16.27
7
16.47
8
16.64
i
y
=
-2 log
(- x )
Sketch
g (x )
9
=
the
-7
10
16.92
Explain
for
- 2) - 1
x ) + 2
solving
graphs
ln
(3 x
of
f
(x )
- 12 ) + 1
=
on
ln x
the
and
same
set
of
axes.
16.79
Describe
a
(x
8
-2 l
 n
(1 -
Problem
5
x )

4
3
5
+ 2)

log

(x
7
ln
(x
15.39
4
the
(km/h)
1
3
in
this
why
kind
a
f
logarithmic
of
model
is
a
good
t
6
to
give
Sketch
the
the
the
transformations
graph
of
g
graphs
of
f
(x )
=
on
log
the
x
graph
of
and
1
data.
g (x ) =
log (2 - x )
on
the
same
set
of
axes.
4
b
Plot
these
values
on
a
graph
and
estimate
the
Describe
value
of
give
c
Find
the
inverse
of
this
function.
State
that
the
inverse
function
the
each
function
in
parts
a
to
of
on
The
data
in
the
the
table
form
y
here
=
a
follow
+
of
f
to
the
natural
domain
an
exponential
k
e:
1
state
graph
x
of
x
i
the
g
represents.
model
For
transformations
graph
the
7
relationship
3
the
a
and
range
of
3
0
1
2
the
2
function
y
ii
nd
any
in
the
inverse
domain
order
for
of
the
given
restrictions
the
inverse
on
to
function,
the
be
a
and
original
4
conrm
found
graphically
is
the
a
y
=
ln ( x
b
y
=
5(7
c
y
inverse
that
of
the
the
- 3)
0.2 x
) + 2
1
=
log
(4 x
- 8) - 5
2
2
log
(x
+ 1)
2
d
y
=
log
3
2
1
e
y
21 4
=
x
4
E10 Change
function
11
state
function
original
7
a
Find
the
inverse
b
Describe
of
this
exponential
function.
function
inverse
iii
5
you
function.
the
dierent
function.
ways
you
can
nd
this
N U M B E R
Review
In
E6.1
of
some
here
in
you
explored
natural
explore
considering
context
the
the
the
logarithmic
phenomena.
graphs
of
The
these
models
on
their
The
by
r
graphs.
=
Richter
pH
is
calculated
using
the
logarithmic
of
energy
can
0.67 log
the
1
amount
earthquake
phenomena
transformations
3
questions
E
be
- 7.6 ,
scale,
and
earthquake
in
released
modelled
where
E
is
the
r
during
by
is
the
the
energy
an
function
number
on
released
by
the
ergs.
formula
+
pH  =  - log
(H
)
a
By
comparing
this
model
to
the
standard
10
logarithmic
Using
transformations
of
functions,
dierence
standard
2
The
in
graphs
logarithmic
formula
for
of
this
model
modelling
y
model
=
the
log
f
(x )
=
log
x,
describe,
describe
using
the
function
and
transformations
of
functions:
the
i
x .
the
eect
the
graph
of
the
of
scale
f ( x ),
factor
and
the
0.67
name
on
of
this
magnitude
transformation
M
of
an
earthquake
using
the
Richter
scale
is
ii
I
the
eect
of
the
parameter
-
7.6,
and
the
c
M
=
,
log10
where
I
is
the
intensity
of
the
c
name
I
of
this
transformation.
n
‘movement’
of
the
earth
from
the
earthquake
b
and
I
is
the
the
intensity
of
the
The
range
for
this
model
is
the
number
on
the
‘movement’
n
Richter
ofthe
earth
on
a
normal
day-to-day
scale,
A
series
of
earthquakes
were
recorded
which
followed
the
standard
series
Explain
of
earthquakes
were
recorded
which
were
4
times
of
these
this
the
graph
the
earthquakes
of
domain
of
this
why
the
graph
State
graph
Height
above
can
that
sea
be
the
of
this
kind
of
would
give
level
and
modelled
is
Australia,
known
Russian
this
to
Reect
be
and
have
Give
specic
you
the
Describe
would
using
How
magnitude
double
quakes.
model
model
the
dier
of
the
how
from
the
the
of
has
no
y-intercept.
the
air
function
p
of
the
graph
of
standard
the
where
per
h
is
square
measured
foot.
transformation

-26 400 
ln

in
feet,
Describe
of
the
2120
and
this
graph

p
in
model
of
y
=
pounds
as
a
ln x
functions.
Statement of Inquiry:
discuss
explored
=
quakes
magnitude
transformations
examples.
a
atmospheric
by

In
model
transformation
functions.
h
b
model.
using
pressure
transformations
the
intensity.
4
Describe
10.
in
on
Mexico
≤
model.
y-intercept.
A
r
in
c
Russia,
≤
basis.
Determine
a
0
statement
of
inquiry?
Generalizing
relationships
and
changes
that
can
in
quantity
model
helps
duration,
establish
frequency
variability.
E10.1 Time for a change
21 5
Meet
E10.2
the
Global
transformers
context:
Objectives
●
Graphing
rational
Inquiry
functions
of
the
form
(x ) =
+ k
x
Finding
technical
innovation
questions
●
What
●
How
is
a
rational
function?
are
asymptotes
linked
to
domain
h
and
●
and
F
a
f
Scientic
asymptotes
of
graphs
of
range?
rational
MROF
●
functions
How
can
understanding
transformations
C
of
●
Transforming
rational
functions
rational
functions
help
you
graph
using
them?
translations,
●
Identifying
reections
and
transformations
dilations
of
graphs
●
Can
●
Does
a
function
be
its
own
inverse?
D
●
Finding
the
inverse
of
a
rational
function
science
solve
problems
or
create
them?
ATL
Critical-thinking
Revise
understanding
based
on
new
information
and
evidence
Statement of Inquiry:
10 2
Representing
helped
change
humans
apply
and
our
equivalence
in
understanding
a
variety
of
of
scientic
forms
has
principles.
11 3
E10 2
21 6
A LG E B R A
Y
ou
●
should
nd
the
already
domain
and
know
range
of
how
a
1
to:
State
function
the
these
domain
and
range
of
functions.
2
a
f ( x )
=
2x
+
5
b
f ( x )
=
x
2
c
f ( x ) =
1
●
write
statements
of
2
Write
propor tionality
b
transform
the
statement
propor tionality
a
●
x
functions
3
y
is
y
is
State
propor tional
inversely
the
to
x
propor tional
combination
transformations
get
of
for:
of
applied
2
nd
an
inverse
function
4
Find
algebraically
x
to
f ( x )
the
−2( x
inverse
+ 3)
relation
algebraically,
whether
or
also
a
not
the
+ 5
of
and
inverse
each
state
relation
function.
3
a
to
2
g( x ) =
,
function
is
x
g( x ):
f ( x ) =
●
to
f ( x ) =
2
x + 2
b
f ( x ) = ( x
− 2)
4
F
A
●
What
●
How
rational
cables
and
( x )
is
the
is
and
is
this
design
forces
is
the
of
together.
g ( x )
on
tension
is
are
≠
0,
real
of
and
so
by
and
a
of
x,
be
and
reciprocal
of
real
excluding
range?
expressed
g(x)
≠
a
quotient
0
function,
numbers
as
is
one
excluding
type
zero,
of
and
zero.
of
the
bridge.
stable,
who
opposing
the
work
Clifton
spans
River
set
of
can
and
reciprocal
layout
the
domain
which
the
the
bridge
which
the
is
compression
photo
Bridge,
called
engineers
that
and
a
the
to
functions
numbers
suspension
keep
bridge
Gorge
Bristol,
a
achieved
This
x
linked
function
domain
design
help
Suspension
Avon
of
any
application
the
used
cables
Its
set
function?
asymptotes
,
practical
function
The
f
function.
range
One
are
rational
function
rational
its
a
function
where
The
is
the
Avon,
in
England.
E10.2 Meet the transformers
21 7
Exploration
1
Complete
x
the
8
1
table
4
of
2
values
Use
x
a
the
table
from
8
Explain
b
.
1
2
4
8
c
In
a
a
you
the
as
x
positive
draw
behavior
of
gets
and
y
as
positive
dierent
color,
draw
Find
b
help
happens
both
the
graph
of
for
values
a
closer
to
negative
x
gets
and
closer
negative
vertical
line
0.
Think
values
and
of
that
closer
values
on
about
x
your
for
to
what
approach
0
on
0.
the
x
graph
that
the
curve
touches.
why
x
can
function
and
to
about
Explain
the
1
Think
never
d
step
8.
for
Describe
graph.
in
to
what
happens
4
function
2
of
3
the
1
y
2
for
the
be
equal
to
0.
Hence,
state
the
domain
of
.
values
negative
Explain
never
why
x
for
which
values
for
x
y
of
can
never
be
y
gets
equal
to
closer
0.
to
How
0.
is
Think
this
about
positive
represented
on
thegraph?
c
In
a
dierent
curve
d
For
●
State
a
never
the
●
The
Reect
●
Find
the
How
of
are
the
21 8
of
x
horizontal
line
on
your
graph
that
the
function
asymptote
x
for
is
approaches
is
discuss
the
of
the
graph
domain
the
horizontal
positive
the
approaches
equation
the
the
asymptote
as
and
asymptote
●
as
vertical
approaches
a
function:
horizontal
approaches
draw
touches.
range
reciprocal
The
color,
0.
or
vertical
(The
line
negative
line
that
that
the
graph
of
f
( x )
innity.
the
denominator
graph
can
of
never
f
( x )
equal
zero.)
1
horizontal
asymptote
and
the
vertical
of
and
asymptotes?
E10 Change
range
of
this
function
related
to
the
equations
A LG E B R A
Exploration
1
a
Copy
x
and
8
2
complete
4
2
this
table
of
values
for
1
1
2
4
8
y
b
Draw
c
State
the
domain
d
State
the
equation
they
e
graph
Explain
g
Write
in
Reect
●
>
and
If
eects
Graph
values
2
Move
of
the
the
of
Generalize
Write
4
for
g
the
and
a
1
and
and
is
0.
8
and
8.
function.
from
for
between
vertical
asymptotes.
How
do
range?
of
<
x
of
1.
Explain
why
3.
only
which
Test
Exploration
in
quadrants
quadrants
your
the
2
and
graph
conjecture
with
a
4.
of
will
GDC.
2
is
g ( x )
=
-f
( x )
transformations
f
or
( x )
is
g ( x )
and
f
(
=
f
x )
(
x )?
How
compare
do
when
3
10
to
to
and
the
explore
eect
the
insert
a
slider
for
the
parameter
a
for
+10.
(positive,
why
this
of
,
two
of
quadrants
graph
and
values
horizontal
conjecture
conclusions
Explain
in
discuss
slider
a
the
graph
function
from
values
3
a
0,
Exploration
1
the
range
domain
only
and
the
ATL
your
down
a
of
at
is
for
and
the
why
for
of
to
back
f
be
graph
relate
Look
that
the
what
negative,
of
for
case
the
fraction,
parameter
when
when
happens
a
a
>
=
0,
0
a
and
is
to
the
graph
for
dierent
zero).
on
for
the
a
ignored
<
in
graph
of
.
0.
step
3
y
The
graph
is
a
of
a
reciprocal
special
hyperbola.
It
case
has
a
of
a
function
curve
horizontal
called
,
a
asymptote
a
at
x
y
=
=
0.
0
and
avertical
Because
the
asymptote
asymptotes
<
0
a
>
0
at
are
x
perpendicular,
rectangular
it
is
also
known
as
a
hyperbola
E10.2 Meet the transformers
21 9
Reect
●
and
Compare
its
eect
discuss
the
eect
of
3
parameter
a
on
the
graph
of
with
on:
a
linear
function
a
quadratic
( y
=
ax)
2
●
For
y
=
on
f
the
rational
(2x)?
the
each
What
graphs
Example
For
function
( y
=
ax
)
function
does
of
this
rational
,
mean
is
about
the
the
eects
same
of
as
these
y
=
f
( x )
or
transformations
functions?
1
function
●
state
●
write
down
●
draw
the
the
a
below:
domain
the
graph
and
range
equations
of
the
of
the
asymptotes
function.
b
a
x
Domain:
;
Horizontal
Vertical
range:
asymptote:
asymptote:
x
y
=
y
=

-4
0.75

-2
1.5
0
0
y
1
3
14
6
10
Make a table of values.
Draw the graph by
12
plotting several points
6
from the table of values.
12
2
4
2
0
2
4
6
x
2
1
3
6
2
1.5
4
0.75
10
14
Continued
2 20
E10 Change
on
next
page
A LG E B R A
b
x
y
0.5
4
Domain:
;
range:
1
2
Horizontal
asymptote:
y
=
0
2
1
Vertical
asymptote:
x
=
0
4
y
10
8
Tip
6
8
When
a
>
0,
2
is
0
5
3
3
in
4
x
1
1
always
5
quadrants
1
and
3.
2
When
1
6
2
1
=
a f
in
x-axis,
the
reected
and
always
( x )
is
a
vertical
dilation
of
f
=
factor
a,
parallel
to
the
2
and
(ax)
is
a
horizontal
dilation
( x ),
f
( x ),
scale
factor
,
parallel
to
y-axis.
the
a f
f
x-axis.
( x )
f
(a x)
y
y
1
f
(x)
=
5
2
x
1
1
f
(x)
f
=
(x)
=
x
5x
1
f
(x)
1
2
=
1
x
0
vertical
dilation
0
x
scale
factor
5
horizontal
x
dilation
is
in
0.5
function
of
scale
the
is
quadrants
y
y
0,
graph
therefore
4
the
<
2
10
For
a
scale
factor
Continued
on
next
page
E10.2 Meet the transformers
2 21
4.
y
y
1
f
(x)
=
1
x
(x)
f
1
1
5
x
=
2
0
x
0
x
1
f
(x)
=
1
x
1
f
(x)
=
2
1
•
x
5
vertical
Reect
dilation
and
●
Explain
●
Based
why
scale
discuss
it
horizontal
factor
makes
dilation
scale
factor
5
4
sense
that
the
vertical
asymptote
of
is
x
=
0.
Rational
on
your
answer
above,
what
do
you
think
the
vertical
with
will
be
for
the
function
When
does
a
rational
?
function
more
vertical
are
●
have
two
vertical
asymptotes?
Give
Example
It
takes
Model
20
of
such
Determine
beyond
14
situation
how
days
with
many
a
days
x
=
number
of
students
Let
y
=
number
of
days
to
eat
all
rational
it
the
snacks
function
would
take
8
in
and
a
vending
the
this
course.
sketch
students
to
its
eat
machine.
graph.
all
the
snacks.
Dene
∝
of
function.
Let
y
one
2
students
this
a
than
asymptote
an
level
example
functions
asymptote
variables.
variable
is
Decide
dependent
which
on
the
other.
⇒
Identify
inverse
students
When
x
=
20,
⇒
k
=
y
=
14,
proportion
increases,
the
–
as
number
the
of
number
days
of
decreases.
so:
280
Use
the
the
initial
constant
condition
of
to
calculate
proportionality
k.
⇒
y
240
When
200
x
=
8
(8
students),
Therefore,
160
x
120
it
would
take
35
cannot
graph
for
80
8students
same
40
0
x
4
2 2 2
8
12
16
20
E10 Change
number
take
negative
values,
so
the
days
to
eat
the
of
snacks.
is
in
the
rst
quadrant
only.
A LG E B R A
Reect
●
In
and
Example
number
If
●
not,
In
ory
what
describe
them.
to
2,
a
kind
how
do
x
each
to
●
draw
a
●
draw
and
value
of
the
domain
graph
of
label
and
the
the
x
your
have
take?
Does
graph
maximum
reasoning.
to
or
Would
Is
it
possible
your
graph
reect
this
minimum
it
make
to
have
represent
any
this?
limitation.
values?
sense
for
If
so,
either
x
0?
range
function
asymptotes.
8
=
y
b
1
=
c
y
length
of
time
it
2
=
x
x
The
y
your
can
fractions?
change
and
4
2
values
function:
state
y
of
1
●
a
5
including
Explain
have
Practice
For
2,
students,
Example
state
1
of
discuss
d
y
=
3x
4 x
takes
to
build
a
community
shelter
is
inversely
A 2016 Design of the
proportional
to
the
number
of
people
working
on
the
project.
It
takes
Year
600hours
for
4
people
to
build
a
to
a
Model
this
situation
with
a
rational
function
and
sketch
its
Determine
how
long
it
a
would
take
12
people
to
build
a
all
shelter.
for
3
Below
is
the
graph
of
the
resistance
in
an
electric
circuit
(R)
at-pack
versus
which
the
(I)
owing
through
it
for
a
constant
shelter
included
tools
required
assembling,
and
the
could
current
awarded
graph.
kit
b
was
shelter.
be
put
together
voltage.
by
y
four
four
people
in
hours.
50
45
40
35
)R(
30
ecnatsiseR
25
20
15
10
5
0
x
0.5
1
1.5
2
2.5
Current
a
Describe
b
Find
c
Hence
the
the
relationship
constant
write
of
down
between
function
3.5
4
4.5
5
(I)
current
proportionality
the
3
using
relating
and
resistance.
points
current
from
(I)
and
the
graph.
resistance
(R).
E10.2 Meet the transformers
2 23
just
4
The
graph
build
a
models
shelter
for
the
number
of
hours
( y)
it
takes
for
aid
workers
( x )
to
refugees.
y
50
40
sruoh
30
fo
rebmuN
20
10
0
x
5
10
15
Number
From
long
this
it
graph,
would
the
1
person
to
build
the
shelter
b
5
people
to
build
the
shelter
c
10
people
Newton’s
force,
a
b
C
the
the
and
law
ii
acceleration
sketch
of
and
ii
acceleration
motion,
versus
and
How
can
and
2 24
how
=
ma,
states
the
relationship
between
relationship
between:
mass.
the
what
rational
functions
transformations
of
rational
functions
help
them?
discuss
function
on
F
understanding
graph
.
your
explain
of :
versus
y
=
graph
demonstrate
Predict
hence
acceleration
Transforming
k
and
mass.
graphs
force
the
function
acceleration
i
hand
rational
shelter.
proportional
force
For
workers
acceleration.
i
Reect
●
build
and
you
●
to
second
Describe
●
35
30
solving
mass
Hence
of
25
take:
a
Problem
5
determine
20
f
(x
of
what
happens
What
6
h)
y
you
to
will
=
f
reasoning.
E10 Change
k,
describe
Use
mean
the
be
+
( x ).
with
graph
the
the
eects
appropriate
an
the
parameters
terms
example.
of
equations
of
mathematical
after
of
its
it
is
transformed
asymptotes?
using
Explain
A LG E B R A
Objective
i.
select
and
complex
In
Investigating
apply
draw
state
your
mathematical
new
4
you
2
Use
conclusions
3
Set
the
values
graph
Describe
Set
the
Sketch
6
Sketch
=
h
and
0
7
a
of
=
1.
8
the
Include
of
Determine
Determine
=
and
your
exploration
ndings
skills
clearly
to
with
function
0.
a.
the
discover
make
new
labelled
the
ndings
sketches
insert
k
sliders
the
at
a
and
for
and
and
eect
and
h
h
explore
on
of
and
and
the
negative
range.
what
fraction).
graph
values
of
of
h.
range.
k
on
fractional
domain
parameters
of
fraction).
and
negative,
fractional
domain
the
to
happens
positive,
domain
time
eect
what
negative,
Include
(positive,
the
negative
the
.
one
explore
positive,
of
negative
state
to
asymptotes,
and
the
slider
(e.g.
demonstrate
and
of
the
h
state
positive,
eects
a
demonstrate
and
to
of
sliders
of
positive,
graphs
and
the
graph
State
values
to
Move
values
on
of
asymptotes
k
the
graph
values
of
of
k.
range.
in
on
the
.
the
,
9
a
graphs
the
k
Move
dierent
several
the
dierent
asymptotes
Generalize
graph
to
k.
and
Include
Label
graphing
Explain
values
several
the
use
graph
eect
value
for
a,
for
the
.
Label
techniques
clearly.
to
h
fractional
happens
5
GDC
parameters
and
4
your
the
problem-solving
4
the
to
will
conclusions.
Exploration
1
patter ns
patter ns
Exploration
and
ATL
B:
horizontal
just
the
by
and
looking
domain
and
at
vertical
the
range
asymptotes
of
a
rational
function,
function.
of
just
by
looking
at
the
function.
Reect
●
In
and
Reect
discuss
and
discuss
might
completely
6,
look.
correct,
7
you
How
what
were
good
should
asked
was
you
to
predict
your
have
how
prediction?
the
If
graph
you
of
weren’t
predicted?
E10.2 Meet the transformers
2 25
The
function
(the
parent
The
domain
Therange
The
,
of
of
Sketch
the
the
of
the
the
is
a
transformation
of
function
function
horizontal
is
x
=
is
.
is
asymptote
is
y
=
k
.
and
the
equation
of
the
h
3
graph
Horizontal
rational
rational
asymptote
Example
,
function).
equation
vertical
where
of
.
asymptote:
y
=
3
Find
Vertical
asymptote:
x
=
the
asymptotes.
1
y
6
Sketch
the
asymptotes
on
the
5
graph
using
dashed
lines.
4
y
=
3
3
x
=
1
1
0
4
3
2
x
1
1
2
3
4
5
6
1
Sketch
2
3
Tip
To
check
value
For
of
the
You
x
For
to
can
then
each
a
the
graph
Practice
1
where
graph
right
in
of
is
Example
sketch
the
located,
the
you
vertical
3:
when
curves
x
of
the
functions
●
write
down
the
equations
●
write
down
through
the
domain
the
value
of
h
and
(x ) =
of
and
the
2;
these
when
x
its
the
=
2,
equation
left.
y
=
4.
points.
asymptotes
range.
b
f
1
(x ) =
c
x
+ 3
+
2
- 11
e
f
+ π
(x ) =
x
x
E10 Change
1
f
(x ) =
+ 8
x
1
1
(x ) =
2 2 6
=
into
to
1
- 7
x
f
y
value
k
1
d
0,
a
below:
nd
f
=
substitute
and
2
●
a
can
axis,
5
a
the
function.
A LG E B R A
2
For
each
function
●
draw
●
sketch
●
state
and
label
the
the
below:
the
asymptotes
function
domain
on
and
a
graph
range.
1
a
y
1
=
- 4
b
y
c
x
1
d
y
1
=
x
y
=
4
+ 3
x
1
=
e
x
y
1
=
+ 5
2
x
+
f
y
=
2
+1
x
3
1
3
Each
of
these
functions
has
an
equation
of
the
form
y
+ k.
=
x
Find
the
equation
a
of
each
graph.
b
y
y
2
x
=
2
x
0
0
8
6
4
h
x
2
2
4
6
=
2
0
8
8
6
4
2
2
x
2
4
6
4
4
y
=
−5
6
6
8
8
10
10
y
c
d
y
x
y
=
=
8
2
=
−7
y
6
10
4
8
2
6
−4
0
4
0
10
8
6
4
x
2
2
4
6
y
=
2
2
2
x
=
3
4
0
6
4
2
x
2
4
6
8
10
6
Problem
solving
1
4
Find
the
equation
of
a
curve
y
=
+ k
x
a
y
=
4
and
c
y
=
-1
x
=
3
b
y
=
2
and
the
x
=
2
d
y
3
and
=
asymptotes:
y-axis
4
and
with
h
x
=
3
4
1
5
Describe
the
individual
transformations
on
f
(x ) =
for
a
each
set
i
h
>
0
ii
h
<
0
of
values
for
h
b
and
1
to
x
obtain
f
(x ) =
+ k
x
h
k.
i
k
>
0
ii
k
<
0
E10.2 Meet the transformers
2 27
Example
For
each
4
function
●
list
●
nd
●
hence
the
below:
individual
the
equations
sketch
the
transformations
of
the
graph
on
asymptotes
of
the
function.
a
b
Use
the
values
a,
h
and
k
from
the
a
rational
h
=
in
3
⇒
the
horizontal
positive
translation
of
3
function
.
units
x-direction
y
10
a
=10
k
=
2
⇒
⇒
vertical
vertical
dilation
scale
translation
of
factor
2
10
units
5
in
the
positive
y-direction
y
Horizontal
asymptote:
y
=
=
2
2
x
5
5
Vertical
asymptote:
x
=
Draw
the
asymptotes
10
3
as
5
dashed
sketching
x
=
lines
the
before
graph.
3
b
Rearrange
factorizing
⇒
horizontal
translation
1.5
units
in
the
positive
the
equation
out
the
to
look
coecient
=
-2
⇒
vertical
stretch
of
factor
2
and
horizontal
y
=
3
⇒
vertical
translation
3
units
8
x
in
the
positive
=
1.5
y-direction
6
Horizontal
asymptote:
y
=
3
4
y
Vertical
asymptote:
x
=
=
3
2
1.5
0
6
4
2
4
E10 Change
x
2
2
2 2 8
x
in
the
denominator.
negative
value
of
a
gives
reection
a
k
of
by
x-direction
A
a
like
4
6
8
reection
in
the
x-axis.
A LG E B R A
The
order
for
performing
1
Horizontal
2
Reection
3
Vertical
dilation
4
Vertical
translation
For
a
rational
multiple
translation
in
the
x-axis
(if
a
horizontal
●
k
determines
the
vertical
●
a
determines
a
the
Practice
0)
:
the
the
is:
(k)
determines
factor
functions
(a)
h
❍
<
function
either
on
(h)
●
❍
transformations
horizontal
reection
in
the
translation;
translation;
dilation
x-axis
(if
of
a
the
the
vertical
asymptote
is
x
=
h
horizontal
asymptote
is
y
=
k
factor
<
or
the
vertical
dilation
of
0)
3
1
1
Perform
these
transformations
on
the
rational
function
f
(x ) =
,
then
write
x
down
the
function
obtained
as
a
single
algebraic
fraction.
1
a
Vertical
dilation
by
scale
factor
,
vertical
translation
of
1
unit
in
the
2
negative
y-direction,
horizontal
translation
of
4
units
in
the
positive
x-direction.
b
Horizontal
vertical
dilation
translation
by
scale
of
2
factor
3,
horizontal
translation
of
3
units,
units.
1
c
Vertical
dilation
by
scale
factor
,
reection
in
the
x-axis,
vertical
,
reection
in
the
y-axis,
vertical
translation
of
4
translation
of
8
units.
3
d
Vertical
dilation
by
scale
factor
2
translation
2
For
each
of
4
rational
units,
horizontal
function
2
units.
below:
1
●
list
●
nd
●
hence
the
transformations
applied
to
f
(x ) =
x
the
equations
sketch
the
of
the
graph
asymptotes
of
the
rational
3
a
f
(x ) =
- 5
x
+
b
f
f
(x ) =
2x
+ 3
2
c
f
x
(x ) =
3x
5
+ 2
8
9
(x ) = -
6
5
d
function.
5
e
f
13
(x ) = -
- 8
2x
+ 1
1
f
f
(x ) = -
- 11
x
E10.2 Meet the transformers
2 2 9
D
The
inverse
●
Can
●
Does
a
function
science
Exploration
1
Using
your
graph
2
Repeat
step
1
solve
rational
own
draw
the
line
y
for
the
graphs
graph
what
a
Find
create
them?
=
x.
and
Describe
what
the
you
line
your
results
.
inverse
y
=
x.
Reect
the
notice.
of :
c
,
the
or
of
b
Explain
function
inverse?
problems
the
function
4
its
in
a
3
be
a
5
GDC,
of
of
tell
Justify
you
why
d
about
this
function
the
inverse
inverse
of
the
relation
relation
is
a
of
a
rational
function.
function
,
algebraically.
b
Determine
c
Summarize
Reect
You
could
because
y
when
x
is
the
●
not
=
How
the
and
say
your
that
in
This
inverse
are
the
and
ndings
discuss
represents
0.
domain
the
the
means
range
about
the
inverse
function
function
multiplicative
that
.
of
,
.
8
reciprocal
function
of
yx
of
y
multiplicative
=
=
1.
inverse
However,
,
(or
the
y
is
the
inverse
reciprocal)
reciprocal
of
x,
of
x,
except
function
x
inverse
(reciprocals)
and
inverse
functions
dierent?
●
What
is
meant
●
What
is
the
inverse
function
of
?
●
What
is
the
inverse
function
of
?
Example
Find
the
by
‘inverse
function’?
5
inverse
of
.
Interchange
the
Isolate
23 0
E10 Change
x
the
and
the
variable
y
y
A LG E B R A
Practice
1
On
4
your
GDC,
graph
each
function
4
a
y
b
y
e
-1
y
y
inverse
of
each
of
b
y
y
y
=
+1
1
x
x
1
+ 6
x
- 3
e
y
c
= -
- 8
x
1
f
x
12
2x
=
x
y
9
3x
=
5
4
function:
=
6
d
+
1
f
each
+ 1
grid.
=
7
=
x
y
x
3
the
same
1
c
= 3 +
5
a
the
3
=
Find
on
x
x
2
inverse
= -
3
y
its
2
=
x
d
and
y
= -
+ 3
x
4
Summary
A
rational
expressed
are
a
the
real
of
For
is
of
x,
and
numbers
a
reciprocal
is
x
≠
is
the
a
f
( x )
can
and
rational
Its
zero
excluding
horizontal
which
be
function
domain
and
its
is
the
range
For
a
rational
as
x
is
h
determines
the
horizontal
●
k
determines
the
vertical
●
a
determines
set
❍
either
the
horizontal
the
vertical
❍
the
horizontal
❍
whether
the
dilation
translation
dilation
of
that
the
is
the
vertical
denominator
function,
line
that
f
positive
the
( x )
approaches
0.
or
never
equal
<
0)
domain
vertical
the
or
graph
not
(a
>
is
reected
approaches
(The
the
isy
=
For
x
=
x-axis
of
the
rational
function
.
The
range
of
the
as
the
rational
function
is
.
denominator
horizontal
asymptote
,
h
and
the
the
asymptote
is
at
x
is
=
at
y
=
k
and
the
h
vertical
horizontal
asymptote
k
Mixed
1
is
the
0.)
function
asymptote
in
0).
asymptote
vertical
For
or
negative
The
can
factor
graph
The
reciprocal
of
a
reection
is
a
factor
horizontal
innity.
For
translation
zero.
line
approaches
:
●
(a
approaches
function
g ( x )
0
function.
function,
the
≠
0,
excluding
numbers
asymptote
function
where
g(x)
,
reciprocal
real
any
quotient
function
called
set
as
functions
The
of
function
practice
each
function
below:
2
Perform
these
transformations
on
the
1
●
rational
state
the
largest
possible
domain
and
f
function
(x ) =
to
obtain
g ( x ).
range
x
Write
●
write
down
the
equations
of
the
down
g ( x )
obtained
as
a
single
algebraic
asymptotes
fraction.
●
sketch
the
graph.
1
a
a
y
The
b
= -
y
=
scale
2
is
vertically
dilated
by
factor
3,
negative
translated
vertically
5
in
y-direction
and
units
horizontally
4
=
- 8
x
(x ) =
x
the
y
f
+1
x
x
c
function
1
5
2
d
y
= -
+ 2
x
5
units
in
the
positive
x-direction.
+ 3
E10.2 Meet the transformers
2 31
by
1
b
The
function
f
(x ) =
is
vertically
dilated
●
sketch
●
nd
the
graph
of
the
rational
function
by
x
1
scale
factor
,
translated
1
unit
vertically
in
its
inverse.
the
2
1
y-direction
and
reected
in
the
+ 9
y-axis.
x
The
f
function
(x ) =
is
reected
in
the
vertically
positive
and
+ 1
by
2
units
horizontally
in
by
3
the
in
m (x ) = -
x
+ 6
8
+
2
1
n (x ) =
the
+1
3x
5
negative
units
4
d
5
e
y-direction
- 3
x
k (x ) =
c
x-axis,
x
translated
h (x ) =
b
2
10
1
c
3
g (x ) = -
a
positive
p (x ) =
f
4 x
4 x
5
x-direction.
Problem
solving
1
d
The
function
f
(x ) =
is
dilated
horizontally
x
by
scale
factor
translated
1
y-direction
12,
unit
and
4
reected
vertically
4
units
in
in
the
the
x-axis,
in
adult
rational
list
the
drug
dosages
pharmaceutical
for
drug
the
mg.
dosage
says
that
the
children’s
dose
is
x

, where
x
+ 12
x
istheageofthechild.

function:
a
●
50
= 50

each
calculates
certain
the

For
is
a
x-direction.
C
3
For
dose
Young’s
positive
dosage
children.
negative
horizontally
Young’s
individual
transformations
Calculate
the
dosage
of
the
medicine
for
a
applied
10-year-old
child.
1
to
f
to
(x ) =
obtain
the
given
a
function
b
x
Rewrite
the
function
in
the
form C
=
+ k
x
●
nd
the
equations
of
the
asymptotes
c
Review
1
in
Catherine
and
then
She
starts
is
to
Explain
the
series
Yen.
travelling
from
New
York
to
with
Tip
1000
spends
USD
40
and
000
converts
Yen
in
this
Tokyo,
her
remaining
money
into
AUD
reaches
reciprocal
Sydney.
Using
functions
and
your
the
knowledge
below,
Catherine
explain
now
has
how
currency
in
much
the
rate,
W
is
the
work
time
(in
this
case
ll
1
reservoir),
t
is
taken.
of
conversion
a
rates
is
when
the
she
r
and
required
converts
transformations.
Tokyo
Sydney.
She
of
context
where
to
h
Let
the
the
fast
hours
needed
to
ll
one
reservoir
from
money
pipe
be
p.
The
rate
of
lling
is
the
AUD.
1
reciprocal
1
USD
=113.4
YEN
=
1.32
Two
in
oil
pipelines
Nigeria.
three
times
pipelines
lled
in
One
14
between
pipeline
faster
are
run
than
a
can
Abuja
ll
second
operational,
an
oil
an
and
oil
b
tanker
pipeline.
reservoir
If
How
you
have
be
c
As
it
both
helped
explored
the
232
apply
and
our
statement
equivalence
of
in
understanding
E10 Change
.
Find
the
reciprocal
inquiry?
a
rate
of
lling
by
the
slow
pipe.
an
takes
the
expression
Give
specic
variety
of
of
scientic
forms
has
principles.
for
the
two
pipes
together.
14
hours
nd
reservoir
discuss
change
humans
the
pipelines,
ll
Statement of Inquiry:
Representing
for
nd
working
both
can
Hence
to
and
=
Lagos
hours.
Reect
r
p
function
2
function
AUD
examples.
to
p.
ll
the
Hence
with
just
reservoir
nd
the
the
slow
with
time
pipe.
taken
Unmistaken
E11.1
Global
context:
Objectives
●
Using
●
Solving
Orientation
Inquiry
trigonometric
expressions
identities
and
identities
solve
equations
to
●
simplify
F
equations
involving
functions
time
How
do
you
solve
trigonometric
How
many
solutions
How
do
you
does
equation
prove
a
have?
trigonometric
identities?
●
How
to
●
D
can
solve
Can
you
use
trigonometric
trigonometric
two
dierent
identities
equations?
solutions
both
be
correct?
●
How
do
we
dene
‘where’
and
‘when’?
Creative -thinking
Make
unexpected
or
unusual
connections
between
objects
and/or
ideas
E7 1
Statement of Inquiry:
Generalizing
help
and
dene
applying
‘where’
relationships
and
between
measurements
in
space
‘when’.
E9 2
E11 1
23 3
S P I H S N O I TA L E R
●
can
and
questions
trigonometric
C
ATL
space
equations?
●
trigonometric
in
Y
ou
●
should
conver t
radians
already
angles
and
from
vice
know
degrees
how
to
1
to:
Conver t:
versa
a
120
b
to
135
radians
to
radians
π
c
to
degrees
6
5π
d
to
degrees
3
●
nd
or
the
value
tangent
of
of
the
any
sine,
angle,
cosine
2
Find
a
special
the
a
missing
sine,
angle
cosine
or
when
given
tangent
3
trigonometric
Find
x
●
use
quadratic
the
side
or
sine
functions
4
sin A
in
sin B
=
=
Describe
0.5
the
graph
of
of
transformations
f ( x ) =
g( x ) =
to
a
nd
a
sin x
to
on
give
the
3 sin ( 2 x ) + 5
2
equations
rule
angle
when:
sin x
graph
solve
)
cos x = 0.321
the
●
3
value
b
transform
(
tan 47
a
●
of :
2π
cos
c
nd
value
sin 45
angles
b
●
the
including
missing
5
Solve 15 + 7 x
6
a
Find
∠
B
in
- 2 x
this
triangle.
=
0
triangle.
B
sin C
=
a
b
c
6
cm
60°
3
b
Find
the
cm
length
of
side
x,
to
3
x
30°
40°
8
Trigonometric
F
In
●
How
do
●
How
many
problems
with
you
solve
trigonometric
solutions
right-angled
equations
does
a
triangles,
equations?
trigonometric
you
solve
equation
trigonometric
1
to
get
a
single
solution,
such
as
solving
sin x
=
Exploration
equation
can
234
1
you
will
investigate
have.
E11 Equivalence
how
many
equations
π
to
get
x
=
30
2
In
have?
or
radians.
3
solutions
a
trigonometric
s.f.
G E O M E T R Y
Exploration
Draw
the
graph
2
Draw
the
line
3
Use
of
graph
State
5
Generalize
how
to
many
8
how
domain
how
many
numbers.
this
●
In
the
angles
Explain
equation
1
and
you
to
8
if
the
the
for
.
angle
that
satises
the
important
it
to
trigonometric
domain,
Trigonometric
identity
is
there
which
this
equation
you
the
domain
other
angle
.
within
when
the
domain
is
the
set
know.
,
is
state
the
how
set
of
to
nd
real
of
,
all
angles
that
numbers.
for
.
1
number
using
generalization
state
within
b
equation
graph
same
Why
An
each
this
domain
discuss
nd
of
if
the
for
Explain.
given
of
equation
,
how
ndings:
negative?
is
value
this
value
satisfy
domain
solving
a
your
steps
your
the
satisfy
the
in
●
angles
within
Would
,
graph.
if
sine
b
Repeat
●
a
domain
Reect
In
many
Generalize
9
ndings:
has
the
satisfy
same
satisfy
within
real
function
determine
a
State
of
the
angles
your
also
7
sine
.
4
State
the
on
equation
6
T R I G O N O M E T R Y
1
1
your
A N D
in
of
the
solutions
unit
Exploration
specify
a
to
sin
x
=
and
cos x
=
circle?
1,
domain
does
for
it
the
matter
if
values
of
b
is
positive
or
when
equations?
may
be
multiple
solutions
to
a
trigonometric
equation.
identities:
true
for
all
values
of
the
variable.
E11.1 Unmistaken identities
23 5
Example
Solve
these
graphically
1
equations
using
a
algebraically
GDC
or
for
dynamic
a
Verify
geometry
your
solutions
software.
b
Rearrange
to
isolate
sin θ.
The
domain
is
a
given
in
radians,
so
give
angles
Find
in
one
radians.
value
of
.
1.1
π
y
2π
3
2
3
1
2
y
=
2
0.5
Graph
the
LHS
and
the
RHS
of
the
x
0
π
π
3π
2
2π
equation
2
0.5
y =
sin
where
and
they
nd
the
x-coordinates
intersect.
x
1
b
Solve
each
equation
separately.
⇒
Use
the
identity
.
⇒
Answer
Use
is
the
outside
the
given
domain.
identity
.
1.1
Use the same identity again, as long
y
4
as the solution is within the given
3
domain.
2
1
(0.322, 0)
(5.96, 0)
Set
0
(2.82, 0)
(3.46, 0)
1
2
y
=
9tan2x
−
1
3
4
23 6
E11 Equivalence
x
your
window
to
the
given
domain.
G E O M E T R Y
Practice
1
Solve
ifnot
A N D
T R I G O N O M E T R Y
1
each
equation
possible,
for
round
0
to
≤ θ
the
≤ 360°.
nearest
Give
exact
degree.
answers
Check
your
where
possible;
answers
graphically.
1
a
sin θ
=
b
4 cos θ
e
tan θ
=
-1
c
tan θ
f
2 cos θ
= 1
2
3
2 sin θ
d
=
-1
= -
=
-
3
4
2
Solve
ifnot
each
equation
possible,
for
round
0
to
≤ θ
0.01.
a
2 cos θ
=
-1
b
d
6 cos θ
=
-5
e
≤
2π .
Give
Check
exact
your
2 cos θ
=
answers
answers
where
possible;
graphically.
2
c
2 sin θ
f
-
=
-
3
5
7 cos θ
= 1
sin θ
= 1
3
2
g
tan θ
=
-1
h
3 tan θ
=
3
θ
sin
i
= 1
parts
the
j
to
q,
factorize
quadratic
rst.
1
2
2
θ
sin
j
+ 2 sin θ
+ 1 =
0
k
2
θ
4 cos
- 1 =
0
tan
l
θ
=
4
2
2
m
2 cos
p
2 tan
θ
+ cos θ
= 1
n
6 cos
q
2 sin
2
a
b
- cos θ
- 1 =
0
o
2 sin θ
cos θ
+ sin θ
=
0
2
θ
- tan θ
Problem
3
θ
- 6
=
0
θ
2 ) sin θ
+ (2 -
-
2
=
0
solving
Find
the
set
of
values
in
the
domain
0≤ θ ≤
2π
such
Find
the
set
of
values
in
the
domain
0≤ θ ≤
2π
that
that
sin
satisfy
=
the
cos
equation:
2
sin
4
x
+ sin x cos x
a
Show
that
b
Hence
Find
all
a
a
0.
6 sin x cos x
solve
Example
=
6 sin x
+ 3 cos x
cos x
+ 4 sin x
+ 3 cos x
+ 2
+ 4 sin x
≡
+ 2
(2 sin x
=
0
in
+
the
1)(3 cos x
domain
+
2).
0≤
x ≤
2π
2
solutions
,
within
the
given
domain.
b
,
Let
Substitute
⇒
a
variable
then
the
for
2x.
domain
If
of
the
u
=
domain
2x
of
x
is
is
.
⇒
Solve
for
u.
or
Substitute
2x
for
u
and
solve
for
⇒
⇒
Continued
on
next
page
E11.1 Unmistaken identities
237
x.
b
If the domain of x is
Let
the domain of u =
then
is
.
⇒
⇒
or
Reject
since
it
lies
outside
⇒
Practice
1
Solve
2
the
equations
within
the
given
domain.
Check

a
sin
(2θ)
=
1,
0
≤ θ
≤ π
θ

your
= −
2
solutions
graphically.
1

cos
b


,
0
≤ θ
≤
2π
2
3

sin
θ
=
c

3
,

sin
0
≤ θ
≤ π
d
cos
(3θ )
=
1,
≤ θ
0
≤
2π
2
e
3
(5θ) −
g
2 sin (3 x )
i
sin
=
= 1,
0
≤ θ
x
≤ π
0 ,
0
≤
≤ π
f
cos
h
sin
(2x)
j
3 cos
2π
2
2
(3θ )
−
sin
(3θ )
=
0,
0 ≤ θ
≤

=
1,
0
sin
θ
2



θ
≤ θ
≤
(2x),
= 0,

2π
0
≤
x
0
≤ θ
≤
2π

+ cos

3
(2θ )
2
≤ π

π
2
k
2 sin
(3 x )
− 3 sin
(3 x )
=  −1,0 ≤
x
≤
3
2
3 co
l
(2x)
−
5
cos
(2x),
0
≤
x
≤
2π
C
●
How
do
●
How
can
you
prove
you
use
trigonometric
trigonometric
identities?
identities
to
solve
trigonometric
equations?
Exploration
Identity
1
For
the
2
1
this
right-angled
trigonometric
triangle,
write
down
ratios:
r
y
a
tan θ
b
sin θ
=
=
θ
c
cos θ
x
=
Continued
23 8
E11 Equivalence
on
next
page
the
domain
.
G E O M E T R Y
2
Rearrange
your
Explain
Identity
4
For
sin θ
of
how
equation
for
cos θ
and
cos θ
equation
terms
3
your
this
to
relates
sin θ
for
make
x
to
how
r
1,
to
make
the
you
y
the
subject.
subject.
Hence
calculate
tan θ
A N D
T R I G O N O M E T R Y
Rearrange
write
using
tan θ
the
in
unit
circle.
2
the
unit
expressions
a
circle
where
=
write
for:
b
c
P(x, y)
5
Using
the
right-angled
triangle,
write
an
r
y
expression
for
r
in
terms
of
x
and
y.
θ
x
6
Rewrite
using
7
Test
your
your
that
values
obtuse
expression
result
your
of
.
result
Make
angles
Reect
from
and
and
is
sure
for
step
4
true
for
you
negative
discuss
dierent
test
some
angles.
2
●
Why
do
you
think
Identity
1
is
called
the
Ratio
●
Why
do
you
think
Identity
2
is
called
the
Pythagorean
Ratio
identity?
identity?
identity:
Pythagorean
Rearranging
identity:
the
Pythagorean
identity
gives
two
equivalent
identities:
and
Once
you
have
trigonometric
Example
Solve
proven
an
identity,
you
can
use
it
to
simplify
and
solve
equations.
3
for
,
and
check
graphically.
Use the Pythagorean identity
to
write
the
Multiply
by
Factorize
Continued
equation
on
1
to
and
next
in
sin θ
avoid
solve
negative
the
coecients.
quadratic
equation.
page
E11.1 Unmistaken identities
23 9
or
or
⇒
or
Find
the
all
the
values
of
in
the
given
domain.
⇒
Reject
as
it
is
outside
the
domain.
⇒
1.1
y
1
2
y =
2 cos
x −
sin x
−
1
(2.62, 0)
(0.524, 0)
0
x
1
2
Practice
1
Solve
3
these
exact
form
trigonometric
(in
radians)
if
equations
possible,
for
0
≤ θ
otherwise
≤
2π .
write
Leave
your
your
answer
answer
to
3
in
In
s.f.
Japan,
use
2
a
2
θ
cos
= 1
b
2
θ
sin
- cos θ
= 1
aid
2
2
θ
c
cos
e
3 cos
cos θ
+
=
2
θ
sin
d
2
θ
3 tan
cos
=
2 + cos
for
θ
to
3
- 6 cos θ
these
s.f.
below
remembering
sides
involved
2
θ
=
θ
sin
- 3
in
Solve
shown
2
θ
the
2
2
students
memory
2
θ
- sin
the
trigonometric
where
equations
for
0
≤
θ
≤
360° .
c
4 sin θ
answer
+ 1 =
cosine
0
b
2 cos
d
4 sin
θ
sin θ
= 1
- cos θ
= 1
+
the
The
triangle
for
and
letter
superimposed
2
θ
- cos
sine,
formulas
tangent.
2
θ
2 sin
your
appropriate.
2
a
Write
the
is
on
drawn
2
=
tan θ
2
e
4 tan
2
θ cos
2
θ
- sin
θ
to
sides
2
θ
= cos
include
Determine
=
is
an
identity.
Justify
your
answer.
2
can
use
in
case.
1
whether 1 +
tan θ
You
two
solving
1
3
involved
θ
each
Problem
the
the
other
identities.
same
as
trigonometric
To
prove
an
sin
in
θ
identities
identity,
you
work
have
on
previously
one
side
until
proved
your
to
prove
result
is
the
os
the
other
side.
an
24 0
E11 Equivalence
G E O M E T R Y
Example
Prove
A N D
T R I G O N O M E T R Y
4
that
.
LHS:
Start
side
on
one
that
side.
seems
Tip:
more
Start
on
the
complex.
=
Use
=
=
Write
the
expression
with
common
the
Ratio
identity.
denominator
cos θ
=
Use

the
Pythagorean
identity.
=RHS
Example
Prove
5
that
.
LHS:
Factorize
Use
=
the
the
Pythagorean
quadratic
identity,
expression
then
=
factorize
in
the
the
Cancel
numerator.
denominator.
common
factors.
=

=
RHS
Objective
iii.
prove,
To
prove
be
if
true.
the
you
graphs
Prove
Investigating
verify
other
If
Practice
1
B:
or
and
identities,
are
are
unsure
the
patter ns
justify
,
make
about
same,
the
general
sure
a
that
r ules
you
statement,
statement
is
only
you
use
can
probably
results
justify
an
that
it
by
have
been
graphing
proven
both
to
sides:
identity.
4
these
identities:
2
1
sin
θ
cos θ tan θ
≡ cos θ
a
b
cos θ
2
c
cos
≡ 1
sin θ
2
θ (tan
+ 1)
≡ 1
d
sin
1

2
θ
θ

≡ 1
1 +

2
tan
θ

E11.1 Unmistaken identities
241
1
4
e
2
θ
sin
+ cos
2
θ
2
sin
≡
2
sin
2
θ
sin
f
+ tan
2
θ
θ
+ cos
≡
2
θ
cos
In
part
g,
multiply
1
+
sin θ
1
the
tan θ
cos θ
≡
1 +
h
cos θ
RHS
by
.
2
cos θ
g
2 sin θ
+ sin
θ
≡
1
1 + sin θ
2
cos
θ
tan θ
cos θ
Problem
Tip
solving
When
2
Use
your
GDC
or
graphing
software
to
graph
these
both
identity
Based
on
the
graph,
conjecture
an
identity
and
justify
your

1
a
+

(1
tan x
sin x )
sin x
+
tan x
in
are
part
work
- cos x
- 1

on
sides
reach
equivalent
ATL
3
●
Explain
how
verifying
an
identity
is
similar
●
Explain
how
verifying
an
identity
is
dierent
The
D
ambiguous
●
Can
●
How
two
do
Exploration
1
Find
2
In
two
the
dierent
we
dene
to
solving
to
an
solving
equation.
an
equation.
case
solutions
‘where’
both
and
be
correct?
‘when’?
3
angles
diagram
where
below,
sin
nd
B
=
,
angle
0
B
<
to
B
3
≤
180
s.f.
B
10
30°
C
A
10√3
3
Hence,
4
Based
step
5
Can
2.
nd
on
your
Write
you
angle
the
B
?
third
answer
down
construct
If
so,
angle
to
the
a
write
of
step
other
triangle
down
the
1,
triangle
there
possible
with
the
(angle
should
A
angle
and
of
C ).
have
solution
angle
third
for
angle
this
the
been
two
solutions
E11 Equivalence
in
B
second
value
for
triangle.
Continued
24 2
you
can
independently
both
you
discuss
complex,
h,
(cos x )

and
an

b
cos x
Reect
of
answer.
as

sides
expressions.
on
next
page
until
an
expression.
G E O M E T R Y
6
Based
(give
on
an
the
diagram
exact
below,
nd
the
value
of
angle
N
in
this
A N D
T R I G O N O M E T R Y
diagram
value).
N
n
=
5√2
45°
M
P
7
Using
the
formed
8
Write
When
are
other
down
using
two
i
word
word
each
why
for
why
the
sine
of
sine
Latin
the
there
is
SSA
is
word
jiba,
for
is
of
for
missing
called
will
the
and
it
is
of
N.
Can
answers?
a
triangle
be
Explain.
triangle.
angle
in
a
give
triangle,
you
ambiguous
cosine
(S)
you
or
ambiguous
to
one
case
sometimes
solution
for
case
of
an
error
in
sinus,
misread
hence
as
the
a
or
a
the
there
as
sine
;
the
rule.
any
given
of
three
combination
whether
missing
or
not
you
side/angle.
these:
SAS
case.
translating
mathematician
was
for
ambiguous
normally
(A)
state
nd
iii
the
are
angles
scenarios,
rule
ASA
Arab
value
these
this
triangles,
sides
following
or
result
fold
possible
each
4
produces
the
5√2
calculator
with
no
=
angles
a
either
rule
famous
is
nd
problems
the
‘sine’
by
triangle
and
For
and
Your
This
ii
Example
In
.
SSS
written
to
information:
use
Explain
The
rule
discuss
solving
other
possible
180
of
Explain
the
is
and
both.
only
the
angle
solutions.
pieces
would
nd
given
sine
When
of
●
the
the
solution
rule,
the
possible
Reect
●
sine
with
m
a
trigonometry
Al-Khwarizmi.
jaib,
word
which
In
means
book
Arabic
‘fold’.
the
And
sine.
6
ABC,
∠B
=
40
,
BC
=
14
cm
and
AC
=
10
cm.
Find
the
other
sides
angles.
B
40°
14
cm
Sketch
the
triangle.
C
10
cm
A
Continued
on
next
page
E11.1 Unmistaken identities
24 3
Find
both
possible
values
of
∠
A
of
∠
A
or
If
∠A
=
64.1
⇒
If
∠A
=
c
Reect
How
Does
In
=
15.7
cm
c
=
then
6.36
the
=
∠C
=
Find
24.1
∠
C
∠
C
and
and
side
side
length
length
c
c
for
for
the
the
rst
second
possible
possible
value
value
of
∠
A
cm
lengths
also
Find
75.9
discuss
opposite
this
triangle
,
and
Example
and
∠C
are
angles
●
then
115.9
⇒
●
,
of
them?
hold
5
the
sides
Why
true
for
in
does
a
triangle
this
triangles
make
related
to
the
sizes
of
the
sense?
produced
by
the
ambiguous
case?
7
XYZ,
∠
X
=
65
x
=
23
cm
and
y
=
18
cm.
Find
the
other
sides
angles.
X
65°
18
cm
Sketch
Z
the
triangle.
Y
23
cm
Find
both
possible
values
of
∠
Y
or
If
∠
Y
=
45.2
,
then
∠
Z
=
Find
69.8
∠
Z
values
and
of
constructed
z
If
=
∠
Y
23.8
=
cm
134.8
Therefore,
24 4
the
,
then
only
∠X
+
∠Z
solution
is
is
already
the
one
E11 Equivalence
more
already
than
180
found.
.
length
∠
Y.
Verify
with
of
side
that
each
a
of
z
for
the
triangle
these
two
can
be
possibilities.
G E O M E T R Y
Reect
Although
and
it
ambiguous
there
are
In
sucient
case,
two
Practice
1
is
for
6
this
you
course
nd
a
without
to
check
general
rule
checking
both
to
them
solutions
determine
of
or
not
both?
ABC,
nd
to
3
all
possible
values
for
the
measure
of
angle
=
38
,
a
=
14,
c
=
22
b
A
=
71
,
a
=
10,
c
=
7
c
A
=
62
,
a
=
12,
c
=
3
d
A
=
53
,
a
=
11,
c
=
13
missing
0.1.
a
r
=
5
c
r
=
20
In
triangle
sides
Make
cm,
=
8
q
=
30
ABC,
and
sure
q
m,
C
s.f.
A
the
the
whether
a
Find
T R I G O N O M E T R Y
5
triangle
nearest
3
could
solutions,
Giveanswers
2
discuss
A N D
all
nd
=
∠R
m,
nd
you
∠R
cm,
angles
=
in
all
triangle
possible
30
22
possible
values
PQR.
all
values
to
the
solutions.
b
r
=
26
d
r
=
7
for
Give
cm,
km,
angle
B
q
q
to
=
=
22
13
the
cm,
km,
∠R
∠R
nearest
=
55
= 18
0.1.
C
15
7
26°
4
In
to
5
In
∆ DEF,
the
∆ RST,
tothe
6
e
=
r
=
Determine
f
=
6
m,
and
25
m,
s
=
30
m,
and
how
many
unique
with
∠ P
=
50
,
p
=
10
b
∆ PQR
with
∠ P
=
90
,
p
=
3
c
∆ PQR
with
∠ P
=
115
,
d
∆ PQR
with
∠ P
=
62
p
the
at
at
point
a
Sketch
b
Determine
A
all
with
m
and
the
B.
is
Find
all
possible
solutions
for
∠F
∠R
=
40
.
Find
all
possible
values
of
t
,
p
=
=
are
possible
m,
q
=
6
m
cm,
q
=
2
cm
9
m,
2.8
q
km,
=
3
m
q
=
3.0
for
each
scenario.
km
the
10.8
Seen
a
Determine
b
Calculate
10
km
from
from
the
diagrams
a
ship,
ship
the
at
point
angle
C
and
between
8
the
km
from
a
lighthouse
for
this
distances
situation.
from
the
ship
to
the
sailboat,
to
the
kilometer.
land
long,
is
enclosed
respectively.
by
a
The
fence.
other
Two
side
sides
forms
of
an
the
fence
angle
of
are
41
side.
how
all
a
of
m
m
is
possible
of
plot
7.6
A
48
possible
tenth
triangular
10.8
point
sailboat
nearest
.
solving
lighthouse
and
= 19
triangles
∆ PQR
A
∠E
meter.
a
sailboat
8
m,
0.1.
nearest
Problem
7
7
nearest
many
possible
lengths
lengths
are
to
possible
the
for
nearest
the
0.1
third
side.
m.
E11.1 Unmistaken identities
24 5
Summary
An
identity
is
true
for
all
values
of
the
variable.
Rearranging
equivalent
Trigonometric
the
Pythagorean
identity
gives
two
identities:
identities:
and
When
in
a
using
solutions.
Ratio
identity:
solution
the
Pythagorean
Mixed
the
triangle,
Your
as
sine
rule
sometimes
;
calculator
the
ambiguous
to
other
case
is
for
nd
there
a
are
will
give
180
the
missing
two
.
sine
angle
possible
you
one
This
is
called
rule.
identity:
practice
2
1
Solve
each
equation
for
0
≤ θ
≤ 360
.
Give
exact
4
a
θ
Find
sin θ
if
=
,
for
0
≤
θ
≤
360
2
answers
where
possible;
if
not
possible,
round
to
b
2
d.p.
Check
a
cos θ
c
5 tan θ
=
your
answers
0
3 tan θ
b
=
-2
cos θ
>
0,
Problem
-1
2
=
If
in
which
quadrants
θ ?
is
graphically.
sin θ
d
c
solving
tan θ
Find
if
cos θ
<
0.
=
7
1
5
Find
tan θ
if
os θ
= -
and
sin θ
>
0 .
1
e
3 cos θ
=
2
sin θ
f
2
= 2
2
Without
using
a
calculator,
nd
these
angles
2
6
Prove
the
identity:
sin x
cos x
tan x
7
Prove
the
identity:
sin x
- sin x
8
Prove
the
identity:
≡ 1 - cos
in
2
radians,
for
0
≤ θ
≤
2π .
Give
exact
3
a
sin θ
2
=
b
cos θ
1 - cos θ
+
=
e
cos θ
=
3
d
sin θ
=
f
tan θ
=
sin θ
sin θ
+ cos θ
sin θ
9
3
Prove
equation
answers
for
0
≤ θ
≤
2π
.
a
in
radians
when
c
2 sin
give
=
θ
cos
Solve
≤ θ
these
answers
in
trigonometric
≤ 360° .
Write
your
degrees
to
appropriate.
the
2
-
(2θ)
3
=
b
1
d
5 cos θ
2
cos
=
2
θ
a
sin
b
2 sin
c
2 sin
d
3 sin
3
(3θ)
+ sin θ  =
cos
θ
2
= −1
θ
=
cos θ
2
+ 2
2
θ
+ 3 cos
θ
=
3
2
24 6
2
θ
sin
E11 Equivalence
equations
answer
possible,
0.1
2 sin θ
1
≡
2
4
θ
sin
for
Give
where
nearest
1
+
identity:
sin
0
otherwise
the
3
10
exact
- 1
2
2
each
x
cos θ
1
Solve
sin
≡
cos θ
-1
1
3
≡
2
1 - sin θ
tan θ
3
x
= -
2
c
cos
answers.
θ
- 4 cos θ
- 4
=
0
to
3
s.f.
θ
x
G E O M E T R Y
11
Solve
0
≤ θ
these
2π
≤
radians
if
.
trigonometric
Leave
your
possible,
equations
answer
otherwise
in
for
exact
accurate
13
form
to
3
in
s.f.
Find
and
both
solutions
sketch
accurate
each
to
1
A N D
for
T R I G O N O M E T R Y
∠ B
solution.
for
each
Write
triangle,
your
answer
d.p.
2
a
2 cos
=
3 cos
a
B
1
= 2 sin θ
b
5
cm
sin θ
2
tan θ
c
θ
sin
2 tan θ
=
30°
2
d
2 cos
e
3 cos
f
(
6
θ
+ 3 sin θ
=
cm
3
B
b
2
θ
- 5 sin θ
- 4
=
0
2
2
- 2
) cos θ
- 2
θ
2 sin
= 1 - 2
4
2
cm
44°
Problem
solving
5
cm
1
2
12
Determine
whether
is
θ
- 1 = tan
an
2
θ
cos
identity.
Justify
Review
1
Anouk
a
in
and
standing
20
m
a
40
away
are
looking
and
answer.
context
Timur
shoreline,
measures
your
angle
the
standing
at
an
between
island.
from
30
island
m
out
where
Kelly
is
on
apart
to
Anouk
Timur
the
N
along
sea.
is
island,
B
O
Timur.
35°
a
Find
the
nearest
two
0.1
)
possible
that
angles
Kelly
(to
could
the
measure
250
400
b
between
Timur
Find
two
the
nearest
0.1
between
2
A
canoe
buoy
it
5
and
the
the
the
Anouk
course
dock,
canoe’s
of
and
the
whether
and
possible
answer.
travels
travels
angle
not
A
toward
the
You
will
than
once.
from
3
the
3
while
Lucas
He
from
a
is
is
lost
knows
his
path,
but
walking
A.
the
current
walking
point
that
after
at
east.
a
from
point
B
but
400
the
is
dock,
more
and
than
the
orienteering
line
the
O.
meters
bearing
In
the
an
position,
Lucas’s
away
on
nish
of
is
he
125
nish
nish
says
line.
line
due
He
diagram,
GPS
use
the
sine
rule
more
km.
a
Find
how
b
Explain
away
one
far
east
Lucas
is
from
point
O
is
Lucas
isn’t
nish
A,
how
far
but
line
could
not
Lucas
be
is
be
250m
located
from
the
at
follows
is
he
of
now
only
at
250
walks
m
to
there.
E11.1 Unmistaken identities
B
actual
line.
course.
realizes
he
the
point
east
instead
he
how
from
Determine
nish
solving
to
buoy
c
Problem
have
Determine
12
there
a
buoy,
another
between
is
canoe
or
the
measure
reaching
position
explain
(to
could
Kelly.
dock
Upon
the
angles
Timur
and
the
away.
distance
Anouk.
possible
that
leaves
km
changes
From
)
and
24 7
4
Both
sun
Earth
(S)
in
distance
149
(E)
paths
from
million
from
and
Venus
that
are
Earth
to
Venus
to
the
sun
orbit
nearly
the
kilometers,
(V)
is
circular.
sun
while
around
is
the
a
The
roughly
distance
108
a
system,
approximately
the
Sketch
diagram
showing
(Draw
the
form
triangle.)
a
of
the
planets
this
part
orbits
and
the
of
of
the
the
sun
so
solar
two
that
planets.
they
million
b
An
astronomer
measures
the
angle
SEV
to
be
kilometers.
20
Determine
Earth
Reect
and
How
you
have
at
space
At
can
explored
the
put
laws
of
turn
the
the
his
physics
statement
of
resulting
in
of
the
a
inquiry?
the
that
forces
Earth
due
at
odds
later,
and
the
there
on
gravity
for
with
the
the
of
a
theory,
relativity.
no
dierence
person
and
law
unied
special
was
Einstein
relativity
Einstein
theory
acting
to
Albert
of
between
‘when’.
framework
was
general
between
the
new
theory
gravity
realized
and
century,
years
Einstein
on
20th
but
relationships
‘where’
special
Some
concepts
applying
dene
provided
of
gravity.
the
of
forward
which
and
help
a
standing
person
−2
travelling
This
is
a
in
a
rocket
simplied
accelerating
description
Equivalence
Principle
studying
eect
It
is
a
common
gravity’
aboard
for
the
for
the
The
free
as
mistake
90%
are
in
force
of
toremain
24 8
fall.
the
by
fall.
term
‘zero
Astronauts
Station
fall,
them
is
is
E11 Equivalence
the
as
surface.
because
which
motion.
(ISS),
but
balanced
ISS
ms
inspired
Earth’s
is
9.81
Einstein’s
free
free
upon
the
at
the
Space
earth
of
orbital
in
experienced
the
velocity
in
on
of
was
use
free
acting
that
towards
tangential
to
constant
weightlessness
fall
objects
describing
gravitational
high
of
which
International
example,
point.
discuss
Give
specic
examples.
Statement of Inquiry:
Generalizing
that
by
the
the
allows
it
measurements
in
how
far
Venus
could
be
from
Go
E12.1
ahead
Global
context:
Objectives
●
Developing
●
Using
and
Orientation
Inquiry
the
laws
of
log
logarithms
in
in
space
and
time
questions
●
What
●
How
can
●
How
do
●
Is
are
the
laws
of
logarithms?
F
the
laws
expressions
of
and
logarithms
solve
to
simplify
equations
you
prove
generalizations?
C
Proving
the
laws
●
Proving
the
change
of
logarithms
of
base
you
solve
logarithmic
equations?
formula
D
ATL
change
measurable
and
predictable?
Critical-thinking
Draw
reasonable
Statement
of
conclusions
and
generalizations
Inquiry:
E6 1
Generalizing
that
can
changes
model
in
duration,
quantity
helps
frequency
establish
and
relationships
variability.
E10 1
E12 1
24 9
S P I H S N O I TA L E R
●
Y
ou
●
should
evaluate
already
logarithms
know
using
how
a
1
calculator
to:
Evaluate:
a
log
5
b
In
3
c
log
6
2
x
●
solve
equations
involving
2
Solve
the
equation
3
3
Solve
the
equation
log
=
20.
exponents
●
solve
equations
involving
logs
●
conrm
algebraically
1

and
4

=
7

7
Show algebraically and graphically
x
graphically
that
two
functions
are
x

that f ( x ) = 3x
4 and
+ 4
g( x ) =
are
3
inverses
●
nd
the
of
each
other
inverse
of
a
inverses
logarithmic
5
Find
function
y
=
of
the
each
inverse
2 log
(x + 3)
other.
of
the
function
11.
5
Creating
F
●
ATL
What
are
Exploration
1
Choose
any
evaluate
log
2
laws
of
logarithms?
value
of
a
1
logarithm
log
a
in
log
3
8
+
log
a
10
+
log
81
product
and
use
decimal
your
calculator
to
places.
1000
9
log
a
the
3
a
log
a
Examine
base,
to
32
log
100
a
+
the
table
a
a
log
for
6
log
4
a
log
this
a
a
log
The
the
positive
each
+
generalizations
a
patterns
rule:
and
log
xy
write
down
a
generalized
rule:
=
a
2
Do
the
log
3
same
-
log
a
log
8
-
log
100
-
-
Examine
10
log
log
10
a
log
a
quotient
table.
a
log
the
this
2
log
4
a
81
in
a
a
a
The
expressions
log
2
a
log
the
a
a
log
for
patterns
rule:
243
a
and
log
write
down
a
generalized
rule:
=
a
Continued
25 0
E12 Generalization
on
next
page
N U M B E R
3
Compare
the
values
in
each
column
to
discover
a
pattern.
Scottish
mathematician
and
astronomer
John
Examine
the
patterns
and
write
down
a
generalized
Napier
rule:
(1550
–
1617),
n
The
power
rule:
log
(x
)
=
most
a
4
Evaluate
log
1
for
dierent
values
of
famous
for
discovering
a
a
logarithms,
Write
down
a
generalized
credited
The
zero
Evaluate
log
a
for
dierent
values
of
The
6
down
unitary
generalized
Write
=
solve
this
Reect
for
as
a
and
How
are
Give
two
●
Explain
●
How
fractions.
rule:
rule:
Rewrite
Hence
●
for
a
the
notation
a
a
Write
also
for
introducing
rule:
decimal
5
is
rule:
the
y
y
as
in
logarithmic
terms
of
generalized
discuss
rules
specic
why
a
the
you
statement.
x
rule.
1
discovered
similar
to
the
laws
of
exponents?
examples.
zero
rule
and
the
unitary
rule
are
true.
lnx
Laws
of
can
you
simplify
e
?
logarithms
+
For
all
logarithm
The
product
The
quotient
The
power
The
zero
The
unitary
rules:
x,
y,
a
∈

rule:
rule:
rule:
rule:
rule:
E12.1
Go
ahead
and
log
in
2 51
Example
Rewrite
6
1
ln x
-
3
ln 3
+
2
ln x
as
a
single
logarithm.
The
power
rule:
=
The
product
rule:
log
xy
=
log
a
The
Example
Express
in
=
ln 1
=
ln 1
=
0
quotient
terms
of
x
and
y,
where
x
=
ln a
and
y
=
The
ln (ab)
The
The
ln (ab)
=
Rewrite
(x
+
zero
log 4
+
c
2 log 5
+
y)
expression
as
a
single
logarithm:
log 3
b
log 5
–
d
3 log x
log 2
+
log 6
1
+
3 log 2
+
log 1
-
log y
+
log z
2
e
2
5 ln x
Express
3 ln 2
in
+
terms
4 ln x
of
ln
f
a
and
ln
log
y
+
4 log x
(5 log 2y
ln ab
b
a
e
c
ln ( a
f
ln ( a
b)

b
1

ln
log 3x)
2
ln

d
+
b:
 a 
a

ln
b )
2


g
3
a

h
b

Problem
3
Write
each

a
3

ln

a

a
ln
i
2

b
ln

b
solving
expression
in
terms
of
x
and
y,
where
log
3
=
x
and
log
a
a
log
18
b
a
d
log
g
log
log
2
9
e
log
a
c
162
a
25 2
E12 Generalization
log
0.5
a
36
6
a
a
a
power
rule:
rule.
rule.
.
ln b)
1
each
a
y
a
rule:
quotient
The
1
log
ln b.
ln
(ln a
Practice
+
2
ln
ln
x
a
f
log
a
27
=
y.
product
rule.
N U M B E R
4
Express
in
terms
of
x
and
y,
where
and
x
=
log a
5

2
a
log (100a
)
b
y
=
log b

a
3
b
and
log
4

3
c
log
6
1000 a
8
d
Find
the
the
10 a
b

3
inverse
algebraically
of

log

Example
1

b

100b
of
the
function
using
the
laws
function
and
its
of
g ( x )
=
ln x
logarithms.
ln (x
State
the
3)
and
conrm
domain
and
it
range
inverse.
Use
the
quotient
Interchange
Solve
for
y,
and
x
rule.
and
simplify.
x
=
ln e
=
x
=
g
1
Since
g ( g
1
( x ))
Domain
of
g
Domain
of
g
is
=
x
x
>
3;
( g ( x )),
range
of
g
they
is
1
y
are
>
inverses.
0.
1
is
x
>
0;
range
of
g
is
y
>
3.
E12.1
Go
ahead
and
log
in
y.
253
Exploration
1
Describe
the
2
transformations
on
the
graph
of
f
( x )
=
log
x
that
give
the
x
that
give
the
3
graph
of
g ( x )
=
log
(9x).
3
2
Describe
the
transformations
on
the
graph
of
f
( x )
=
log
3
graph
of
h ( x )
=
2
+
log
x
3
3
Graph
down
4
a
the
functions
what
you
Describe
the
the
graph
f,
g
notice.
and
h
on
Explain,
horizontal
the
same
using
dilations
coordinate
laws
that
of
axes,
and
write
logarithms.
transform
the
graph
of
f
into
of:
i
y
ii
=
log
(27x)
3
b
By
using
that
the
laws
transform
of
the
logarithms,
graph
of
f
describe
into
the
i
the
horizontal
graph
of :
ii
log
y
=
translations
(27x)
3
5
Summarize
what
transformations
Practice
1
Find
the
its
inverse
and
c
f
learned
and
in
this
Exploration
translations
with
about
the
logarithmic
functions.
of
each
function
algebraically.
and
State
prove
the
that
domain
it
is
and
an
inverse,
range
of
both
the
function
inverse.
x
a
have
dilations
2
graphically
and
you
of
(x )
h( x )
=
=
3
2
4 log (1 + 2 x )
b
g (x )
=
d
j (x )
=
ln ( 2 x )
ln ( x
+ 2) -
ln ( x
- 1)
n
2
Use
the
vertical
laws
of
logarithms
dilation
of

3
Given
f (x ) =
the
9
to
graph
explain
of
f
( x )
=
why
the
log x
graph
for
any
of
g ( x )
positive
=
log x
is
a
n.

,
log
use
properties
of
logs
to
nd
an
equivalent
function.
3

Describe
the
x

transformations
on
the
graph
of
y
=
log
x
that
give
the
graph
3
of
the
equivalent
Problem
4
a
Use
solving
properties

i
g (x ) =
function.
of
8
logs
to
nd
an
equivalent
expression

ii
h( x )
=
log
2
b
Then
describe
x
function:
(x
+ 3)
2

the
each
2
log

for
sequence
of
transformations
on
the
graph
of
y
=
log
x
2
that
give
25 4
the
graphs
of
g
and
h
E12 Generalization
N U M B E R
Proving
C
●
How
can
the
you
generalizations
prove
generalizations?
c
If
a
and
b
are
equivalent
to
positive
log
b
=
real
c.
numbers,
Does
this
then
mean
the
that
exponential
the
exponent
statement
rules
are
a
=
also
b
is
related
a
to
ATL
the
logarithm
rules?
Exploration
1
Proving
Let
log
the
x
=
3
product
p
rule
and
.
a
a
Write
both
b
Multiply
c
Write
d
Substitute
these
the
expressions
two
expressions
as
exponential
together
to
statements.
obtain
an
expression
for
xy.
Internet
this
expression
in
logarithmic
engines
and
every
.
to
You
should
have
the
product
Proving
the
quotient
rank
website
determine
rule.
an
2
search
form.
approximate
measure
rule
of
important
Let
log
x
=
p
and
is.
Write
both
these
expressions
as
exponential
Divide
c
Write
to
obtain
an
expression
d
expression
Substitute
in
logarithmic
and
form.
should
have
the
a
5
Proving
the
power
is
100
with
site
quotient
rule.
The
5
3
this
2
Raise
3
Write
4
Substitute
You
should
Reect
●
this
How
=
2
expression
expression
of
on
as
an
exponential
to
in
the
power
order
statement.
n.
logarithmic
an
the
scale,
magnitude
power
of
is
a
10.
form.
.
have
the
power
discuss
you
use
a
rule.
2
graphing
tool
with
a
slider
to
demonstrate
the
2
power
●
What
rule
for
f
( x )
=
log (x
transformation
takes
)?
the
graph
of
f
( x )
=
log
x
to
the
graph
of
2
f
( x )
=
log (x
)?
of
of
signies
logarithmic
expression
and
could
than
rank
rule
of
this
a
dierence
orders
where
Write
rank
.
.
1
a
times
a
magnitude
Let
a
scale,
with
popular
two
3
site
more
3.
You
is
for
of
this
Ranking
logarithmic
statements.
so
b
site
.
a
a
how
the
Explain.
E12.1
Go
ahead
and
log
in
25 5
Example
4
Express
as
a
single
logarithm.
The
The
power
rule.
product
rule.
Expand.
Practice
1
Express
3
as
a
4 log
c
3 ln ( x
a
single
logarithm:
p + 2 log q
b
n log
p
d
ln x
f
4 ln x
3 log q
1
- 2) -
ln x
- 2 ln
( x
- 2)
2
e
2
1
Let
ln x
a
=
ln x,
x

a
b
=
ln (x
1)
and
c
=
ln 3.
Express

x
1
ln
(3 x
terms
- 1)
of
a,
b
and
c :
ln
(x
x )

x
2
c
( x
2
b
ln

in
+ 2 ln
)
d
+ 1
ln
x
2
e
ln
3x
+ 6x
Problem
3
Express
as
+ 3
solving
a
sum
and/or
dierence
linear
logarithms:

2
2
a
of
ln ( x
b
9)
ln ( x
+ 5x
+ 6)
c


d

+ 1

can
2
x
4
use
a
2
x
- 5x
-
6 

ln
e

You
x
ln
f
x
3
x
+

calculator
to
nd
x

4
the
of

4
x
3
- 3x
5x

logarithms
4
2
- 10 x

20

ln
2


ln
any
number
with
any
base.
You
Until
quite
recently,
calculators
could
not
do
this.
Books
of
log
tables
the
logarithms
to
base
10,
and
mathematicians
used
the
change
of
base
may
formula
change
Any
are
logs
positive
the
(base
most
10)
simplifying
large
25 6
other
value
useful
and
Logarithms
very
to
the
date
bases.
can
for
be
‘natural’
back
to
astronomical
numbers
the
base
practical
of
logarithm
1614,
a
logarithm,
applications
when
e).
John
Napier
E12 Generalization
which
but
the
(base
calculations
together.
are
often
the
two
bases
‘common’
took
an
interest
involved
that
logarithm
in
multiplying
of
used
base
to
formula
calculate
have
gave
in
E6.1.
N U M B E R
ATL
Exploration
4
How
log
to
calculate
5
using
logs
to
base
10
2
To
nd
when
Let
you
,
1
Write
all
2
Equate
3
Substitute
4
Equate
,
three
the
know
for
a
in
expressions
to
get
:
and
expressions
two
and
an
exponential
for
b
to
identity
get
an
form.
identity
involving
z
c
x
involving
and
a
c
x
and
y.
y.
In
the
exponents
to
nd
an
identity
linking
x
y
and
z.
step
3,
use
exponential
the
form
of
statement
5
Rearrange
6
Substitute
the
7
to
the
change
Use
values
Change
of
make
of
the
base
.
subject.
logarithmic
from
base
x
expressions
for
x
y
and
z.
You
should
have
formula.
the
log
table
below
to
nd
the
value
of
.
formula:
x
log
x
10
Practice
1
Use
a
the
log
log
table
to
nd
each
value
7
log
b
to
2
0.301029996
3
0.477121255
4
0.602059991
s.f.
5
0.698970004
6
0.77815125
7
0.84509804
8
0.903089987
9
0.954242509
8
5
10
log
3
log
d
log
3
e
0
4
2
c
1
5
9
2
f
log
2
5
4
10
Verify
your
Example
Given
that
a
express
b
express
results
with
a
1
calculator.
5
ln
a
=
4:
as
a
simple
as
natural
a
single
logarithm
logarithm.
a
The
power
Change
ln a
Continued
E12.1
on
Go
next
=
4
of
is
rule.
base.
given.
page
ahead
and
log
in
2 57
b
=
Practice
1
Given
5
that
ln a
=
5,
express
as
a
simple
natural
logarithm:
3
a
log
(x
2
)
b
log
a
(x
1
x

c
log
x
d
Given
a
that
ln a
=
1,
express
as
a
single

log
a

2
2
)
a
3

logarithm:
3
a
ln ( x
) + 2 log
x
b
2 ln x
3 log
a
1
c
3 log
x
+
log
a
x
d
A
●
How
●
Is
do
3
2
log
of
Example
both
x
equations
you
change
technique
logarithms
3
a
Solving
useful
x
2 ln
a
2
D
x
a
for
solve
logarithmic
measurable
solving
and
equations?
predictable?
equations
involving
exponents
is
to
take
sides.
6
Solve
by
taking
logarithms
of
both
sides.
The
power
rule.
Isolate
x
=
2.32
(3
s.f.)
Reect
and
Show
you
the
that
original
prefer?
Use
discuss
can
obtain
equation
as
a
3
the
same
result
logarithmic
Explain.
25 8
E12 Generalization
in
Example
statement.
6
Which
by
rst
rewriting
method
do
you
a
calculator
to
x
evaluate.
N U M B E R
Example
A
7
mathematician
who
have
been
very
dicult
25%
of
has
,
Find
the
the
determined
exposed
to
to
a
where
appoint
population
maximum
is
an
has
that
news
the
city
unbiased
read
number
of
the
story
the
number
after
t
population.
jury
to
of
days
people
is
A
given
P
by
lawyer
determine
guilt
in
a
the
knows
in
a
city
function
that
crime
it
is
after
news.
days
available
to
select
a
jury.
25%
of
the
population
Take
There
is
a
maximum
of
9
days
available
to
select
a
natural
implies
logs
solve
equations
that
contain
logarithms,
you
often
need
to
use
the
laws
a
the
x
log
8
value
+
log
2
b
log
of
3
x
=
log
each
log
+
1)
x
+
+
log
log
the
argument
the
answer.
and
equation.
9
(x
3
=
2)
log
2
log
=
1
9
2
(3x)
=
log
2
9
The
(x
+
product
rule.
2
3x
=
9
x
=
3
The
log
has
base,
4
2
b
the
2
4
a
in
2
(x
logarithm
parts:
rst.
Example
Find
sides.
of
three
logarithms
both
jury.
Each
To
of
1)
+
log
4
(x
2)
=
1
2)]
=
1
2
=
4
6
=
0
2)
=
0
x
=
3
arguments
of
the
logarithms
must
be
equal
to
each
other.
4
[(x
log
+
1)(x
The
product
rule.
4
2
x
1
x
–
2
x
(x
x
≠
-2
since
x
3)(x
the
+
logarithm
or
log
4
Therefore,
x
=
2
(
2
2)
=
log
(
4)
doesn’t
exist.
4
3
E12.1
Go
ahead
and
log
in
25 9
Practice
1
Solve
6
these
equations
by
taking
logs
x
2
=
25
x
b
7
b
4
b
7 × 4
5x
=
41
c
12
=
5x
2 x
3
=
Solve
45
these
equations
for
=
25
Solve
9
2 x
=
9
these
exponential
+ 1
c
3
=
c
5 × 3
60
x :
2 x
3 × 4
a
4
sides:
Solve:
a
3
both
x
3
a
of
+ 1
3x
=
89
1
=
52
equations:
Look
2x
a
5
x
2
7
Find
a
the
log
x
×
+
value
+
log
6
of
log
5
b
2 x
2
(3x)
10
log
+
log
+
1)
–
1)
x
A
=
+
1
A:
use
state
day
log
log
t
12
log
(2x
+
+
4)
=
-2
(x
+
6)
=
3
(x
5)
1)
and
understanding
correctly
rules
you
have
in
a
variety
found
to
of
solve
contexts
these
context
problems
involving
equations.
surveys
school
children
years
the
to
trac.
At
the
end
of
2010,
there
were
25 000
cars
per
school.
number
of
cars
C
was
modelled
by
the
exponential
0.07 t
equation
C
a
that
Show
=
children
b
By
7
The
to
taking
taking
time
25 000 × e
by
the
end
of
2015,
there
were
35 477
cars
per
day
taking
school.
logs
of
children
taken
both
to
for
sides,
school
an
calculate
reaches
online
video
the
time
until
the
number
of
cars
50 000.
to
go
viral
can
be
modelled
by
an
0.5 t
exponential
watched
a
Describe
b
By
By
is
is
logs
rewriting
the
Determine
2 6 0
the
of
the
H
= 15e
number
the
watched
calculate
d
t
what
taking
video
c
function,
and
value
both
,
of
15
500 000
which
of
H
is
since
represents
calculate
the
the
in
this
the
number
rst
of
people
times
the
video
shared
the
video.
is
model.
number
of
hours
before
the
times.
exponential
number
where
hours
sides,
statement
hours
method
(b
before
or
c)
E12 Generalization
quadratic
equations.
equation:
(x
Knowing
the
taking
After
0
7
log
problems
exponential
6
=
20
Objective
can
+ 24
2
log
You
=
for
x
= 10 × 4
3
(x
solve
log
12
20
iii.
each
4
5
2
e
in
=
3
d
b
7
(x
log
x
0
5
7
c
=
is
as
the
a
logarithmic
video
most
is
statement,
watched
ecient.
1 000 000
times.
N U M B E R
Problem
solving
Rearrange
8
to
get
all
Solve:
the
5x
2
2 x
x
2
a
=
5
1
x
3
b
=
x
+ 1
9
x
2 x
If
log
x
=
p,
show
that
rewriting
it
as
a
+
2
3
d
=
2 x
=
terms
side,
4
to
c
x
−
then
isolate
on
one
factorize
x
2
9
logarithmic
statement
and
solving
a
the
resulting
equation
proves
the
power
rule
for
logarithms.
Summary
Laws
of
logarithms
The
zero
rule:
The
unitary
+
For
all
logarithm
The
product
The
quotient
The
power
rules:
x,
y,
a
∈

rule:
Change
Mixed
1
of
base
formula:
rule:
rule:
practice
Rewrite
in
terms
of
log a,
log b
and
log c :

a
rule:
log (abc)
b
c
log ( a
e
log
2
b
4
c
)
Express
in
and
log b .
y
=
terms
d
log
b

(a
b
a
log (a
given
that
x
=
log
b)
b
log (10
c
log (0.01a
d
log
4
)
2
3
ab
)
4
b
2
)

c

Express
as
a
2 log 5 +
b
3 ln 5 −
c
3 log
a
single
log
4 −
logarithm:

ln
4 +
Find
the
ln 8
5
b
5 +
log
4

the
inverse
its
1

inverse.


2
log
4 −
1
each
function
of
Prove
2 x
a
a

and

log
a

of
range
both
they
the
are
and
state
function
inverses
both
8

+
a
and
4
graphically
2 log

domain
and
5 − 2 log
4
7
50 a
log 8
5
d
y,
a

2
and
5

b
x

3

of
log

3
ac
4
f
(x )
=
algebraically.
+ 1
6 (5
)

4
3
b
e
6 ln x
f
log
2 ln (x
+
1)
3 ln (2x
+
g (x )
=
log
x
− 2
6)
2
c
2 + 3 log
4
1
g
16 +
log
x
− 2
log
7
2
d
y
=
log
(7 x ) −
log
4
6
that
log
3
=
x
and
log
a
expression
log
0.25
4
=
y
write
(x
+ 5)
4
terms
of
x
and
b
y
Use
properties
equivalent
describe
log
of
logarithms
expression
the
of
for
sequence
the
graph
y
the
equivalent
=
log
of
x
to
each
nd
an
function.
Then
transformations
that
give
the
3
48

1
12
for
f
and
g

d

expressions

log
144
a
a
f
( x ) =
( 27 x )
log
3
4

e
27
b
g
( x ) =
( x
log

− 9)
3
log
a

16

E12.1
Go
ahead
and
log
in
on
graphs
a
log
a
down
a
in
a
c
k (x )
7
3
Given
a
− 1)
1
log
each
ln(3 x
3
7
3
h( x ) = −
3
4
2 61
of
a
7
Let
a
=
ln x,
b
=
ln (x
+
2)
and
c
=
ln 4.
13
Let
log
3
=
x
and
log
a
in
Express
4

a
in
terms
of
a,
b
a
+
2

c
ln ( 4 x

x
and
y,
=
y.
Find
expressions,
for:
7
b
log
3
+ 8x )
3
c
log
7
49
3
Given
that
ln a
=
8:
2
x
ln
+
4 x
+
4
3
a
ln
d
express
log
(x
)
as
a
simple
natural
a
16 

x

14
3
log
2
b
x
of
c :

ln

and
7
a
terms
4 x
logarithm
2
4
e
ln
b
3
x
express
3 log
x
ln( a
)
as
a
single
a
+ 2x
logarithm.
Problem
solving
15
Reproduction
in
a
colony
of
sea
urchins
can
0.03 x
be
8
Express
as
a
sum
and/or
dierence
modelled
by
the
N
equation
= 100e
of
where
N
is
the
number
of
sea
urchins
and
x
is
logarithms:
the
ln ( x
c
ln
25)
b
ln( x
d
ln
- 3x
introduced
- 28)
a
x


+ 1
x
8
2


x
-
4 x
days
since
the
original
100
were
the
the
reef.
number
of
days
it
takes
for
the
- 5 
to
double.
2
x

+
6x
+ 8 
b
x
Find
to
population

2

e
of
2
2
a
number
- 10 x
Find
how
many
sea
urchins
were
born
on
24 
-
ln
day
19.
3
x

9
Solve
these
2 x
a
4 x

exponential
+ 1
Problem
equations:
x
3
+ 3
16
= 10 × 3
A
of
2 x
b
solving
mathematician
people
calculated
infected
after
the
that
rst
the
case
number
of
a
x
2
- 3 × 2
- 4
=
0
disease
was
identied
followed
the
model
0.001t
10
Solve
these
logarithmic
equations:
P
=
P
(1 - e
)
where
P
is
the
number
of
0
infected
a
log
x
+
log
4 -
a
b
log
log
a
x
-
log
7
4
5
=
log
a
=
log
x
+
log
9
(x
2
3
) +
log
9
(x
and
t
is
was
identied.
1 -
log
( x
+ 3) =
log
log
(x
)
=
5
a
9
( x
log
- 5) -
( x
log
Calculate
log
2
=
x
and
log
a
and
+ 4
) =
5
expressions
=
log
y.
Find,
in
terms
In
a
of
2
city,
b
log
40
a
400
d
a
A
game
log
log
by
taking
logs
of
both
sides:
x
19

=
2
5x
2
2 6 2
1

= 25
b

c
city,
take
for
case
it
be
would
5%
2
2

5x
=
80
of
infected.
people
the
are
infected
population
after
of
in
in
Swaziland
2000,
the
and
number
26
of
recorded
15
rhinoceros
rhinoceros
in
in
2011.
the
2.5
nd
x
a
58 187
park
follows
0.4
Solve
the
rst
a
a
12
long
to
Calculate
park
rhinoceros
game
e
of
the
city.
Assuming
log
after
x
the
5
c
population
days
for :
17
a
the
of
-1
a
y,
how
population
21days.
11
is
number
x
b
e
the
4
) +
9
the
d
P
0
4
2
c
people,
12
a
d
3
6
= 111
E12 Generalization
the
function.
an
exponential
function,
N U M B E R
Review
In
E6.1
and
modelling
the
in
E10.1
the
Richter
context
you
used
magnitude
M
this
of
formula
an
for
2
earthquake
using
The
pH
value
whether
scale:
value
than
I
a
of
7
7
is
of
a
solution
solution
is
is
neutral,
basic.
To
basic
less
is
used
to
(alkaline)
than
calculate
7
is
the
determine
or
acidic.
acidic,
pH
of
a
and
A
pH
more
liquid
you
c
M
=
log
10
need
I
to
know
the
concentration
of
hydrogen
ions
n
+
(H
where
I
is
the
intensity
of
the
‘movement’
of
)
in
moles
per
liter
(mol/l)
of
the
liquid.
the
c
earth
from
the
earthquake,
and
I
is
the
intensity
of
The
n
the
‘movement’
on
a
normal
day-to-day
basis.
pH
is
then
calculated
using
the
logarithmic
mol/l.
Determine
formula:
+
You
can
also
write
this
as:
pH
=
- log
(H
)
10
M
=
log
I
10
7
+
a
Milk
has
H
1.58
=
× 10
I
c
where
I
is
the
intensity
ratio
whether
milk
is
acidic
or
basic.
I
n
b
1
a
The
San
Francisco
earthquake
of
Find
the
pH
of
+
Shortly
afterward,
America
greater
Find
magnitude
had
than
the
an
8.3
an
on
San
magnitude
Richter
earthquake
intensity
the
the
ratio
Francisco
b
A
recent
7.5
on
more
in
on
South
the
in
four
South
c
scale
Richter
intense
the
sample
H
earthquake.
Richter
A
of
pool
=
6.3 × 10
sample
of
water
has
concentration:
7
+
times
of
mol/l.
pool
Determine
the
pH
of
the
water.
the
Some
chemicals
are
added
to
the
water
because
America.
earthquake
the
concentration:
mol/l.
scale.
d
earthquake
with
3
= 1.58 × 10
H
measured
vinegar
1906
in
Afghanistan
scale.
San
Find
how
Francisco
measured
many
times
the
answer
the
water
is
taken
in
is
c
indicates
not
after
that
optimal.
adding
the
A
the
sample
concentration
is
H
Determine
the
level
of
chemicals,
+
earthquake
pH
of
water
and
now
the
8
= 7.94 × 10
mol/l.
was.
c
Find
how
much
magnitude
is,
if
larger
it
is
20
an
earthquake’s
times
as
levels:
intense
on
7.0
if
<
pH
water
<
is
now
within
optimum
7.4.
the
+
e
Richter
scale
as
another
Find
the
range
d
Determine
how
earthquake
whose
much
more
intense
is
scale
than
magnitude
an
for
H
is
7.8
earthquake
on
7.0
<
pH
<
that
give
a
pH
in
the
7.4.
an
3
Richter
values
earthquake.
The
volume
of
(db)
and
the
sound
D
is
measured
in
decibels
the
I
is
intensity
of
the
sound
measured
whose
2
in
magnitude
is
Compare
the
,
using
intensities
of
two
An
anti-theft
car
13
in
2008
with
magnitude
6.0
Sichuan
in
2008
with
magnitude
Jill’s
scream
Find
scream
How
you
have
alarm
has
an
intensity
Find
the
volume
of
the
was
measured
how
than
much
56
db,
more
and
Jack’s
intense
the
statement
Jack’s.
of
inquiry?
Give
specic
examples.
Statement of Inquiry:
Generalizing
that
can
changes
model
in
duration,
quantity
helps
frequency
establish
and
alarm
relationships
variability.
E12.1
Go
ahead
and
log
in
was
Jill’s
discuss
explored
of
2
W/m
decibels.
48db.
and
I
7.6.
b
Reect
= 10 log
and
in
Eastern
D
formula
earthquakes:
5.8 × 10
Nevada
the
10
a
e
Watts/m
6.2.
2 63
Are
we
very
similar?
E13.1
Global context: Personal and cultural expression
Objectives
●
Idntifying
Inquiry
ongrunt
triangls
asd
●
on
questions
What
onditions
ar
nssary
to
F
standardritria
stalish
●
●
Using
ongrunt
and
similar
triangls
Why
is
ongrun?
‘Sid-Sid-Angl’
onditionfor
provothr
gomtri
not
a
to
ongrun?
rsults
CIGOL
●
How
an
ongrun

usd
to
prov
C
othr
●
rsults?
Whih
is
mor
usful:
stalishing
D
similarity
ATL
or
stalishing
ongrun?
Communication
Undrstand
and
us
mathmatial
notation
13 1
Statement of Inquiry:
Logi
our
an
justify
appriation
gnralizations
of
th
that
inras
asthti.
13 2
E13 1
2 6 4
G E O M E T R Y
Y
ou
●
should
use
already
Dynamic
(DGS)
to
Geometr y
draw
basic
know
how
Software
1
a
Use
shapes
of
DGS
side
Use
reections,
rotations
2
translations
=
Describe
that
a
A
d
D
to
with
∠
ACB
and
to
construct
lengths
DGS
ABC
recognize
T R I G O N O M E T R Y
to:
b
●
A N D
4,
80°
=
and
triangle
7
cm.
triangle
40°,
BC
single
a
and
construct
∠
ABC
the
6
=
5
cm.
transformation
takes:
to
to
B
b
A
E
e
B
to
to
C
c
D
f
C
E
to
to
D
B
y
6
B
A
4
C
2
x
0
6
4
2
2
4
6
2
4
6
●
nd
angles
in
parallel
lines
3
Find
of
the
the
marked
and
●
nd
lengths
in
similar
triangles
4
size
110°
angles
a
Triangles
Find
b
a
b
ABC
lengths
and
AC
DEF
and
are
similar.
ED
B
E
5
3
Establishing
F
●
What
●
Why
onditions
is
Exploration
congruent
ar
nssary
‘Sid-Sid-Angl’
not
a
to
cm
cm
triangles
stalish
ondition
ongrun?
for
ongrun?
If
1
you
ass
1
Us
dynami
gomtry
softwar
to
do
to
triangl
A
with
sids
5
b
triangl
B
with
a
c
triangl
C
with
angls
m,
6
m
and
7
dynami
onstrut
m
triangls
sid
of
8
m
and
all
angls
60°
and
softwar,
ths
auratly
60°
using
45°,
hav
onstrut:
gomtry
a
not
75°
pnil,
ompasss
and
protrator.
d
triangl
D
with
angls
40°,
60°
and
80°
and
a
sid
of
5
m
Continued
on
next
page
E13.1 Are we very similar?
2 6 5
e
triangl
E
with
sids
f
triangl
F
with
on
nd
g
2
G,
of
a
siz
your
xatly
mor
Two
4
m
than
shaps
on
ar
sid
4
triangl
triangls
th
m
and
m,
8
m,
and
and
with
an
angl
angls
of
of
50°
30°
and
45°
at
ah
sid
right-angld
Compar
drw
sam.
with
with
othrs.
Disuss
possil
on
sid
4
m
Idntify
whih
of
th
and
whih
hypotnus
triangls
dsriptions
a
5
m.
vryon
to
g
lad
to
triangl.
congruent
if
thy
hav
th
sam
sid
lngths
and
th
sam
angls.
Reect
●
th
6
Ar
and
th
discuss
triangls
in
ah
1
pair
ongrunt?
a
b
3.5
2.5
2
2.5
3.5
2
2.5
3.5
2
3.5
c
d
60º
2
2
60º
40º
70º
45º
●
●
Look
at
th
diagrams
you
3
drw
rat
triangls
whih
ar
rat
triangls
whih
might
What
sam
Two
do
you
all
40º
70º
45º
in
always
triangls
not
that
Exploration
ongrunt,

3
1.
and
Whih
whih
dsriptions
dsriptions
ongrunt?
hav
th
sam
angls,
ut
ar
not
th
siz?
shaps
rtion,
ar
congruent
rotation
or
if
on
an
translation,
or

a
transformd
omination
into
of
th
othr
ths.
y
a
Th
word
oms
Latin
working
with
ongrunt
triangls
you
nd
to
idntify
sids
and
angls:
C
F
A
2 6 6
E13 Justication
D
‘I
agr’
th
‘I
corresponding
th
Congruō,
maning
Whn
congruence
from
om
togthr’.
or
G E O M E T R Y
Δ
DEF
D
is
DE
is
th
is
th
∠
DEF
ATL
Th
th
≅
is
symol
so
Using
To
th
A
this
and
of
of
notation
and
AB,
of
≅
rst
shap
nd
th
sids,
Example
or
DE
lttrs
Δ DEF
to
hlps
into
in
th
ar
and
so
D
AB
of
In
th
triangls
ar
ongrunt.
vrtis.
∠
ABC
ar
onvntion
‘triangl
your
you
sond,
ABC
to
E ,
using
is
sids.
orrsponding
for
ongrunt
angls.
triangl
rotation,
an
C
to
triangl
orrsponds
to
F ’.
larly.
idntify
You
ongrunt
and
working
an
vrtis.
largst/smallst
two
orrsponding
and
orrsponds
show
th
th
T R I G O N O M E T R Y
important.
vrtis,
th
imags
so
ar
∠
DEF
mans
B
you
is
lin,
orrsponding
ongrun.
orrsponding
th
so
A
∠
ABC,
th
Δ ABC
Δ
ABC
D
rprsnts
ordr
thn
longst
of
so
orrsponds
idntify
taks
A,
rtion
statmnt
DEF,
of
rtion
th
notation
Th
rtion
imag
A N D
th
transformation
rtion
also
math
or
that
translation,
pairs
of
shortst/
angls.
1
N
Δ
PQR
and
Δ
LMN
orrsponding
ar
ongrunt.
Idntify
th
sids.
Q
L
P
M
R
Δ
PQR
LN
a
rotation
orrsponds
∠
L
MN
LM
is
orrsponds
orrsponds
The
to
to
∠
P.
QR.
The
side
opposite
shortest
side
of
each
triangle.
The
obtuse
angle
in
each
triangle.
the
obtuse
angle
in
each
triangle.
PQ.
The
last
pair
of
sides.
1
qustions
1
Δ
LMN
PR.
to
orrsponds
Practice
In
to
of
1
to
4,
idntify
B
th
orrsponding
sids
ah
pair
of
triangls.
I
2
E
in
L
4
4
K
3
5
5
F
A
C
D
G
H
J
3
N
3
O
X
4
Q
T
V
R
U
M
P
S
W
E13.1 Are we very similar?
2 67
5
Idntify
symol
th
pairs
notation,
of
ongrunt
with
th
triangls
in
orrsponding
this
diagram.
vrtis
in
th
Us
ongrun
orrt
ordr.
L
W
K
N
V
O
J
X
P
Q
M
B
C
S
F
T
R
D
H
I
A
G
To
show
on
of
that
th
two
triangls
onditions
Exploration
Th
two
1
aim
of
this
triangls
Choos
Now
try
original.
2
Rord
a
to
ongrunt,

is
to
disovr
thr
sid
triangl
lngths
with
onstrut
anothr
whthr
rsults
from
sides
3
angles
2
sides
angle
show
that
thy
satisfy
what
information
guarants
that
or
stp
ould
sid
triangl
not
1
it
in
of
is
a
a
triangl.
that
is
not
ongrunt
to
th
possil.
tal
the
mak
lngths.
two
lik
this:
triangles
and
angles
an
and
and
between
Now
look
did
on
triangls.
3
Always
elements)
th
tal.
YES
or
congruent?
NO
them
angle
them
the
between
angles
not
the
the
between
side
3
and
between
sides
not
2
to
elements
3
2
that
ths
(indicate
2
nd
ongrunt.
Sketch
3
you
2
Disuss
th
ar
ongrun.
xploration
will
any
Construt
for
them
a
side
them
at
th
othr
rows
masurmnts
Rord
of
for
whthr
or
th
not
In
lmnts
it
is
th
and
possil
sam
try
in
to
th
way
as
in
onstrut
stp
right-hand
olumn.
Continued
2 6 8
E13 Justication
1,
dirnt
on
next
page
G E O M E T R Y
4
Basd
to
5

on
your
Compar
th
two
sids
B,
two
th
oth
in
in
two
Reect
for
two
triangls
drivd
rsults
from
Exploration
1.
th
sam
th
angl
two
ongrun
B,
and
A
two
ar
is
th
twn
th
is
th
triangls
sids
lngths
lngths
triangls
(ASA
thr
triangls,
sam
as
th
thr
sids
of
(SAS):
ar
th
triangl
th
ongrunt.
and
in
or
sam
a
thos
ar
pairs
qual
of
sids
is
th
ongrunt.
siz
as
triangls
lngth
orrsponding
AAS):
ongrun
hav
two
two
orrsponding
two
sam
as
ar
orrsponding
sid
lngth
is
(RHS):
hypotnuss,
in
whos
ongrunt.
oth
and
triangls,
on
th
of
th
two
ongrunt.
and
how
discuss
th
using
Right
ah
2
Angl-Hypotnus-Sid
of
th
othr
ongrun
ongrun
ondition
an
onditions.
3
Draw:
a
a
lin
b
a
irl
c
a
lin
In
Δ ABC,
sgmnt
of
AB
radius
from
AB
A
=
th
7
that
6
of
lngth
m
m,
BC
m
ntrd
maks
position
6
=
of
an
7
on
angl
m
point
of
and
C
B
40°
with
∠
CAB
on
your
=
AB.
40°
diagram.
Draw:
a
a
lin
b
a
irl
c
a
lin
In
Δ DEF,
Explain
your
5
B,
A
ar
ar
ongrun
oth
ar
Dtrmin
4
A
triangls,
Exploration
3
onditions
(SSS):
triangls
right-angld
triangls
2
nssary
your
Angl-Hypotnus-Sid
rmaining
Explain
to
triangl
triangl
triangl
sam
Right
If
1
of
two
of
angls
angls

th
triangl
in
hr
ongrun
Angl-Sid-Angl
If
th
congruence
sids
sids
of
sam
●
rsults
Sid-Angl-Sid
If
●
for
thr
triangl
●
your
Sid-Sid-Sid
If
disuss
T R I G O N O M E T R Y
ongrunt.
Conditions
●
rsults,
A N D
sgmnt
of
radius
from
DE
why
DE
D
=
it
4
that
6
is
of
m
EF
6
m
ntrd
maks
m,
not
lngth
=
an
4
angl
m
possil
on
to
E
of
and
40°
with
∠
FDE
dtrmin
=
th
DE.
40°
position
of
point
F
on
diagram.
Explain
why
Sid-Sid-Angl
(SSA)
is
not
a
ondition
for
ongrun.
E13.1 Are we very similar?
2 6 9
Example
Prov
that
2
in
this
diagram,
Δ ABC
≅
Δ DEC.
C
E
Givn:
AB
and
ED
ar
D
paralll
Separate
AB
Prov:
=
are
ED
Δ ABC
≅
the
given,
question
and
the
into
part
you
the
part
need
you
to
prove.
Δ DEC
Proof
∠
BAC
=
∠
EDC
(altrnat
angls)
∠
ABC
=
∠
DEC
(altrnat
angls)
AB
=
DE
Give
a
reason
(givn)
for
Both
sign
Thrfor
each
Δ ABC
≅
Δ DEC
th
(=)
statement.
quals
and
ongrun
ar
usd
that
Example
3
ar
to
qual.
that
in
a
rtangl
ABCD,
Δ ABC
≅
Δ CDA
us,
sign
or
angls
Both
in
though
ar
gnral
som
xaminations
A
(≅)
indiat
lngths
aptal
Prov
th
(ASA)
or
B
shools
you
to
instad
D
might
us
of
xpt
on
th
othr.
C
Draw a diagram.
Givn:
ABCD
is
a
rtangl.
Separate
Prov:
Δ ABC
≅
the
question
into
the
part
you
Δ CDA
are
given,
and
the
part
you
need
to
prove.
Proof
∠
ABC
ar
=
∠
CDA
right
angls.
Δ ABC
and
aus
AB
≅
90°
Δ CDA
oth
CD
=
ar
hav
aus
Give
right-angld
hypotnus
th
opposit
triangls
with
qual
≅
Δ CDA
for
each
statement.
hypotnuss
sids
of
rtangl
ar
(RHS)
are
qual.
could
equal
opposite
27 0
reason
AC
You
Δ ABC
a
E13 Justication
also
(SSS)
sides
prove
or
by
and
this
result
using
the
the
right
by
showing
two
angle
all
congruent
in
between
the
sides
pairs
of
them
(SAS).
G E O M E T R Y
Objective
v.
C:
Organiz
In
Practice
you
1
4,
possible
using
a
organize
logial
your
str utur
answers
whthr
onditions
for
th
triangls
in
ah
pair
b
c
d
Prov
to
help
make
sure
that
that
ar
ongrunt.
If
thy
ar,
stat
ongrun.
a
a
systematically
triangles.
2
Dtrmin
th
2
question
all
Practice
T R I G O N O M E T R Y
Communiating
information
2,
include
A N D
Δ PQR
is
ongrunt
to
Δ STR
S
Q
R
T
P
b
Prov
that
Δ GHK
≅
G
Δ JHK.
H
K
J
c
In
of
triangl
AB.
ABC,
Prov
CD
that
is
th
Δ ADC
prpndiular
is
ongrunt
to
C
istor
Δ BDC
A
3
A
rgular
to
Δ CDE
pntagon
has
vrtis
ABCDE.
Explain
why
D
Δ ABC
B
is
ongrunt
In
3,
lal
hlp
Problem
4
A
rgular
ongrunt
a
and
diagram
you.
solving
hxagon
to
draw
has
Δ ADE.
vrtis
Stat
th
ABCDEF.
ondition
List
usd
all
to
th
triangls
stalish
that
ar
ongrun.
E13.1 Are we very similar?
2 71
to
Example
Cirls
C
4
and
C
1
hav
ntrs
O
2
rsptivly,
and
and
O
1
intrst
at
C
points
X
X
C
1
2
and
2
Y
O
O
1
Prov
that
Δ
O
XO
1
is
ongrunt
to
Δ
O
2
2
YO
1
2
Y
Givn:
irls
and
C
C
1
with
ntrs
O
2
and
O
1
2
Separate
rsptivly
Prov:
Δ O
XO
1
≅
intrst
Δ O
2
C
points
X
and
into
‘given’
and
‘prove’.
Y.
YO
1
X
at
2
C
1
2
O
O
1
Draw
the
triangles
Δ O
2
XO
1
≅
2
Δ O
YO
1
on
the
diagram.
2
Y
Proof
Give
O
X
=
O
1
O
X
=
O
2
O
O
=
radii
of
C
XO
(oth
radii
of
C
O
(sam
lin
sgmnt)
2
Δ O
2
YO
1
Practice
)
2
1
≅
)
1
Y
O
2
1
ATL
(oth
2
1
Δ O
Y
1
a
(SSS)
2
3
A
1
Triangl
M
is
th
Prov
ABC
is
isosls
midpoint
that
of
Δ

ABM
with
AB
=
AC
BC.
and
Δ

ACM
ar
ongrunt.
C
2
Points
Th
A,
B,
hords
Prov
that
C
and
AB
D
and
Δ

ABO
li
CD
and
on
ar
a
irl
qual
Δ
CDO
with
in
ar
ntr
M
B
A
O.
lngth.
ongrunt.
C
O
B
D
3
PQR
is
atan
4
Points
Prov
5
an
isosls
angl
Cirl
A,
B
that
C
of
90°.
and
C
Δ

AOB
has
triangl
Prov
li
on
and
ntr
with
that
a
irl
Δ
AOC
A,
and
PQ
=
Δ
QPX
≅
with
ar
QR.
A
lin
from
Q
mts
PR
at
X,
Δ
QRX.
ntr
O
∠
ABC
=
∠
ACB.
ongrunt.
irl
C
1
has
2
D
C
1
C
2
ntr
of
C
of
C
.
B
whih
Points
D
lis
and
on
E
th
ar
irumfrn
th
intrstions
1
and
C
1
Prov
B
A
2
that
Δ
ABD
and
Δ
ABE
ar
ongrunt.
E
272
E13 Justication
reason
for
each
statement.
G E O M E T R Y
6
HIJK
is
a
squar.
H
and
J
li
on
a
irl
with
ntr
A N D
T R I G O N O M E T R Y
F
H
Prov
that
Δ
FKH
and
Δ
FKJ
ar
ongrunt.
I
J
7
Prov
that
if
lins
PQ
and
RS
ar
paralll,
and
X
is
th
Q
midpoint
of
PS,
thn
Δ
PXQ
and
Δ
SXR
ar
ongrunt.
P
X
S
R
Using
C
●
All
How
an
mathmatial
known
usd
as
to
formalizd
Th
rst
axioms.
prov
of
In
y
Grk
gomtry
othr
othr
Thr
start
ar
gomtry;
th
th
ongrun
systms
axioms.
dsri
congruence
words,

y
usd
in
this
to
dirnt
most
prov
topi
assums
som
th
to
agrd
of

around
fats.
axioms
300
and
Ths
whih
gomtry,
ongrun
tru,
results
rsults?
Eulidan
Eulid
taks
thm
is
other
othr
systms
famous
mathmatiian
it
prove
stalishing
many
th
to
ar
an

whih
was
bce.
onditions
thn
uss
to
thm

Eulid’s
to
famous
rsults.
is
his
The
Exploration
puliation
Elements
In
th
ist
that
diagram,
ah
PS
othr
Δ
STU
≅
and
at
U.
TQ
ontaining
P
dnitions,
thorms
Δ
PQU
cm
12
th
lngth
of
ST .
9
rasoning.
Idntify
Δ
STU
th
and
ongrunt
Δ
PQU.
angls
Justify
Q
known,
a
you
an
singl
him
R
S
If
prov
two
triangls
ddu
using
aout
Corrsponding
this
as
part
of
a
SSS,
th
SAS,
othr
parts
of
proof,
of
th
th
ASA/AAS
ongrunt
an
of
Eulid’s
thm
in
arnd
modrn
“th
ar
orrsponding
you
alrady
pla
niknam
fathr
ongrunt
ut
gathring
in
your
rasoning.
4
wr
U
cm
T
3
Many
Justify
rsults
your
and
cm
proofs.
Find
of
13
mathmatial
Prov
10
2
–
4
ooks
1
most
ahivmnt
or
triangls
writ
RHS,
angls
ar
CPCTC
xplain
and
sids
ongrunt.
for
what
in
th
you
of
gomtry”.
an
triangls.
Whn
using
short.
E13.1 Are we very similar?
273
Reect
and
discuss
Th
3
isosls
triangl
Rad
and
disuss
this
proof
to
mak
sur
you
undrstand
was
Isosceles
triangle
theorem
pons
known
Δ ABC
is
isosls
with
AB
=
AC,
thn
∠
ABC
≅
∠
ACB.
as
th
asinorum,
‘ridg
If
thorm
it.
of
or
asss’,
A
aus
studnts
Proof
in
AB
=
AC
mdival
who
(givn)
ould
not
undrstand
∠
BAC
=
∠
CAB
(th
sam
=
AB
≅
Δ

ACB
(SAS)
wr

∠
ABC
≅
∠
ACB
(CPCTC)
B
Copy
4
this
proof
and
add
th
rasons.
Theorem
If
ABCD
is
a
rhomus
thn
∠
DAC
≅
∠
BCA
A
Proof
AB
=
CD,
AC
=
CA
Δ ABC
BC
=
DA
∠
DAC
≅
aus
aus
≅Δ CDA
aus
∠
BCA
B
aus
D
C
2
Δ
ACE
B
lis
Copy
is
an
on
isosls
AC
and
and
D
omplt
Theorem
right-angld
lis
th
on
C E
proof
triangl
suh
that
that
with
BE
∠
CEB
≅
=
th
right
angl
at
C.
DA
∠
CAD.
____________________
Proof
A
BE
=
DA
∠
BCE
=
aus
∠
DCA
=
90°
aus
B
X
CE
=
CA
Δ BCE
aus
≅Δ DCA
aus
E
C
∠
CEB
2 74
≅
∠
CAD
aus
E13 Justication
was
a
ndd,
thought
to
This
C
of
1
why
ignorant.
proof
Practice
or
(givn)
proof
Δ

ABC
th
angl)
proof,
AC
tims
D
is
y
Pappus
Alxandria.
G E O M E T R Y
Problem
3
Copy
and
A N D
T R I G O N O M E T R Y
solving
omplt
this
sklton
proof.
Theorem
If
Δ ABC
is
isosls
prpndiular
to
with
AB
=
AC
whr
th
angl
thn
th
angl
istor
of
∠
BAC
is
BC
Proof
Lt
X

Show
AB
=
th
that
AC
∠
BAX
∠
Hn
⇒
×
∠
BAC
mts
BC.
Draw
Δ ACX :
Give
a
reasons
diagram,
to
prove
showing
congruence.
.
Δ ABX
∠
AXC
Δ ACX
AXC.
∠
≅
∠
___
≅
=
=
180°
(
).
(
Give
).
the
condition
Explain
aus
for
the
congruence.
reason
for
this.
.
180°
⇒
Thrfor
4
Copy
and
th
angl
omplt
istor
this
is
prpndiular
sklton
proof.
to
Inlud
BC.
a
suital
diagram.
Theorem
If
AB
OM
is
is
a
hord
to
a
irl
prpndiular
to
C
with
ntr
O,
and
M
is
th
midpoint
of
AB,
AB
Proof
OA
=
AM
OB
=
aus
BM
is
Hn
∠
∠
+
2
.
aus
.
ommon.
Thrfor
⇒
X.
.
aus
∠
+
2
≅
of
ommon.
Thrfor
∠
Δ ABX
istor
aus
≅
is
point
Δ OAM
≅
∠
×
≅
Δ OBM
(
).
∠
=
aus
.
=
⇒
Thrfor
OM
is
prpndiular
to
AB
E13.1 Are we very similar?
2 75
5
Cirl
C
has
ntr
O
L
is
a
tangnt
to
C
at
X
L
1
L
and
L
1
mt
at
is
a
tangnt
to
C
at
Y
2
Z.
2
a
Skth
L
,
L
1
b
Prov
c
Hn
and
C.
Mark
points
O,
X,
Y
and
Z
on
your
diagram.
2
that
triangls
show
that
Δ OXZ
XZ
=
and
Δ OYZ
ar
ongrunt.
YZ
Tip
6
A
paralllogram
is
dnd
Prov
that
opposit
Prov
that
th
sids
as
ar
having
qual
two
in
pairs
of
paralll
sids.
lngth.
In
6,
divid
th
paralllogram
7
diagonals
of
a
kit
ar
two
8
Prov
9
Squar
X
is
that
th
diagonals
TUVW
th
is
midpoint
of
a
intrstd
of
AD
and
paralllogram
y
th
also
straight
th
into
prpndiular.
ist
lin
midpoint
W
ah
othr.
ongrunt
triangls.
ABXCD.
of
UW
T
D
C
X
B
A
V
Prov
that
∠
AWB
Problem
10
Th
28
km
th
uildings.
of
∠
DUC
solving
Pntagon,
largst
≅
U
hadquartrs
It
orridors,
taks
and
th
has
of
th
shap
an
of
US
a
intrnal
military,
rgular
is
among
pntagon,
ourtyard
aout
th
world’s
ontains
th
siz
of
ovr
Tip
thr
Rmmr
Amrian
A
is
statu
footall
(K )
is
quidistant
Prov
(A
that
and
it
plad
from
is
also
dirtly
th
two
in
front
narst
quidistant
of
on
ornrs
from
th
fa
of
nxt
of
th
uilding,
th
uilding
two
ornrs
(D
of
and
th
so
that
it
E ).
uilding
C ).
C
K
E
B
A
Using
●
similarity
Whih
is
mor
usful:
to
prove
stalishing
other
similarity
or
results
stalishing
ongrun?
You
hav
gomtri
to
sn
how
rsults.
important
27 6
stalishing
Somtims,
mathmatial
ongrun
simply
rsults.
E13 Justication
an
lad
stalishing
to
th
similarity
proof
is
of
othr
nough
to
is
quidistant
A
and
that
D
D
that
‘P
lds.
lad
B ’
AP
=
from
mans
BP
G E O M E T R Y
Exploration
1
Copy
and
sgmnts
of
a
omplt
irl
point
X
If
T R I G O N O M E T R Y
5
ratd
Theorem:
A N D
A,
suh
this
y
B,
C
that
intrior
to
proof
to
intrsting
and
th
th
D
ar
hord
irl,
disovr
th
rlationship
twn
hords.
points
on
ABmts
th
th
irumfrn
hord
CD
at
thn
C
A
B
X
Reproduce
the
diagram
showing
Δ
AXC
and
Δ
DXB
O
D
∠
AXC
=
∠

aus
.
Show
∠
CAB
∠
ACD
Sin
=
∠
CDB
=
all
know
aus
aus
∠
thr
that
angls
Δ
AXC
is
that
corresponding
angles
are
equal.
.
ar
.
qual,
similar
to
w
Δ
DXB
AX
=
Corresponding
sides
of
similar
triangles
are
proportional.
BX
Rarranging
Reect
●
Ar
all
Justify
●
Whih
givs:
and
discuss
similar
your
is
.
triangls
This
the
‘intersecting
chords
theorem’.
4
ongrunt?
Ar
all
ongrunt
triangls
similar?
rasoning.
mor
usful:
stalishing
similarity
or
stalishing
ongrun?
Summary
Two
sid
shaps
ar
lngths
and
congruent
sam
siz
if
thy
hav
th
●
sam
angls.
Angl-Sid-Angl
If
two
two
Two
shaps
ar
congruent
if
on
an
into
th
othr
y
a
rtion,
translation,
or
a
omination
of
for
orrsponding
a
orrsponding
oth
triangls,
Sid-Sid-Sid
Right
th
thr
ongrun
as
sids
th
of
thr
triangl
sids
of
A
ar
th
triangl
If
two
th
ar
is
th
two
sids
ongrun
of
triangl
orrsponding
angl
oth
sid
th
whos
two
lngth
triangls
right-angld
and
sam
triangls
is
thr
th
sam
ar
ongrunt.
ongrun
on
of
triangls
th
hav
rmaining
qual
two
ar
lngth
in
oth
triangls,
th
ongrunt.
twn
triangls,
sids
thos
th
A
ar
of
pairs
two
th
sam
triangl
of
parts
of
ongrunt
triangls
ar
(SAS):
ongrunt.
two
as
and
ongrunt.
Sid-Angl-Sid
If
B,
two
Corrsponding
●
triangl
AAS):
siz
sam
B,
two
triangls
or
sam
(SSS):
sids
lngths
in
(ASA
th
Angl-Hypotnus-Sid
hypotnuss,
If
angls
ar
congruence
(RHS):
●
A
ths.
●
Conditions
ongrun
triangl
rotation
in
or
in

is
transformd
angls
sids
triangls
B,
is
ar
lngths
and
th
You
an
writ
CPCTC
for
short.
as
th
sam
in
ongrunt.
E13.1 Are we very similar?
277
Mixed
1
Ar
If
th
so,
practice
two
giv
prov
triangls
th
in
ah
onditions
for
diagram
ongrunt?
ongrun
usd
4
to
ABCDE
of
it.
CD.
is
a
rgular
Prove
istor
that
pntagon.
AX
is
th
X
is
th
midpoint
prpndiular
ofBE.
A
a
b
Y
X
B
A
W
C
D
B
E
Z
c
d
T
R
E
S
D
To
prov
AX
is
X
th
C
prpndiular
istor
of
F
G
I
BE,
show
that
AX
is
prpndiular
to
BE
and
U
that
V
5
X
Part
is
of
joining
a
In
th
diagram,
givn
that
AD
=
BC
an
midpoint
origami
paralllogram
H
2
th
th
and
two
=
∠
ACB,
prove
that
∠
ADC
≅
onstrution
thr
most
fold
distant
onsists
lins:
on
ornrs
of
of
a
diagonal
and
two
short
paralll
th
lins
∠
ACB
joining
A
BE
and
paralllogram,
∠
DAC
of
th
othr
ornrs
to
th
diagonal,
as
B
illustratd.
D
b
In
is
th
C
diagram,
paralll
to
givn
IGH,
that
prove
E F
that
ists
IG,
and
E I
Prove
∠
FEI ≅ ∠
GFH
that
th
short
paralll
lins
ar
qual
in
lngth.
E
6
Th
diagram
shows
points
A,
B,
C
and
D.
G
F
Prove
I
that
triangls
ABD
and
ACD
ar
ongrunt.
H
A
c
In
th
MK
diagram,
ists
givn
∠
JML,
that
prove
∠
J
=
that
∠
L
and
Δ
JMK
≅
that
Δ
LMK
J
D
K
M
C
B
L
7
In
th
and
3
ABCDEFGH
is
a
rgular
otagon.
By
diagram,
CAB
=
60°
AB
is
paralll
Prove
that
to
DE
CDE
is
and
an
AB
=
AC
isosls
onsidring
triangl.
Δ
AEF
and
isosls
Δ
ADE,
prove
that
Δ
ADF
is
an
A
triangl.
F
B
E
D
G
C
C
H
A
278
B
E13 Justication
D
E
G E O M E T R Y
8
A
small
larg
frams,
mast
of
lvl
is
formd
frams
illustratd
lngths
on
radio
triangular
mtal:
low,
AC,
y
togthr.
is
AD,
joining
thr
On
th
mad
BD
of
from
and
CE.
11
thr
xd
four
It
Railway
traks
ountris
stands
Th
ground.
Th
in
th
Limrik,
th
rails
ar
paralll
gaug.
uss
1600
shows
Irland.
an
T R I G O N O M E T R Y
a
lins
mm
railway
viw
of
a
gaugs.
Irish
Gaug,
apart.
juntion
Thdiagram
ovrhad
at
Dirnt
national
Irland
photograph
rprsnts
in
th
dirnt
of
two
st
alld
hav
Rpuli
whr
A
ar
distan,
A N D
low
th
it
layout
of
lins.
E
B
C
Givn
and
that
BD
AD
and
thatlngths
9
A
King
and
Th
or
is
AB
king
post
that
AC
ar
for
ar
qual
and
Truss
suital
mtal
and
CE
Post
D
AE
ar
Bridg
short
PQ
is
hlps
a
qual
in
in
lngth,
lngth,
prove
qual.
has
a
simpl
strutur
spans.
singl
prvnt
pi
th
of
ti
wood
am
B
ABfromsagging.
P
C
A
F
D
G
E
Q
Prove
that
kingpost
thn
AP
∠
PAQ
if
PQ
and
=
Q
is
is
th
midpoint
vrtial,
BP
must
and

of
AB
qual
AB,
is
in
H
th
horizontal
lngth
and
Assuming
∠
PBQ.
and
10
all
of
th
Irish
that
Δ

ABD
b
prove
that
Δ
BAC
c
hence
show
d
prove
that
of
Th
thr
qual
ah
an
lads
lngth
pair.
and
Prove
quilatral
of
a
wind
hav
that
th
th
triangl.
turin
sam
tips
of
ar
angl
th
all
of
twn
lads
form
ars
at
laddr
opn
ars
Use
ongrunt
and
paralll,
to
Δ
CDB
Δ
BCA
ABCD
≅
is
a
Δ
EFG
ar
to
rhomus
givn
that
th
paralll.
a
stpladdr
idas
of
on
oth
sam
mak
ongrun
positioning
hight
that
th
on
th
how
sam
nsurs
out
is
≅
Δ

ABC
larly
th
straight
solving
stal.
explain
that
traks
horizontal
mor

Gaug:
prove
Problem
Th
to
a
twosts
12
traks
oth
sids
th
horizontal
sids
of
th
of
th
laddr
angl.
E13.1 Are we very similar?
it
to
279
13
Similar
Angl
triangls
Bistor
an

usd
Thorm.
to
Copy
dvlop
and
th
Proof:
omplt
Draw
this
PT
paralll
to
SR.
proof.
T
P
Theorem:
a
triangl
An
angl
divids
sgmnts
that
twosids
of
ar
th
istor
th
of
opposit
an
angl
sid
proportional
to
in
th
of
two
othr
Q
triangl.
P
S
R
Q
S
∠SQR
=
∠PQT
∠RSQ
=
∠
aus
aus
R
Givn:
SQ
is
th
istor
of
angl
S
∠
SR
=
∠TPQ
aus
QR
=
Prov:
SP
QP
Sin
w
all
thr
know
angls
ar
that
qual,
.
SR
=
QP
∠PST
Hn
=
∠QSR
∠PST
Thrfor,
PS
SR
SP
E13 Justication
∠PTS
=
PT
QR
=
Hn,
2 8 0
=
aus
QP
aus
aus
G E O M E T R Y
Review
Th
anint
ultur
to
and
rat
in
art
A N D
T R I G O N O M E T R Y
context
of
origami
involvs
modls.
has
folding
Som
roots
at
modls
in
shts
ar
Japans
of
Th
rsulting
papr
dsignd
rass
ar
illustratd
A
in
this
B
diagram.
C
to
E
rsml
just
animals,
lrat
th
owrs
auty
or
of
uildings;
gomtri
othrs
dsign.
D
F
H
G
Th
Z
artist
thinks
lngths
FH
and
DY
ar
qual.
1
b
Explain
why
EX
=
EG.
2
1
An
origami
and
folds
it
artist
in
taks
half,
a
squar
sht
of
c
Hence
d
Explain
why
Th
and
to
artist
folds
th
runs
top
up
h
sht
at
mt
as
from
folds
th
a
th
ongrunt
unfolds
to
dg,
straight
Finally,
th
thn
it
folding
two
th
th
of
sht
sht,
ornr
ras
hight
nd
∠
EGX
Δ GED
≅
Δ GZD
Find
f
Show
that
GF
g
Show
that
∠
GFH
h
Hence
and
∠
DGZ
∠
YED
show
and
DY
=
EY
that
ar
=
∠
EYD
th
artist’s
qual
–
is
laim
–
that
lngths
orrt.
half
taks
lin,
Th
to
on
not
fold
th
paralll
th
in
∠
EGD
rtangls.
ntr
illustratd.
on
why
nd
e
FH
produs
to
papr,
Fold
Explain
trigonomtry
asshown.
Hn
a
use
to
foldd
ornr
quit
ratd
opposit
th
top
sid.
dg
of
ornr.
Reect
and
How
hav
you
Giv
spi
discuss
xplord
th
statmnt
of
inquiry?
xampls.
Statement of Inquiry:
Logi
our
an
justify
appriation
gnralizations
of
th
that
inras
asthti.
E13.1 Are we very similar?
2 81
A
E14.1
world
Global
context:
Objectives
●
Solving
of
dierence
Identities
Inquiry
quadratic
and
rational
inequalities
and
relationships
questions
●
What
●
How
is
a
non-linear
inequality?
F
both
●
algebraically
Solving
other
and
graphically
non-linear
inequalities
do
you
inequalities
solve
non-linear
graphically?
graphically
MROF
●
Using
mathematical
non-linear
models
inequalities
to
containing
solve
●
How
●
Which
does
a
model
lead
to
a
solution?
C
real-life
problems
is
more
ecient:
a
graphical
D
solution
●
ATL
Use
Can
or
good
an
algebraic
decisions
be
one?
calculated?
Critical-thinking
models
and
simulations
to
explore
complex
systems
and
issues
11 1
14 3
E14 1
Statement
Modelling
of
Inquiry:
with
representation
2 8 2
equivalent
can
forms
improve
of
decision
making.
A LG E B R A
Y
ou
●
●
should
solve
linear
sketch
already
know
how
inequalities
graphs
of
linear,
1
quadratic,
to:
Solve
2
−3 x
Sketch
− 2 < 5 − 4 x
the
graphs
of :
2
rational,
radical,
logarithmic
exponential
and
y
a
=
x
x
+ 3x − 4
y
b
=
2e
functions
− 3
2
c
y
=
y
d
4 log 2 x
=
4
x
e
y
2 x
3
3x
+ 6
y
f
=
=
x
− 5
2
●
solve
quadratic
completing
the
●
What
●
How
Objective
apply
D:
the
by
3
Solve
x
+
6x
+
3
=
0.
square
Non-linear
F
iii.
equations
is
a
do
non-linear
you
Applying
selected
inequalities
solve
in
one
variable
inequality?
non-linear
mathematics
mathematical
in
inequalities
real-life
strategies
graphically?
contexts
successfully
to
reach
a
solution
In Exploration 1, when you have selected your solution to the quadratic inequality, determine
2
graphically if
the range of
Exploration
In
ski
‘y’
jumping,
as
far
as
the hip angle guarantees a lift area of
at least 0.5 m
1
athletes
possible.
descend
The
a
angle
take-o
of
the
ramp,
skier’s
jump
hip
a
o
the
end
determines
of
the
it
and
‘lift
ATL
2
area’
A
(in
between
1
Use
m
hip
this
)
which
angle
determines
and
function
to
lift
area
write
a
how
can
be
far
the
skier
modelled
mathematical
jumps.
by
the
model
The
relationship
function:
(inequality)
of
the
2
2
3
range
of
Use
GDC
a
the
hip
to
angle
graph
both
sides
of
this
a
lift
area
inequality
of
at
and
b
the
region
of
the
graph
that
models
the
set
of
points
c
the
region
of
the
graph
that
models
the
set
of
points
write
least
0.5
m
nd:
point
graph,
graphs
ensure
the
your
both
would
a
From
where
that
intersect
where
where
.
down:
2
a
the
range
b
the
range
of
the
skier’s
hip
angle
to
ensure
a
lift
area
of
0.5
m
or
more
2
4
Explain
ranges
of
how
in
the
you
step
the
skier’s
can
test
hip
angle
whether
that
or
gives
not
you
a
lift
have
area
less
selected
than
the
0.5
m
correct
2
Rearrange
6
Graph A(a) , and explain how you would use the inequality to answer step 2
7
Explain
how
inequality
you
could
in
use
step
this
1
to
the
graph
to
form
A(a)
5
check
≥
your
0.
answer.
E14.1 A world of dierence
2 83
Reect
●
How
do
graphs
●
and
method
original
inequality?
To
●
solve
●
a
Graph
or
f
the
0
the
●
Choose
a
test
your
point
this
x-values
satisfy
to
x-intercepts,
you
prefer:
and
the
include
as
in
one
compare,
x-values
step
where
or
the
6)
in
you
one
where
your
two
solution
graph
where
the
each
you
set?
side
of
rewrite
the
of
the
graphically:
inequality,
or
rearrange
into
the
form
f
( x )
>
0
graph.
points
solution
point
in
the
Example
Method
not
of
intersection
of
both
functions,
or
nd
the
zeros
function.
check
the
or
inequality
sides
and
To
Solve
did
the
inequality
both
<
whether
(or
quadratic
( x )
1
Explain.
Determine
of
know
intersect
Which
the
you
discuss
in
in
this
algebraically:
the
the
region
you
inequality.
region
inequality,
satisfy
then
the
test
think
If
it
satises
satises
the
inequality.
the
x-values
If
in
the
inequality,
inequality,
your
the
test
other
and
then
point
the
does
not
region(s).
1
inequality
.
Check
your
solution
algebraically.
1
1.1
Graph
both
sides
of
the
inequality
and
60
2
f
(x)
=
16
+
12x
−
x
2
determine
the
points
of
intersection.
50
40
(1, 27)
(12, 16)
20
f
(x)
=
28
−
x
1
10
The
x-values
than
0
those
of
on
points
the
on
the
quadratic
line
(LHS)
(RHS)
for
are
1<
x
lower
<
12.
x
2
Test
a
point
with
x-value
between
1
and
12:
Select
When
x
=
an
check
that
original
26
The
≤
x-value
36
✔
solution
is
it
the
satises
inequality
endpoints
(x
=
solution
1
the
was
interval,
and
x
E14 Models
on
next
inequality.
“less
=
.
Continued
2 8 4
in
and
2:
page
12)
than
are
or
Because
equal
included.
the
to”,
the
A LG E B R A
Method
2
Rearrange
to
f ( x )
.

⇒
1.1
y
Graph
the
quadratic
function.
35
2
f
(x)
=
−12
+
13x
−
x
1
30
25
20
15
10
5
(12, 0)
(1, 0)
x
0
4
2
6
8
10
12
14
The
The
solution
is
zeros
of
f ( x )
for
the
Select
Check:
When
x
=
2:
Solve
the
Method
quadratic
are
x
=
1
and
x
=
12.
an
x-values
x-value
between
in
the
these
solution
zeros.
interval,
.
and
Example
the
.
check
that
the
inequality
is
satised.
2
inequality
Check
your
solution
algebraically.
1
1.1
y
2
f
(x)
=
x
x −
6
2
Graph
2
the
both
points
sides
of
of
the
inequality
intersection
of
the
and
determine
graphs.
x
3
2
1
1
2
3
4
2
(2.47,
2.35)
4
(
0.808,
4.54)
6
8
2
f
(x)
=
−2 x
+
4x
1
The
x
<
−0.808
or
x
>
blue
curve,
for
lies
below
the
2.47
red
curve
in
the
regions
x
<
−0.808,
x
>
2.47.
Check:
When
6
<
x
=
−1:
=
3:
LHS
=
−6,
RHS
=
−
4
−4
When
x
LHS
=
−
6,
RHS
=
0
Continued
on
next
page
E14.1 A world of dierence
2 8 5
Method
2
Rearrange
to
f ( x )
>
0.
⇒
1.1
y
2
f
(x) =
3x
− 5x −
6
1
6
4
2
(
(2.47, 0)
0.81, 0)
Graph
0
2
the
quadratic
function.
x
1
1
4
6
8
x
ATL
<−0.808
Practice
1
Solve
or
x
>
2.47
The
quadratic
is
greater
than
0
above
each
inequality
graphically,
and
check
your
answers
algebraically.
Use
x
d
x
2
− 2x
− 9
>
2x
+ 3
whichever
2
2x
b
− 5
≤
x
+ 1
c
5x
− 10
≥
23 x
method
2
2
For
2
+ 3x
− 8
<
astronaut
4 − 2x
−
training,
2
x
2
2x
e
− 7x
weightlessness
is
+ 7
>
8x
− 2x
simulated
in
a
+ 1
reduced
gravity
2
aircraft
that
models
the
of
the
ies
in
a
parabolic
relationship
aircraft
in
aircraft’s
height
a
Write
an
b
Determine
aircraft
how
in
The
above
inequality
ies
between
meters.
is
The
the
ight
the
time
t
(t ) =
in
−3t
+ 191t
seconds,
experience
and
+ 6950
the
weightlessness
height
when
this
astronauts
situation.
experience
weightlessness
when
A
rectangle
Write
4
A
and
is
6
cm
solve
gardener
has
longer
the
arc.
an
24
than
inequality
meters
of
it
to
is
wide.
nd
fencing
its
to
Its
area
possible
build
a
is
greater
than
216
cm
dimensions.
rectangular
enclosure.
2
Theenclosure’s
inequality
5
For
to
a
to
drivers
visual
area
nd
aged
should
possible
between
stimulus
be
16
(such
less
lengths
as
and
a
than
of
the
70,
trac
24
m
.
Write
and
solve
an
enclosure.
the
reaction
light)
can
be
time
y
(in
modelled
milliseconds)
by
the
function
2
y
=
0.005 x
inequality
than
25
− 0.23 x
to
determine
milliseconds,
Problem
6
A
baseball
time
t
+ 22 ,
(in
where
the
and
ages
x
is
the
for
driver’s
which
a
age
in
driver’s
years.
Write
reaction
time
an
is
more
solve.
solving
is
thrown
seconds)
from
and
a
the
height
height
h
of
1.5
(in
m.
The
meters)
of
relationship
the
baseball
between
while
in
the
the
2
air
is
modelled
a
Write
b
Determine
2 8 6
an
by
the
function
inequality
how
that
long
h
(t ) =
describes
the
E14 Models
baseball
−4.9t
how
is
in
+ 17t
long
the
+ 1.5
the
air.
baseball
is
in
the
air.
h
the
m.
describes
parabolic
h
function
astronauts
9600
that
long
a
arc.
2
3
x-axis.
1
2
a
the
you
prefer.
A LG E B R A
7
In
a
the
right-angled
other.
The
Determine
8
A
2
the
cm
the
possible
square
sides
are
triangle,
one
hypotenuse
is
cut
then
is
of
lengths
from
folded
for
each
up
the
more
to
perpendicular
than
the
the
shortest
corner
form
twice
of
an
a
is
of
2
cm
the
longer
shortest
than
side.
side.
square
open
sides
length
box.
piece
Pia
of
card,
needs
a
and
box
with
3
maximum
Example
Find
the
volume
40
cm
.
Find
the
smallest
possible
dimensions
of
the
box.
3
set
of
values
such
that
in
the
domain
and
check
The
your
solutions
Examples
3
to
and
4
answer.
can
also
using
1.1
be
the
found
same
Method 2
approach
y
as
described
in
4
Examples
1
and
2.
1
3
f(x)
=
x
2
f(x)
= √x
Graph
both
sides
of
the
inequality,
1
and
nd
the
point
of
intersection.
x
0
1
2
3
4
5
6
Select
in
the
interval
0
<
x
<
an
interval
Check:
When
Example
x
=
0.5,
x-value
in
the
solution
1.
and
≈
0.707
<
and
test
it
in
the
inequality.
2
4
Solve
,
and
check
your
answer.
1.1
y
10
x
= 5
2x −
8
f
(x)
7
=
1
x
−
5
6
Locate
4
f
(x)
= 3
2
(blue)
2
0
where
the
LHS
function
(8, 3)
RHS
function
(red).
(3.5, 0)
x
2
4
6
or
Continued
on
next
page
E14.1 A world of dierence
2 87
Check:
Select
When
x
=
10,
and
and
When
x
=
0,
Practice
1
Find
the
,
an
and
check
.
2
set
of
values
that
satisfy
each
non-linear
1
inequality:
1
a
− x
<
0
b
x
− 4
≥
0
x
x
+ 6
c
>
d
2
ln x
<
( x
log
− 2)
2
x
+1
1
4
2
x
e
>
2x
+ 3
<
f
x
+ 5
2x
+ 3
3x
1
≥
h
<
x
x
x
When
two
resistors,
R
and
R
+ 4
are
connected
modelled
by
the
as
shown,
the
total
resistance
2
1
is
x
solving
1
R
3+
−1
+ 3
Problem
2
x
1
g
1
equation
=
1
+
R
R
R
1
2
+
R
R
1
a
R
2 ohms
=
and
R
must
be
at
least
2
1
ohm.
Find
R
2
b
R
1
=
2.2
ohms
and
R
must
be
at
least
1.7
ohms.
Find
R
1
C
2
Non-linear
●
How
does
a
inequalities
model
lead
to
a
in
two
variables
solution?
When
you
solve
an
inequality
in
one
variable,
When
you
solve
an
inequality
in
two
variables,
the
solution
the
is
solution
a
is
set
a
of
set
x-values.
of
points.
2
For
this
graph
of
y
=
x
− 2x
− 3,
the
blue
shaded
area
represents
the
set
of
2
points
that
region
by
satisfy
y
<
substituting
x
− 2x
the
− 3.
You
coordinates
can
of
a
test
that
point
in
you
this
have
the
region
correct
into
the
original
2
inequality.
correct
For
region
example,
is
for
the
point
(0,
4):
x
− 2x
− 3
=  −3 > −4,
shaded.
y
8
6
2
y
=
x
−
2x
−
4
3
2
x
10
8
6
4
2
2
4
6
8
2 8 8
x-value
in
each
region,
.
E14 Models
so
the
in
the
original
inequality.
A LG E B R A
The
unshaded
region
of
the
graph
on
the
previous
page
represents
the
set
of
2
>
with
a
dashed
solution
solve
●
Graph
●
Select
show
non-linear
the
a
that
values
point
is
region
(e.g.
on
−
the
.
The
line
quadratic
are
not
graph
included
is
in
drawn
the
that
is
inside
the
not
If
by
on
the
the
in
the
the
is
variables:
equality.
curve
inequality
curve).
point
two
not
If
the
(e.g.
and
is
substitute
true,
shade
inequality
outside
the
is
the
the
not
values
region
true,
into
where
shade
the
the
the
curve).
5
sketching
the
dened
inequality.
where
Example
inequalities
region
point
original
Shade
to
x
set.
To
By
line,
−
a
suitable
region
graph,
containing
solve
the
the
inequality
possible
solutions
.
on
your
graph.
Factorize to nd the roots of f ( x ) = 0.
⇒
⇒
Use
x-coordinate
of
the
formula
coordinates
Vertex
=
(3.5,
to
nd
the
vertex
of
the
vertex.
2.25)
y
3
(3.5, 2.25)
2
Draw
1
line,
(3, 0)
the
as
it
parabola
with
represents
a
a
dashed
true
inequality.
x
0
1
1
2
y
<
−
x
+
7x
−
10
2
Check:
For
(3,
0):
,
hence
Test
the
region
shaded
is
a
then
Practice
point
in
the
original
shade
3
the
correct
In
Sketch
a
suitable
graph
and
shade
the
region
on
the
graph
that
shows
of
the
>
x
+ 3x
− 1
b
y
≤
− x
(
d
y
>
ln
( x
a
line
− 2) + 3
e
y
≤
e
x
the
is
dotted
graph
as
above.
2
2
2
y
of
inequality.
in
a
case
inequality,
the
the
solution
region.
the
strict
1
inequality,
correct.
+ 3x
− 2
c
y
≥
+ 1)
− 3
f
4 − 3x
x
+ 3
x
3
− 2x
In
≤,
the
the
case
of
graph
≥
of
or
the
y <
quadratic
a
solid
E14.1 A world of dierence
would
line.
2 8 9
be
2
Write
an
inequality
to
describe
the
shaded
region
y
a
in
each
graph.
y
b
10
10
2
y
=
−x
−
4x +
5
8
8
6
4
4
2
2
2
y
4
3
2
1
14
2
1
0.5 x
+
2x −
1
12
10
8
6
4
2
y
y
d
8
8
6
6
2x
y
=
e
+
1
−
4
x
4
y
4
=
x
−
2
2
2
0
8
6
Problem
3
For
each
i
nd
ii
write
4
4
4
4
10
2
2
2
c
x
0
x
0
6
=
2
0
x
2
4
6
8
10
10
8
6
4
2
x
2
2
2
4
4
6
6
8
8
4
6
8
10
solving
graph:
the
quadratic
an
function
inequality
that
for
the
describes
parabola
the
shaded
y
a
region.
b
y
3
4
2
3
1
1
0
4
3
2
1
x
1
1
0
5
4
3
2
1
1
3
2
3
c
y
20
16
12
4
0
2
1
x
1
7
4
8
2 9 0
E14 Models
x
1
2
A LG E B R A
4
A
parabolic
dam
Themaximum
and
the
length
is
built
height
of
the
of
across
the
dam
a
river,
dam
across
as
above
the
shown
the
river
in
water
is
90
the
graph.
level
is
3
m,
m.
y
5
4
3
(45, 3)
2
1
0
x
10
a
Write
b
State
c
Write
a
quadratic
the
domain
an
20
function
and
inequality
that
range
that
30
of
40
50
60
models
the
denes
70
the
region
90
parabolic
quadratic
the
80
arch
of
the
dam.
function.
between
the
dam
and
the
river.
D
inequalities
●
Which
●
Can
is
good
Exploration
1
Factorize
two
2
For
the
linear
the
can
two
the
intervals
the
the
or
an
algebraic
one?
in
the
inequality
to
You
the
is
into
be
less
need
to
inequality
the
solution
than
solve
in
to
zero,
the
step
the
1
one
two
and
factor
cases
solve
original
must
be
separately.
each
one.
inequality.
graphically.
quadratic
inequality
algebraically
by
rst
nding
the
zeros
function.
function
zeros
by
a
for
case
answer
solve
quadratic
hence
in
also
cases
which
quadratic
solution
.
factors
and
Checkyour
The
two
negative.
2:
Determine
the
of
one
Case
4
graphical
calculated?
function
and
the
be
a
that
1:
Create
of
such
Case
3
You
decisions
quadratic
factors
and
ecient:
2
product
positive
more
of
the
roots
inequality
to
in
Exploration
function
of
the
<
x
=
equation.
determine
x
are
if
it
2
factorizes
4
and
Choose
satises
−1
x
1
<
x
<
=
into
-1.
an
( x - 4 ) ( x + 1) ,
The
x-axis
x-value
in
is
each
divided
into
interval,
test
it
theinequality.
x
4
>
4
x
3
2
1
0
1
2
3
4
5
6
E14.1 A world of dierence
2 91
Inter val
Test
x
<
point
−1
1
<
in
4
x
− 3 ×
− 2 − 4
=
0
6
4
2
− 3 ×
0 − 4
=
−4
−
× 7 − 4
=
24
24 > 0
6 > 0
4
No
Is
>
7
2
2
( −2 )
<
0
2
Substitution
x
<
0
Yes
No
2
The
values
Written
in
satisfying
set
Reect
●
How
●
Do
●
Which
notation,
and
would
you
you
prefer
two
To
solve
●
Rearrange
●
Factorize
●
Place
●
Select
original
the
the
inequality.
If
Repeat
To
check
●
Choose
a
test
point
for
your
this
x-values
Solve
the
in
all
If
− 3x
:
1
<
x
− 4
<
<
0
−1 <
are
the
need
in
graphical
to
of
<
4.
and
nd
its
the
graphically?
this
to
for
this
question?
solve
inequality?
algebraically:
either
of
2
approach
consider
set
that
on
Exploration
a
line,
intervals
inequality
inequality
is
( x )
<
0
or
f
( x )
>
0.
zeros.
number
the
f
is
false,
dividing
and
true,
then
it
substitute
then
it
is
that
into
it
intervals.
into
the
interval
is
part
of
not.
solution
in
in
this
algebraically:
the
the
region
you
inequality.
region
satisfy
think
If
the
it
satises
satises
the
the
inequality,
inequality,
and
then
the
inequality.
6
quadratic
inequality
and
check
your
answer
graphically.
Continued
2 9 2
x
4}.
intervals.
point
in
the
one
{x
so
x-values
x-value
or
solution
inequality
quadratic
is
inequality
you
the
inequality
two
●
is
set
2
the
would
x
inequality
solution
algebraic
What
the
solution.
Example
the
quadratic
an
the
solve
cases
these
quadratic
discuss
algebraically?
a
the
E14 Models
on
next
page
A LG E B R A
Method
1
For
a
product
to
be
positive,
the
and
factors
⇒
and
must
have
the
same
signs.
,
⇒
Because
both
the
less
value
than
4
of
and
x
needs
less
to
than
be
3,
and
using
⇒
and
x
<
-3
covers
all
possibilities.
,
⇒
As
Check
your
solutions
by
sketching
a
of
graph.
before,
the
using
x
>
inequalities,
4
x
covers
>
4
and
both
x
>
-3.
y
10
2
y
=
x
−
x −
12
From
the
graph,
solve
the
original
you
can
see
inequality
that
both
cases
since
5
when
0
5
4
2
x
<
-3
or
when
x
>
4.
x
1
1
3
4
5
5
15
The
solution
Method
The
is
x
<
-3
or
x
4.
2
zeros
of
are
Inter val
Test
>
x
x
=
<
point
Is
>
Hence,
f
Which
0?
table,
( x )
Reect
=
4.
-3
3
<
x <
4
x
0
>
0
for
and
your
>
0
12
Yes
the
algebraic
Explain
x
>
4
5
in
8
the
and
4
Substitution
From
-3
solution
x
<
-3
is
or
discuss
method
for
x
x
>
<
8
0
No
<
-3
or
x
>
>
0
Yes
4.
4.
3
solving
quadratic
inequalities
is
more
ecient?
reasons.
E14.1 A world of dierence
2 9 3
Practice
1
Solve
4
these
quadratic
inequalities
algebraically
2
− 2x
− 3
≤
0
b
x
d
x
2
c
12 + x
− 10 x
+ 16
>
0
≥ 0
+ 6x
≤
40
2
2x
+
5x
< 3
3x
f
− 1 >
2x
2
g
2
graphically.
2
− x
2
e
conrm
2
x
a
and
6x
One
the
+ 4 x
of
the
other.
+ 1 <
5x
+ 3
perpendicular
Determine
the
sides
of
a
possible
right-angled
lengths
of
triangle
the
shorter
is
3
side
cm
longer
that
give
than
a
2
triangle
with
Problem
3
Each
area
a
−1 <
c
x
x
than
4 cm
.
solving
inequality
quadratic
less
is
the
solution
set
of
a
quadratic
inequality.
Find
a
possible
inequality.
< 1
b
x
<
−1
or
x
>
2
2
1
≤
2
or
≤
x
≤
x
≥
10
x
d
<
x
or
>
−
3
2
1
e
6
3
Exploration
1
By
using
square’,
2
From
the
the
quadratic
solve
the
Conrm
4
Write
5
Check
the
roots
solution
3
to
your
out
3
the
that
formula
of
the
the
the
method
steps
of
‘completing
equation
in
step
1,
explain
a
how
and
you
each
to
solve
a
non-factorizable
steps
work
when
quadratic
inequality.
solving:
5
inequality
algebraically,
accurate
to
2
1
a
x
c
2x
3
5x
+
1
>
0
b
x
+
2
x
3x
–
x
3
≤
0
d
3x
5x
5
≥
0
f
<
–
7x
+
2
0
>
2x
+
4x
+
2
2
–
2x
1
<
0
2
i
2
0
2
+
2
g
6x
2
+
2
e
s.f.
2
–
3x
2 9 4
h
x
j
x
2
–
2x
+
1
<
x
+
10x
–
5x
≥
−8
2
+
x
obtain
graphically.
b
Solve
would
b
a
Practice
the
.
inequality
solutions
your
or
equation
–
1
E14 Models
≤
−1
≤
0
A LG E B R A
Problem
2
Solve
of
the
each
solving
inequality.
quadratic
Explain
your
result
by
x
c
x
the
discriminant
∆
2
2
a
considering
function.
+ 2x
+ 1 ≤
+ 2x
+ 4
0
b
x
− 4 x
+ 4
≥
0
2
2
Reect
How
∆
2
can
>
0
and
the
Example
d
discuss
discriminant
(x
− 1)
− 3
<
=
b
4ac
0
4
help
you
solve
quadratic
inequalities?
7
Solve
,
and
conrm
your
answer
graphically.
Rearrange and simplify.
A
fraction
and
Method
Case
1:
when
have
the
and
signs.
,
hence
,
hence
and
Then
and
Solution
or
x
<
−1.
2
Find
and
the
the
x-intercept
asymptote
by
by
setting
setting
the
the
numerator
x
<
−1
1
<
x
<
x
>
to
the
2
Substitution
Yes
x
<
1
No
Yes
−1
or
x
on
next
right,
these
and
values.
page
E14.1 A world of dierence
0.
intervals
>
Continued
0,
to
into
Is
Solution:
0
to
equal
the
left,
between
point
equal
denominator
Consider
Inter val
Test
numerator
opposite
and
2:
Method
negative
1
Then
Case
is
denominator
2 9 5
Check:
1.1
From
the
graph,
x-values
greater
6
4
f
(x)
= 1
2
2
(0.5, 1)
x
6
4
0
2
2
2
f
(x)
−
x
=
1
x
+
1
4
Reect
●
is
●
more
Does
of
Solve
x
1
x
the
type
of
and
5
method
explain
inequality
algebraic
method
for
your
solving
reasons
(quadratic,
to
rational,
inequality
algebraically
and
check
0, x
≠ 1
> 0, x
3x
you
feel
aect
your
choice
8
x
x
2
5x
+ 1 < 0, x
Determine
5, x
≥
x
Problem
2, x
≠ −1
≠ 2
8
f
≠ 0
graphically.
6
≤
1
solution
+ 1
d
≠ 4
x
2
+
≥
4
e
etc.)
your
b
+1
x
inequalities
6
x
c
rational
why.
use?
x
<
a
x
algebraic
ecient,
each
1 +
discuss
which
which
Practice
1
and
Decide
2, x
≠ 5
5
solving
the
numbers
that
can
be
added
to
the
numerator
and
2
denominator
of
the
fraction
so
that
the
resulting
algebraic
fraction
is
3
1
always
less
than
2
3
Gretchen
currently
would
scores
consecutive
bring
her
2 9 6
f
( x )
is
below
1
y
free
like
to
about
throws
average
up
improve
9
to
free
she
would
70%.
E14 Models
her
throws
free
in
throw
20
need
average
trials.
to
in
basketball.
Determine
successfully
how
make
in
She
many
order
more
to
than
f
( x )
for
2
0.5
or
less
than
1.
A LG E B R A
Summary
To
solve
●
Graph
into
●
a
quadratic
both
the
functions,
to
●
a
Choose
f
of
( x )
the
or
Remember
the
sides
form
Determine
inequality
check
point
If
x-values
this
in
the
in
it
inequality,
or
f
of
( x )
the
<
of
region
this
satises
region
or
solve
●
Rearrange
graph.
of
both
or
think
in
quadratic
( x )
Factor
●
Place
>
the
inequality
inequality
algebraically:
so
that
either
f
( x )
<
0
0.
the
quadratic
satises
these
dividing
●
the
inequality,
the
f
a
●
algebraically:
you
satisfy
To
and
nd
its
zeros.
function.
point
the
rearrange
and
the
solution
test
0
intersection
zeros
your
and
inequality.
0
points
nd
inequality,
the
>
graphically:
Select
it
an
the
it
inequality
inequality.
the
x-values
several
x-value
substitute
then
two
into
into
is
in
true,
solution.
If
one
the
a
of
number
the
original
then
the
on
line,
intervals.
that
intervals
and
inequality.
interval
inequality
is
is
false,
If
the
part
of
then
it
isnot.
To
solve
●
Graph
●
Select
and
non-linear
the
a
region
point
If
the
where
the
Mixed
Solve
the
the
point
is
not
on
values
point
is
is
not
not
in
by
into
is
(e.g.
●
Repeat
To
check
●
Choose
for
all
intervals.
your
solution
algebraically:
curve
original
true,
(e.g.
variables:
equality.
the
is
true,
two
the
the
inequality
the
inequality
where
1
If
dened
that
substitute
inequality.
region
inequalities
shade
inside
shade
the
the
outside
the
the
curve).
region
the
a
point
inequality,
inequality.
If
x-values
this
in
it
in
the
and
region
test
this
satises
region
you
think
point
the
in
inequality,
satisfy
the
satises
the
then
the
inequality.
curve).
practice
each
answers
or
inequality
accurate
graphically,
to
3
giving
exact
4
A
rectangular
of
s.f.
100
m
is
playing
to
have
eld
an
with
area
of
a
no
perimeter
more
than
2
500m
(x
+ 2)
> 1
b
x
− 3x
d
2x
− 1 ≥
−11x
−2 x
+ 15
< 3x
possible
5
− 3 > 4, x
≠ 0
+ 1 < 3x
the
playing
eld.
+ 1
A
tourist
agency’s
prots
P
(in
dollars)
2x
f
< − 1, x
x
of
2
− 3x
5
2
e
maximum
+ 4
length
2
2
c
the
2
2
a
Determine
5
≠ 5
can
be
modelled
by
thefunction
x
2
( x ) =
P
2
Shade
the
region
representing
the
graph
of
−25 x
number
inequality.
Check
you
have
shaded
the
using
a
test
2
y
>
x
of
tourists.
where
Determine
how
x
is
the
many
are
needed
for
the
agency’s
prot
to
be
point.
at
a
− 3000 ,
correct
tourists
region
+ 1000 x
each
least
$5000.
2
−
4 x
+ 2
b
y
≥
−2 x
+ 6x
Problem
solving
2
c
y
<
2x
− 8x
− 3
d
y
<
log ( x
+ 1)
6
(
e
y
≥
x
2)
x
e
+
5
f
Orange
cans.
y >
x
Solve
algebraically
and
To
conrm
+ 5x
c
2x
+
− 6
<
0
b
x
sold
in
down,
2-liter
the
cylindrical
surface
must
be
less
than
1000
cm
area
of
Determine
graphically.
d
(x
− 10 x
2
x
− 24
>
cans,
0
values
for
− 4 x
+ 3)
<
25
7
A
packing
at
< 7x
f
5x
− 13 x
≥
3
3x
> 0, x
+ 5
1
the
radius
and
height
of
the
d.p.
least
company
600
cubic
designs
inches.
boxes
with
Squares
are
a
volume
cut
from
−6
the
g
to
2
2
2x
accurate
2
≤ 35
of
x
be
2
can
possible
x
x
to
costs
2
2
a
e
is
keep
+ 4
a
3
juice
2
≠ −5
of
a
20
inch
by
25
inch
rectangle
of
5
h
> 4, x
x
corners
≠1
cardboard,
and
the
aps
are
folded
up
to
make
an
1
open
box.
should
be
Determine
cut
from
the
the
size
of
the
cardboard,
E14.1 A world of dierence
squares
accurate
to
2 97
that
3
s.f.
Review
1
A
small
in
context
satellite
dish
has
a
parabolic
cross-section.
2
A
parabolic
microphone
uses
a
parabolic
ATL
Its
diameter
is
30
cm
and
it
is
15
cm
deep.
reector
onto
a
Sketch
b
When
the
parabola
on
the
coordinate
a
decide
where
receivesa
is
region
place
the
a
inside
the
State
its
dish,
receiver
number
inequality
wherethesignals
strongest.
satellite
maximum
Determine
the
to
the
cross
are
of
that
so
that
of
a
Draw
signals.
b
The
be
which
How
you
have
Statement
Modelling
2 9 8
of
cm
a
the
a
to
of
the
state
On
its
your
needs
statement
of
inquiry?
Give
specic
the
equivalent
forms
E14 Models
of
a
waves
maximum
depth
this
of
20
cm.
information.
representation
can
improve
be
placed
that
in
the
cross-section.
mathematical
describes
model
this
region
and
domain.
graph
check
your
from
examples.
decision
a,
shade
solution.
Inquiry:
with
to
parabola’s
discuss
the
sound
has
maximum
represent
microphone
(inequality)
domain.
explored
focus
cross-section
and
graph
Determine
and
and
50
interior
c
Reect
Its
then
it
represents
to
receiver.
and
engineers
section,
likely
collect
axes.
width
designing
to
making.
the
region
in
b,
Branching
E15.1
Global
context:
Objectives
●
Drawing
tree
out
Identities
Inquiry
diagrams
to
represent
conditional
and
relationships
questions
●
What
●
How
is
conditional
probability?
F
probabilities
●
Calculating
diagrams,
Venn
Determining
probabilities
diagrams
whether
two
and
from
two-way
events
are
tree
can
you
probability
tables
●
C
independent
In
which
What
a
ways
probability
●
represent
on
be
aects
tree
can
conditional
diagram?
conditional
represented?
the
decisions
we
make?
D
ATL
Communication
Understand
and
use
mathematical
notation
4 4
15 1
E15 1
Statement of Inquiry:
Understanding
resultfrom
health
using
and
logical
making
healthier
representations
choices
and
systems.
2 9 9
CIGOL
●
conditional
Y
ou
●
should
draw
tree
already
diagrams
know
1
how
Caesar ’s
the
He
new
1980s
Draw
a
playlist
and
selects
choosing
a
to:
eight
two
the
songs
songs
same
tree
contains
at
from
songs
the
random,
song
diagram
ten
from
1990s.
without
twice.
to
represent
this
situation.
b
Find
the
both
●
use
the
addition
multiplication
and
2
You
probability
from
have
a
the
that
the
two
songs
are
1980s.
standard
deck
of
52
playing
cards.
rules
a
You
take
one
you
take
that
card. What
either
a
is
red
the
probability
card
or
a
face
card?
b
You
take
one
card,
another. What
take
c
You
a
5
without
one
calculate
of
probabilities
mutually
3
exclusive
A
box
15
events
a
use Venn
set
diagrams
and
4
notation
A
be
There
7
5
card?
then
take
a
are
a
12
take
is
5
soft-center
from
event ‘pick
33
a
the
a
another,
the
followed
by
chocolates
One
center ’ and
taking
copy
of
of
taking
Ar t
both
subjects.
3 0 0
E15 Systems
let
are
Represent
diagram.
the Venn
this
diagram
below
question.
the
par t
B
of
the Venn
diagram
that
represents:
B
A
is
and
Biology. There
U
d
and
chocolate
center ’
.
A
a
a
box.
hard
soft
students
taking
a Venn
par t
Shade
take
you
P(A  ∪ B )
with
each
face
then
that
P(A).
students
Make
you
event ‘pick
students
this
a
and
chocolates.
random
the
the
Find
28
at
be
Find
b
●
contains
picked
B
that
and
card?
hard-center
Let
by
card
it,
probability
replacement. What
probability
●
replace
the
followed
take
face
is
b
′
e
B′
( A ∩ B )′
c
A ∩ B
for
S TAT I S T I C S
Conditional
F
●
What
●
How
is
Exploration
The
probability
you
that
it
will
rain
Amanda
the
that
she
1
A
be
Copy
the
and
conditional
probability
on
a
tree
diagram?
1
that
Let
probability?
represent
theprobability
probability
P R O B A B I L I T Y
probability
conditional
can
A N D
event
plays
‘rains
complete
tomorrow
plays
tennis
tennis
is
tomorrow’
this
tree
is
is
0.25.
0.1.
If
If
it
it
rains
tomorrow,
doesn’t
rain
tomorrow,
0.9.
and
diagram
B
be
for
0.1
the
the
event
events
‘plays
A
B
P (A
∩
B)
B
P (A
∩
B
B
P (Aʹ
∩
B)
B
P (Aʹ
∩
B
and
tennis’.
B
A
0.25
)
0.9
A
2
3
State
Find
whether
the
A
and
probability
B
are
independent
)
If
the
of
an
outcome
events.
event
aect
a
b
Amanda
Amanda
plays
does
tennis
not
tomorrow,
play
tennis
given
that
tomorrow,
it
is
given
the
Amanda
plays
d
Amanda
does
tennis
tomorrow,
given
that
it
that
is
In
step
3
Amanda
In
of
Exploration
playing
tennis
mathematical
that
the
Labelling
is
you
A
has
conditional
tennis
tomorrow,
should
have
conditional
notation,
condition
the
1
play
P (B|A)
occurred’,
on
is
probabilities
found
whether
‘the
or
it
not
is
raining
the
outcomes
of
the
event
or
the
not
A)
tree
it
is
not
raining.
probability
it
of
probability
the
P (B
that
is
B
of
diagram
in
the
experiment,
raining
that
probability
‘the
on
given
the
events
are
independent
of
raining.
occurring
B
given
looks
B
P (A
∩
B)
B
P (A
∩
B
B
P (Aʹ
∩
B)
B
P (Aʹ
∩
B
given
A’.
like
this:
A
P (A)
P (B
P (B
P (A
A)
A
)
A
)
)
)
A
P (B
not
probability
of
then
not
one
does
raining
second
c
in
experiment
that:
)
E15.1 Branching out
3 01
Multiplying
along
the
branches
for
P (A ∩
This
rearranges
A
and
B
gives:
P (A
∩
B )
=
P (A)
×
P (B | A).
B )
to: P( B | A) =
ATL
P ( A)
The
You
conditional
can
From
use
MYP
Theorem
For
any
1
A
A
and
can
and
B
The
theorem
use
B
are
outcome
of
in
A
and
the
Let
A
B
A
as
other.
B )
=
rule
B,
a
P ( A) + P ( B ) - P ( A ∩
test
events
if
for
B )
P ( A) × P ( B )
independent
events.
.
outcome
using
axiomatic
=
then
the
this
B )
is:
P (A ∩
events,
Writing
probability
recall:
is:
and
2
probabilities.
of
one
probability
does
not
notation
aect
gives
independent
that
that
a
events
particular
Minnie
misses
probability
that
probability
misses
a
Draw
b
Find
P(
c
Find
P(B ).
d
Show
tree
that
her
system.
then
that
to
train
her
she
the
dentist
diagram
P (B | A)
=
P (B | A′)
.
is
late
dentist
misses
train
is
is
.
If
the
appointment
her
dentist
late
and
train
is
.
is
If
late,
the
appointment
let
B
be
the
the
train
is
is
.
probability
appointment.
represent
this
situation.
).
the
appointment
Theorem
train
are
being
not
late
and
independent
Minnie
events
missing
using
i
her
dentist
Theorem
2
and
3.
Continued
3 0 2
the
another
1
thatMinnie
ii
rule
independent
are
the
a
Standard,
conditional
3
the
be
calculate
P (A ∪
independent
probability
late,
B,
events
probability
not
to
axiom:
multiplication
are
Example
The
and
Theorem
the
Theorem
If
addition
A
independent
You
If
The
2
formula
Mathematics
events
Theorem
For
this
probability
E15 Systems
on
next
page
S TAT I S T I C S
a
A N D
P R O B A B I L I T Y
3
P (A
B
∩
B)
5
A
1
4
2
B
5
B
(missing
the
conditional
appointment)
on
A (train
being
is
late).
1
P (Aʹ
B
∩
B)
5
3
A
4
4
B
5
b
P (A
c
P(
B )
=
)
=
P (A
,
P
)
B
=
occurs
These
P (B )
d
so
=
either
two
add
the
by
events
and
P (
are
)
d
ii
and
P (B | A)
=
P (B | A′)
=
P (B )
=
P (B | A)
≠
P (B | A′)
≠
P (B )
and
events
therefore
A
and
B
events
are
A
not
and
B
or
=
P (A
P
mutually
).
exclusive,
Use
Theorem
2.
Use
Theorem
3.
independent.
are
not
independent.
In
Example
happened.
Aand
B
1
the
probability
So
the
outcome
are
not
Reect
●
How
is
the
dierent
discuss
another
independent.
●
A
B
diers
aects
depending
the
outcome
on
of
whether
B
and
or
thus
not
the
A
has
events
independent.
and
Describe
of
of
type
Explain
tree
from
of
how
diagram
the
one
1
problem
in
you
for
the
in
probability
know
they
are
probability
Example
where
not
events
are
not
independent.
problem
you
B).
probabilities.
i
So
B)
thought
of
1?
E15.1 Branching out
3 0 3
Practice
For
A
1
questions
and
B
One
Jar
5
X
Jar
1
is
and
and
let
There
are
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takes
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a
3
be
10
B
and
be
two
the
in
‘the
and
A
be
bag:
and
rst
thrown.
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5
a
doughnut
is
be
the
takes
‘taking
events
‘coin
from
a
Y
lands
contains
shoelace
brown
and
takes
toasted
not
3.
event
Jar
a
a
shoelace
he
or
sixes’.
coconut
then
whether
Theorem
shoelaces.
event
brown
it,
A
Norina
toasted
eats
or
two
black
the
2
Let
‘throwing
‘taking
a
determine
Theorem
shoelaces.
Let
event
doughnut
event
are
event
black
Y.
doughnuts
one
dice
the
and
either
shoelaces
4
Jar
the
diagram
using
be
brown
from
B
tree
by
and
let
shoelaces
shoelace
X’
draw
ipped
up’
contains
a
Let
4,
independent
side
brown
and
3
are
coin
tails
2
1
5
jar
and
Jar
X
from
Y’.
Boston
second
coconut’
from
shoelace
creme.
doughnut.
let
B
be
As
the
you
life,
event
‘the
second
doughnut
is
toasted
my
ramble
The
probability
that
Mickey
plays
an
through
coconut’.
Whatever
4
on
friend,
online
room
escape
game
given
that
it
Keep
be
your
your
eye
goal,
upon
the
doughnut,
is
Saturday
is
60%.
On
any
other
day
of
the
week,
the
probability
is
50%.
And
Let
B
be
the
event
‘playing
a
room
escape
game’
and
let
A
be
the
event
‘the
–
not
In
is
questions
Justify
5
Saturday’.
There
two
are
the
rst
the
event
10
Let
B
is
to65%.
the
rst
to
1
token
Let
is
or
not
A
a
be
selected,
a
12
A
the
two
events
are
independent.
be
is
placed
it
is
event
second
a
put
4
boys.
Your
name
back
girl.
of
‘a
boys
boy
in
a
box
put
Let
that
eat
that
the
a
in
B
in
a
the
be
eats
a
two
in
rst
the
teacher
hat.
hat.
the
will
are
selects
When
Let
event
is
and
drawn
an
even
breakfast.
exercise
likelihood
healthy
drawn
box,
token
selected
healthy
he
the
and
back
number
breakfast,
event
and
not
is
not
likelihood
healthy
the
it
girls
everyone’s
selected
the
60%
the
6
A
be
that
the
girl.
are
the
statistics,
eat
class:
putting
drawn,
that
breakfast,
your
person
to
is
Let
event
that
that
he
If
a
is
at
random.
second
number
number.
a
day
boy
is
exercises
breakfast’.
Let
B
eats
90%.
be
falls
the
‘exercises’.
Problem
8
whether
after
is
selected
selected.
be
doesn’t
event
name
the
rst
in
random
numbered
ahealthy
he
at
that
According
If
students
person
the
token
determine
person’s
Tokens
After
7
8,
answer.
students
second
6
5
your
Blood
is
solving
classied
in
a
variety
of
ways.
It
may
or
may
not
contain
Bantibodies.
Its
rhesus
status
can
be
positive
The
likelihood
of
someone
The
likelihood
of
being
Rh+
The
likelihood
of
being
B +
Let
X
be
3 0 4
the
event
‘having
(Rh +)
having
is
B
antibodies
B
is
antibodies
antibodies’.
E15 Systems
negative
(Rh
).
85%.
80%.
(having
B
or
Let
Y
and
be
being
the
Rh +)
event
is
‘being
68%.
Rh+’.
4.
the
hole.
The modied Optimist’s
Creed
day
upon
S TAT I S T I C S
C
Representing
●
The
In
In
Venn
this
The
diagram
scenario,
sample
This
which
is
it
space
can
represents
is
is
represented
ways
taken
in
the
conditional
two
that
reduced
conditional
to
Venn
A
events,
has
the
and
already
elements
diagram
at
B.
be
Find
represented?
the
probability
The
shading
region
of
on
B
represents
‘B
represents
this
event
that
given
is
the
A.
A
B
right
3
in
Here,
A,
or
the
A ∩
darker
B ,
A
shading
B
region.
4
numberofelementsinA ∩
P (B | A)
B
n (A ∩
a
Venn
sample
diagram
space
Example
In
a
group
to
can
given
5
=
7
n ( A)
calculate
conditional
probability
by
reducing
the
condition.
2
of
30
chocolate,
6do
a
Draw
Venn
b
Given
a
you
the
3
3
B )
=
=
numberofelementsin  A
In
5
A
also
A’.
P (B | A).
occurred.
of
4
bythe
P R O B A B I L I T Y
probabilities
probability
A
A N D
that
girls,
not
12
like
girls
diagram
she
likes
like
milk
chocolate.
to
nd
One
the
chocolate,
chocolate,
girl
is
probability
nd
the
15
like
selected
the
at
girl
probability
dark
random.
likes
she
chocolate.
likes
milk
chocolate.
c
Find
the
milk
d
Test
whether
using
Let
x
probability
that
she
likes
dark
chocolate,
given
that
she
likes
chocolate.
be
i
liking
Theorem
girls
who
2,
milk
and
like
and
ii
both
dark
chocolate
Theorem
milk
and
are
independent
events
by
3.
dark
chocolate.
Draw
Use
24
=
12
24
=
27
x
=
3
+
M
15
a
Venn
diagram.
Theorem
1
to
nd
n
.
x
x
D
D
9
3
is
the
event
‘likes
dark
chocolate’.
12
M
is
the
event
‘likes
milk
chocolate’.
6
Continued
on
next
page
E15.1 Branching out
3 0 5
a
P ( likes
chocolate)
b
P (M | likes
=
Reduce
chocolate)
=
given
the
M
6
the
sample
condition
people
space
(likes
who
do
to
only
the
chocolate).
not
like
Ignore
chocolate.
D
9
3
12
6
c
P (D | M )
=
Reduce
M
the
sample
space
to
M.
D
9
3
12
6
d
i
M
and
D
are
independent
,
events
ATL
M
and
D
are
P (D | M
)
=
P (D | M
)
=
P (D | M
)
Practice
and
only
if :
independent
P (D | M ′)
,
=
therefore
events
if
they
and
are
only
not
independent
From
if :
P (D)
P (D | M ′)
P (D | M ′)
=
,
P (D),
therefore
they
are
not
independent
events.
2
a
2
the
a
eld
kayaking,
32
Venn
diagram
at
the
P (B | A)
On
b
trip,
50
caving,
students
caving,
and
chose
4
a
Represent
b
Find
the
students
neither
this
26
chose
kayaking
ii
chose
caving,
iii
chose
caving
iv
chose
kayaking,
given
a
on
they
that
E15 Systems
a
chose
activity.
Venn
student
that
given
do
both.
students
neither
information
that
to
or
they
diagram.
picked
chose
at
random:
kayaking
chose
B
15
nd:
P (A | B )
choose
chose
probability
right,
activity,
kayaking,
students
i
3 0 6
Theorem
2.
Using
Theorem
3.
events.
A
1
Using
,
,
ii
if
caving.
11
21
S TAT I S T I C S
3
In
a
and
group
8
of
study
a
Represent
b
Find
the
A
this
Art
ii
studies
Drama,
iii
studies
either
iv
studies
both
group
of
and
25
a
does
not
sing
b
does
not
dance
c
sings,
The
Venn
Drama
a
on
a
Venn
student
that
Drama
5
or
Art,
15
study
Drama,
the
diagram.
picked
student
at
random:
studies
Art
Art
and
that
are
in
students
Determine
Problem
that
given
students
given
study
Art,
given
the
student
studies
either
Art.
dance,
atrandom.
students
information
studies
or
12
the
the
a
show.
do
not
sing
probability
student
Of
does
these,
or
dance.
that,
not
10
in
the
of
A
the
students
student
show,
the
is
sing,
selected
student:
dance.
solving
diagram
shows
35
students’
homework
subjects.
Ar t
a
Find
the
P R O B A B I L I T Y
subject.
i
6sing
5
students,
probability
Drama
4
30
neither
A N D
probability
that
a
student
selected
at
random
(A)
5
i
Art
ii
Biology
iii
Chemistry
Biology
(B)
had:
2
4
homework
7
homework
7
1
b
Find:
i
c
Chemistr y
P (A | B )
ii
Determine
the
homework
in
Exploration
In
42
homework.
the
rst
probability
all
were
three
that
given
of
asked
a
to
European
Cup
predict
10
●
6
●
15
females
predicted
Italy
●
11
females
predicted
Spain
males
1
Represent
2
A
student
was
predicted
predicted
this
was
student
subjects,
●
males
a
iii
picked
that
the
P (C | A′)
at
random
student
had
had
homework.
2
semi-nal
students
P (B | C )
(C)
the
tournament,
Italy
play
Spain.
winner.
Italy
Spain
information
selected
at
in
a
two-way
random.
Find
table.
the
probability
that
this
student
male.
Continued
on
next
page
E15.1 Branching out
3 07
3
4
By
altering
the
sample
probability
the
student
Find
the
probability
predicted
5
space
as
you
predicted
that
the
did
Italy,
student
in
the
given
was
Venn
that
female
diagrams,
the
student
given
that
nd
was
the
the
male.
student
Spain.
Find:
a
P (M )
b
P (F )
c
P (S )
d
P (I )
e
g
P (M | S )
h
P (S | M )
i
P (F | I )
j
P (I | F )
k
P (M | I )
l
P (I | M )
In
the
M
P (F | S )
f
F
I
S
6
other
shows
how
Male
semi-nal,
42
predicted
England
(E )
(F)
Total
a
Find
the
b
Find:
Reect
Compare
make
a
i
the
The
Germany
(G)
12
16
28
18
24
42
P(M
probabilities
about
a
table
Total
14
that
two-way
winner.
8
student
selected
| E )
discuss
statement
the
Germany.
6
probability
and
play
students
(M)
Female
England
ii
at
P(M
random
is
male.
| G )
2
for
the
which
two
semi-nals
events,
if
any,
are
in
Exploration
dependent
2.
and
Hence
which
are
independent.
Practice
1
There
male
3
are
28
players
18females
a
Use
this
participants
and
12
are
in
a
golf
competition.
professional
female
Of
these,
players.
5
are
professional
There
are
a
total
of
playing.
information
to
complete
Male
Professional
the
two-way
(M)
table.
Female
(F )
Total
(P)
Amateur(A)
Total
b
Use
i
c
the
table
to
P (M | P )
Hence,
calculate
ii
using
a
the
following
P (M | A)
theorem,
determine
professional
are
independent
3 0 8
E15 Systems
probabilities:
iii
P (M
whether
events.
=
Male
P (S | F )
)
being
male
and
being
a
=
=
=
Female
Italy
Spain
S TAT I S T I C S
2
100
students
andwhether
Use
not
the
were
or
results
theevents
asked
not
they
below
are
to
whether
were
on
they
the
complete
a
were
left-
school
or
P R O B A B I L I T Y
right-handed,
soccer
two-way
A N D
team.
table
and
decide
whether
or
independent.
4
●
P (left-handed)
=
10
3
●
P (left-handed
and
not
on
the
soccer
team)
=
10
9
●
P (right-handed
and
not
on
the
soccer
team)
=
20
Problem
3
From
of
the
gender
solving
table,
determine
(male
Justifyyour
or
Male
Female
Making
●
What
Objective
iii.
move
You
will
D:
need
new
the
test
test
and
75
on
of
positive
A
negative
Here
are
teacher
use
can
test
test
Dr
not
kidney
Total
Other
this
is
independent
data
came
from.
teacher
25
16
50
the
dierent
the
table
with
decisions
we
probability
make?
forms
and
also
of
a
mathematical
tree
diagram
in
representation
order
to
solve
the
problem
in
3
quickly
140
kidney
Does
teacher
where
8
detect
patients.
do
She
kidney
knows
disease.
that
65
of
Dr
Julia
them
Statham
have
performs
kidney
disease
not.
result
result
indicates
indicates
Statham’s
kidney
no
disease
disease.
kidney
disease.
results:
Positive
Has
science
3.
them
A
a
school
Communication
Exploration
A
the
decisions
aects
between
Exploration
being
in
answer.
Science
D
if
female)
test
result
Negative
test
result
Total
30
35
65
15
60
75
45
95
140
have
disease
Continued
on
next
page
E15.1 Branching out
3 0 9
Let
T
Let
D
1
A
be
the
event
the
patient
a
The
be
event
is
is
b
disease
kidney
disease.
2
the
Use
3
use
Construct
the
4
5
the
not
For
or
this
the
After
and
it
gives
Statham
in
a
the
to
disease’.
P (D)
positive
result
will
positive’.
Calculate:
c
negative
for
use
table
the
and
decide
d
result
for
people
test
if
your
whether
tree
diagram
doctor
in
that
your
should
problem,
tree
which
diagram?
having
an
claimed
it
Patient
data
Using
did
this
on
represents
tree
use
the
2
was
T
is
the
eye
have
this
for
an
the
diagram
eye
caused
for
same
to
accurate.
of
doctor
information
conrm
do
you
by
as
whether
prefer:
this
condition,
the
your
is
the
table
eye
condition.
the
table,
The
sight
lost
table
Did
his
sight
shows
not
lose
25
225
75
675
determine
whether
the
the
sight
patient’s
claim
answer.
of
the
represents
‘plays
operation
in
a
tree
representation
validity
diagram
event
in
patient
operation.
diagram
and
use
it
to
determine
if
the
justied.
which
the
eye
a
the
operation
information
claim
Determine
Venn
the
information
Represent
The
the
have
test.
representation
treatment
Justify
illustrates
ATL
had
not
the
valid.
patient’s
c
90%
not
with
not
Explain.
operation
that
Patient
b
is
do
knowledge
or
Lost
is
people
who
4
hospital’s
a
P (D ′)
test.
information
the
Practice
1
a
kidney
random.
probability
the
has
is
table.
Use
or
Dr
if
a
information
conditional
should
and
at
result
P (T ′)
successful
kidney
test
‘patient
selected
P (T )
test
‘patient
tennis’
(the
patient’s
65
and
table
claim.
members
G
is
the
of
or
the
Give
a
event
tree
sports
‘plays
T
diagram)
evidence
for
best
your
decision.
club.
golf ’.
G
If
30
10
Represent
b
Use
all
golf
are
c
three
information
in
representations
independent
Explain,
310
this
with
a
to
two-way
table.
determine
if
playing
tennis
and
playing
events.
reasons,
which
E15 Systems
are
using
tree
diagram,
and
T ′
pair
of
a
put
T
15
10
a
you
representation
was
the
easiest
to
work
with.
on
the
rst
branches.
S TAT I S T I C S
Problem
3
The
L
is
tree
the
A N D
P R O B A B I L I T Y
solving
diagram
event
shows
‘studies
information
Law’
and
E
is
about
the
100
event
students
‘studies
at
a
university.
Economics’.
L
0.8
E
0.4
0.2
L
L
0.4
0.6
E
0.6
L
a
b
Use
the
tree
i
P (E )
v
P( E
diagram
ii
∩
L ′)
Determine
whether
●
Can
●
How
do
Practice
1
The
use
cafeteria
had
From
b
Hence,
d
Verify
e
Using
2
70
b
What
are
2
the
L)
vii
Law
or
probabilities.
P (L | E′)
P( E ′
∩
studying
diagram
diagram
waes
one
10
to
is
iv
L ′)
viii
Economics
represent
most
at
any
useful
morning
morning.
students
and
4,
Of
had
using
is
a
iii
table
results
use
who
that
)
2
to
are
and
P( E
∩ L)
P (L)
are
for
situation?
a
given
situation?
break.
these,
33
There
students
were
had
apples
(A),
neither.
Theorem
a
the
the
1,
draw
a
Venn
diagram.
represent
the
same
3,
)
the
for
test
iv
P (W | A)
v
P (W | A′)
information.
each
form
whether
of
representation.
having
waes
and
having
events.
most
worked
their
P (A | W
Theorem
independent
about
worked
Using
and
P (W
two-way
worked
Given
and
cafeteria
(W),
ii
students
a
apples
Theorem
questions
rest
iii
following
3
which
information,
that
In
surveyed
studying
the
calculate:
a
apples
The
the
P (A)
Make
c
in
the
c
∩
probability
decide
sells
waes
a
i
nd
5
51students
18
P( E ′
discuss
any
you
you
events.
and
you
help
P (L | E )
vi
independent
Reect
to
in
work.
appropriate
the
12
representation
holidays,
boys
and
30
28
of
whom
girls
to
are
worked
solve
boys,
the
problem.
were
full-time.
part-time.
student
worked
probability
that
part-time,
a
person
what
is
chosen
the
at
probability
random
is
she
male
is
female?
and
full-time?
theorem
‘working
as
part-time’
a
justication,
are
determine
independent
whether
‘being
male’
and
events.
E15.1 Branching out
31 1
3
The
and
probability
95%
if
it
Gregoire
clear.
The
arriving
Calculate
the
probability
of
b
Calculate
the
probability
that
c
Determine
Out
of
There
150
whether
are
the
30
are
left-handed
a
Find
P (F
|
L).
b
Find
P (L
|
F ).
c
Determine
whether
independent.
it
is
of
events
is
60%
if
there
is
fog,
at
school
that
school
on
time.
Gregoire
on
time’
arrives
and
‘a
late.
clear
answer.
(L)
are
‘being
time
20%.
given
to
your
who
is
on
arriving
‘getting
left-handed
Justifyyour
fog
clear,
Justify
students
the
school
Gregoire
events
independent.
students,
are14
at
probability
a
morning’
4
is
of
and
80
female
male’
are
(F
and
male
(M
).
).
‘being
left-handed’
are
answer.
Summary
In
mathematical
probability
condition
of
B
of
A
given
B
notation,
occurring
has
P (B | A)
given
occurred’,
or
is
‘the
that
‘the
If
A
A
and
B
conditional
probability
events,
then
B
are
does
independent
not
If
this
formula
the
events
outcome
if
the
of
outcome
the
other.
A
and
3
B
P (B | A)
use
aect
axiom:
P (B | A)
can
independent
probability
Theorem
You
are
A’.
one
The
and
the
to
calculate
are
=
independent
events
then:
P (B | A′)
conditional
probabilities.
You
as
a
can
test
for
Mixed
1
A
lab
at
2
are
random
type
to
45
B.
rule)
events.
check
for
Find
the
probability
have
the
same
the
probability
A,
tree
type
Two
b
type
multiplication
samples
are
Draw
Find
a
blood
these,
a
c
(the
independent
100
Of
15
Theorem
practice
has
tested.
and
use
samples
to
are
are
represent
that
blood
that
need
40
to
be
type
to
be
2
the
rst
From
the
Venn
diagram
Find
not
the
a
P (A)
b
P( A ∩ B )
c
P (A | B )
d
P (B | A)
e
P (B )
f
P (A | B ′)
g
P (B | A′)
this.
two
samples
group.
that
the
type
312
probability
O,
given
nd:
drawn
the
rst
second
one
was
sample
type
is
A
B
B.
20
d
below,
A
contamination.
diagram
given
that
O,
that
that
the
the
second
rst
E15 Systems
sample
onewas.
10
15
is
20
of
S TAT I S T I C S
3
The
probability
that
it
will
be
sunny
tomorrow
6
4
is
.
If
it
is
sunny,
the
probability
that
In
a
netball
court:
8
There
are
match
have
there
brown
A N D
are
hair
14
and
P R O B A B I L I T Y
players
6
have
on
blue
the
eyes.
the
5
4
students
whohave
neither
brown
2
cafeteria
will
sell
ice
creams
is
.
If
it
is
not
hair
nor
blue
eyes.
Let
H
be
the
event
‘has
3
brown
sunny,
the
probability
that
the
cafeteria
will
hair’
and
B
be
the
event
‘has
blue
eyes’.
sell
1
ice
creamsis
a
Draw
a
Venn
b
Determine
diagram
to
represent
this.
3
a
Represent
this
information
in
a
tree
diagram.
these
mathematical
b
Find
the
c
probability
Determine
of
4
the
cafeteria
S
is
ice
the
event
It
is
P (S
will
if
the
creams
event
‘it
is
=
‘Tim
ice
sun
it
will
sunny
shining
and
wearing
the
selling
i
P (H )
iii
P (H
shorts’
and
R
is
v
P ( H ′)
vii
P (H
∪
B ′)
ix
P (H
∪
B )′
that
P (R)
=
0.5,
P (S | R)
=
0.2,
Represent
b
Use
both
c
Hence
this
situation
in
a
tree
Theorem
events
is
related
S
trying
as
to
iv
P (H
vi
P ( B ′)
∩
B )
viii
P( H
∩ B ′)
2
and
and
to
being
liking
70students,
30
T
are
Theorem
not
determine
180
cm
or
basketball.
of
which
if
all
the
correct
conditional
mathematical
probabilities,
notation.
3
to
A
survey
was
carried
out
regarding
the
school
verify
code.
The
two-way
table
shows
the
results:
independent.
being
more)
She
were
state
the
diagram.
dress
dened
B )
and
0.8.
a
Sahar
∪
P (B )
the
7
5
ii
events.
using
that
correct
and
creams.
independent
is
be
using
raining’.
known
| R′)
are
that
sell
probabilities
notation:
is
tall
Should
(here
have
somehow
180
have
code
surveyed
over
Should
dress
not
Total
dress
code
cm.
Middle
48
students
surveyed
like
basketball,
29
30
of
45
school
whichwere
under
180
cm.
High
a
Draw
a
two-way
table
to
represent
40
this.
school
b
Find
P(tall).
c
Find
P(not
d
Are
Total
the
tall
events
basketball’
|
likes
‘being
55
110
basketball).
tall’
independent?
and
a
Copy
b
Use
and
complete
the
table.
‘liking
Justify
Theorems
2
and
3
to
determine
whether
your
type
of
school
and
wanting
a
dress
code
are
answer.
independent
Review
in
1
shows
The
200
table
events.
context
the
occurrence
of
diabetes
in
a
From
the
table
nd:
people.
Diabetes
Not
overweight
No
10
35
Let
D
be
the
event
‘has
diabetes’.
Let
N
be
the
event
‘not
overweight’.
P (D)
iii
P (D | N )
ii
P (N )
iv
P (D | N ′)
90
b
Overweight
i
diabetes
65
Determine
being
whether
overweight
having
are
diabetes
independent
E15.1 Branching out
and
not
events.
3 13
2
The
has
a
probability
a
test
bone
for
disorder
this
condition
not
there
Let
B
that
is
randomly
is
0.01.
condition
there,
(a
a
false
and
is
The
selected
probability
positive
0.05
if
person
the
is
0.98
if
4
that
survey
is
positive).
cancer.
Of
survey,
981
lifetime.
were
be
be
the
a
Draw
the
event
event
‘test
‘has
is
the
bone
disorder’
and
of
tree
to
a
represent
cancer
at
who
some
in
the
smoking
non-smokers
smokers
people
at
the
battled
in
point
and
the
in
survey,
their
there
cancer.
T
positive’.
diagram
the
looked
cigarette
125 000
had
Of
1763
people
between
the
Find
the
selected
a
200 000
relationship
the
condition
A
probability
from
the
that
study
an
individual
who
battled
cancer
these
was
a
smoker.
probabilities.
b
b
Calculate
the
probability
of
success
for
Find
the
cancer
test.
Note
positive
the
for
negative
test
is
people
for
successful
with
people
the
if
it
the
and
c
disorder.
Let
the
a
3
Michele
is
making
drinks
to
sell
at
the
She
makes
a
‘Super
green
B
be
the
event
former
event
‘being
not
baby
grapes
a
with
‘Triple
milk,
spinach,
a
cucumber,
caloric
chocolate
chocolate
content
milkshake’
a
caloric
syrup
and
content
of
kiwi
140
with
370
kcals,
and
d
kcal
Does
records
the
number
of
MYP
your
people
two
buying
each
product.
26
super
green
cancer’
and
(includes
Demonstrate
A
be
being
are
whether
independent.
and
MYP
smoothies
answer
mathematically.
in
c
choosing
encourage
to
smoke?
or
discourage
Explain
The
following
information
was
extracted
from
a
cup.
cancer
website.
DP
1:
About
69%
of
skin
cancers
are
students
associated
purchase
‘having
smoker’
events
answer
from
Statement
students
had
ice-cream
per
skin
She
who
and
chocolate
chocolate
of
someone
smoothie’
apple,
5
with
a
smoker).
these
Justifyyour
with
that
smoked.
school
or
play.
never
tests
disorder
without
probability
the
and
48
with
exposure
to
indoor
tanning
triple
machines.
chocolate
13super
milk
milk
green
shakes.
DP
smoothies
students
and
24
purchase
triple
chocolate
Statement
people
shakes.
atotal
Determine,
using
whether
age
the
independent
two-way
to
table
Reect
both
of
the
their
to
and
theorem
student
choice
answer
of
this
2
and
(MYP
drink.
theorem
vs
DP)
Use
have
you
Give
specic
In
year
a
are
United
treated
population
Statement
tanning
of
330
States,
for
skin
3.3
million
cancer
out
of
million.
3:
10
million
US
adults
use
indoor
machines.
question.
Let
A
Let
B
be
be
the
the
event
event
‘developing
‘exposure
to
skin
cancer’.
indoor
tanning’.
discuss
explored
the
statement
of
inquiry?
Write
down
probability
statements
1,
2
and
3
in
notation.
examples.
b
Determine
cancer
and
whether
exposure
independent
Statement of Inquiry:
Understanding
health
and
making
healthier
choices
result from using logical representations and systems.
31 4
the
3,
is
a
How
a
2:
E15 Systems
events.
developing
to
indoor
skin
tanning
are
Practice
Answers
1
a
5
4112
b
114 333
5
c
340 030
5
5
E1.1
2
Y
ou
1
a
c
2
a
c
should
already
11 204
know
b
2996
1161
d
999 538
Fifty
b
Five
hundred
d
Five
tenths
Five
Practice
how
×
1
2
3
4
5
10
1
1
2
3
4
5
10
2
2
4
10
12
14
20
3
3
10
13
20
23
30
4
4
12
20
24
32
40
5
5
14
23
32
41
50
10
10
20
30
40
50
100
to:
thousand
1
1
a
23
b
25
c
109
2
a
191
b
226
c
107
a
b
4533
112 024
6
3
1215
=
6
299
6
d
92
3
1011
4
a
e
=
31
3
,
52
100 011
10
Every
f
=
35
2
digit
has
,
1010
10
moved
one
=
68
,
1111
10
to
the
=
b
has
been
lost.
So
the
number
10
×
23
10
=
21
10
×
35
6
6
156
5
right,
and
Mixed
10
the
has
practice
nal
1
zero
13
70
4
place
=
a
13
b
54
c
69
halved.
d
52
e
135
f
g
14 559
h
2418
i
a
1 100 100
e
1 101 100
2387
107
10
Practice
2
2
b
10 201
2
c
a
1 111 100 111
b
1 101 000
d
12 444
2
3
33 213
4
i
j
a
472
=
314
8
b
472
10
=
a
a
223
=
63
5
c
=
214
=
82
1741
8868
=
=
=
2468
a
b
c
c,
6542
b
is
at
1011
a
=
=
=
h
b
=
48
b
25.25°
e
=
a
10 110
e
22 211
45
2
=
9
231
10
the
3BEE
16
7,
=
6
largest
the
and
d
12″
58.205°
c
10 001 111
2193
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j
10 000
h
1081
3
1 011 323
4
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12
d
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4
k
12
11A
12
16
16
written
because
c
31
2
20 010
base,
because
his
number
5
was
a
10 110
b
11 110
2
d
out.
=
e
contains
a
1111
2
2120
3
number
c
2
2101
f
34 535
4
34 535
h
6
29B1
6
7,
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f
1 001 010
f
l
g
=
c
62.5083°
b
i
4
3464
when
9
3
8
used
90°
2
1725
i
3FFE
12
16
6.
a
11:20
b
13:15
c
14:25
9
d
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1319
10
101 101
3
4
15
10
=
4
22 222 022
=
11
110 213
d
least
8,
l
16
2
6
d
419
12
100 011 001
2
f
10
a,
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8
=
1844
smallest
b,
281
10
d
67
10
=
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2
=
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the
1 010 010
55
9
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k
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431
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9
i
b
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09:26
e
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729
10
7
a
1100
b
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2
d
Practice
a
210
=
300
12
c
b
301
10
BAA
=
e
=
1714
d
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10
10
=
30
a
190
=
c
b
b
10
c
47 806
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d
g
16
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2313
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4
g,
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ACE
f
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16
BEEF
d
a
The
possible
=
codes
256
run
from
00
g.
+
weight
system
231 232
1
g,
4
=
is
equivalent
to
counting
in
base
4.
1 223 111
4
g,
16
×
12
g,
3
×
64
4
g,
2
×
256
g,
2
×
1024
g,
1
×
4096
g
16
to
FF
,
or
0
16
10
to
255
1330
4
,
=
23 220
4
4
10
2
are
1
4
FACE
10
16
sothere
the
331 213
16
e
3
g,
10
Because
10
CAFE
10
1
183
16
48 879
16
51 966
=
10
10
e
2766
16
5
320
20
BE
10
4244
5
4
2
f
769
16
12
a
1 000 001
2
432
3
3
8
1
c
2
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×
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2
×
16
g,
3
×
64
g,
2
×
256
g
codes.
10
b
16
=
1 099 511 627 766
(Roughly
1.1
trillion)
9
a
It
might
ve
Practice
4
be
times
between
b
540
=
thought
the
the
a
445
2
a
222
b
360
b
144
8
c
2003
c
425
8
cent
a
6
137
b
c
6
1554
8
732
c
11 717
b
Since
216
8
=
1441
165
8
=
could
be
was
worth
relationship
highly
confusing.
5
10
12
+
dollar
dollar
the
5
1441
=
3421
5
5
BA99
12
5
the
9
5
a
the
that
redening
notes.
10 412
10
8670
7
eight
=
10412
4
and
but
so
8
246
3
coin,
idea
5
Hence
6
largest
good
4130
10
1
a
403
,
8
So
216
8
165
8
=
31
d
It
ten
notes.
would
require
a
lot
of
notes
to
pay
amounts
that
were
8
close
to
the
next
bill,
such
as
$990.
Answers
31 5
Review
1
a
2,
3,
in
5,
context
7,
Practice
11
1
a
2
A
B
C
g
f
b
13,
17,
19,
23,
29,
31,
37,
41,
43,
47
1
c
10,
11,
101,
11 101,
2
“
”
a
11 111,
represents
1,
4,
111,
9,
16,
1;
1011,
1101,
100 101,
no
25;
101 001,
number
the
10 001,
starts
square
10 011,
101 011,
with
a
=
|
— — |
— —,
49
=
101 111
leading
1,
1,
2,
=
3,
4,
8,
13,
21;
|
— — — |
—,
55
the
=
,
64
=
1,
3,
729
3
a
=
27,
|
81;
|

|
16487.5…
b
c
9,
659.5…
Fibonacci
|

|
|

|

|,
Jupiter:
Saturn:
powers
|

|
— —
,
6
A
89
=
of
2187
3;
=
243
=
2,
{1,
2,
2,
3},
B
|
13},
C
=
{
2,
2,
6}
=
Domain
of
g

is
f
A.
Range
of
g

is
f
C.
|

|
— —
a
h
g
|

|

|

|
— —|

|,
|
— — — |
— — — |
0
2
1
5
1
5
2
4
3
|

|
days
Domain
10 100 010 011 111
of
h 
is
g
{1,
2,
4};
range
of
h 
is
g
{
1,
3,
5}.
101 101 110 110 010
f
h
1 010 100 101 011 000
Neptune:
10 110 000 010 111 010
A
Venus
days
9,
numbers;
b
day
{5,
days
Uranus:
e
=
6
hours
1 010 010 100
d
the
2
13
|
— — — — — —
2
c
9
3
numbers;
|

|
— — —
d
34
2
zero.
c
b
2
10 111,
b
36
5
on
per
year
is
is
longer
less
than
than
its
year,
so
the
number
of
1.
1
2
3
3
5
4
4
7
6
Domain
of
f
is
 h
{2,
5,
7};
range
of
f
 h
is
{3,
4,
6}.
E2.1
c
g
f
Y
ou
should
1
already
know
how
x
to:
a
b
x
b
c
y
c
d
z
y
5
4
3
2
Domain
2
of
g

is
f
{b,
c,
d };
range
of
g

f
is
{x,
y,
z}.
0
0
3
1
a
4.5,
8,
12.5
2
b
2
Domain
is
{6,
8,
10};
range
is
{4.5,
8,
12.5}.
3
4
2
a
b
c
d
Domain
and
Domain
Domain
is
is
Domain
e
Domain
a
f
x
x
is
Range
≥
≠
x
is
0,
0;
≠
2
both
Range
2;
≤
are
Range
is
is
Range
x
≤
2;
y
y
is
≠
y
a
0.
≠
Range
(4)
=
5
b
f
(
1)
=
0.
is
y
≥
1
Domain
−5
c
f
(3x)
=
6x
2
+
5
0
3
9
4
5
13
8
Range
of
Practice
3
g

f
is
{
2,
0,
4,
8}.
9
1
of
k)(a)
Domain
h
is
domain
,
=
of
k
is
;
1
of
2a
h
a
is
2;
domain
domain
,
of
k
of
is
h
+
k
is
a
12
b
e
1
f
h)( x )
=
−x
7
c
18
d
2.25
h
11

;
36
g
6
k
x
2
i
3
( g
2
1
3
3
(h
3
0.
5
Practice
0
0.
b
3
g
f

≥
3x
2
+
6
j
3x
+
1
+
2
5
=
l
x
+
2x
+
2
+
5x
x
+
5;
domain
of
( g
h)
is

2
2
a
x
5
1
3
Domain
of
h
is
domain
;
of
k
is
x
≥
;
2
b
(2x
+
2
1)
5
+
1
=
(4x
2
+
4x
+
1)
4x
+
4x
4
3
1
2
(h
+
k)( x )
=
2x
−
x
+
3x
+ 1 ;
domain
of
(h
+
k)
is
x
2
≥
3
a
f
( x )
=
x
,
Domain
is
,
range
is
x
≥
0
3
2
4
(k
h)(c)
=
2 − 3c
−
2c
+ 5;
domain
of
(k
c)
is
c
2
≤
1

f
b

g (x )

=
,
x
≠ 1
3

1
5
( g
+
h −
k )( x )
=
6x
− 8 +
−
x
+
4 ;
domain
is
x
≥
−4,
x
≠
x
1

0;
4
5
2x
1
(k
−
h −
g )( x )
=
x
+
4
−
5
−
6x
+ 8;
domain
is
x
≥
−4,
x
≠
6
a
b
Domain
Domain
of
of
f
is
f
+
x
g
≠
is
0;
x
domain
≠
0,
x
≠
of
g
is
x
≠
2.
a
3.5
b
e
2
f
a
x
6
represents
commission
2
earns
7
Students’
own
2%
answers
b
7
316
Answers
2
c
0
d
0
f
( x )
0.
2x
x
f
=

1,
g ( x )
3
the
0
amount
paid,
on.
g
and
g
of
sales
3.5
in
the
week;
represents
the
commission
is
the
she
3
Practice
b
4
Domain
of
f
is
domain
;
of
g
is
x
≥
1;
f

g (x )
=
x
− 1
;
2
1
a
b
15
cents;
210
g;
45
265
domain
cents
x
c
33
cents.
of
f

g
is
g
;

f
(x )
=
x
− 1;
domain
of
g

f
is
g
An
orange
of
radius
5.1
cm
weights
250
≥
1
or
x
≤
−1.
g,
2
c
and
costs
33
Domain
of
f
is
domain
;
of
g
is
f
;

g (x )
=
9x
− 15 x
+
6;
cents.
2
d
f
g
( g ( x ))

f
(x )
=
−3 x
− 3x
+
2;
Both
domains
are

1
2
2
a
3
Time
t
(seconds)
0
1
2
3
4
0
2.5
5
7.5
10
p (q (t ))
a
=
4t
− 15;
domain
is
b

q (r (t ))
=
−
r
(cm)
≠
0
; t
≠
±2
1
4
Radius
4; t
2
t
c
p (r (t ))
=
+ 1;
t
≠
0
r
d
(q (t ))
=
2
t
t
b
f
(t)
=
4
2.5t
4
a
g (
f
( x ))
b
h ( g ( x ))
2
c
g (r)
πr
=
a

2
d
A
=
g (
f
(t))
=
2
πr
π (2.5t)
=
2
5
6.25π t
=
a
g (
f
( a ))
=
2
a
v (t)
=
10t
6
k
7
a
=
f
v
πr
=
h
h
b
g (
( g ( − b ))
− 1
=
3
4
2
=
f

3
v
2
b
b
+ 1

3
−2b

2
( g (x
− 1))
=
4( x
− 1)
2
− 3
=
4 x
− 8x
+ 1
2
πr
2
f
( −3 x ))
=
2 ( −6 x
− 3)
10t
c
h  v (t )
=
2
2
πr
The
height
8
of
the
water
in
the
cylinder
as
a
function
f
2
(3 x
− 1))
=
4 (3 x
−
2) + 3
of
8
9
time
g (
a
7
b
2
c
t.
3
d
t
=
5π
a
r
=
2t
=
15.8
seconds
2
10
4
b
Area
a
B (
T (t ))
a
as
a
function
of
the
radius
as
a
function
of
time.
=
20 ( 4 t
function
of
+
the
2)
− 80 ( 4 t
amount
of
+
2 ) + 500 ;
time
food
Bacteria
is
outside
as
the
refrigerator.
2
c
5
a
45 289
cm
b
i
Fuel
cost
is
a
ii
Volume
of
a
iii
Volume
of
a
iv
Size
function
tank
is
circle
a
of
vehicle
function
velocity.
of
is
a
function
of
a
regular
11
time.
of
each
angle
a
s
c
d
a
i
=
1.5w
=
polygon
is
a
number
of
sides
of
a
f
Interest
earned
is
a
g
Students’
own
1.2w
b
d
d
$86.40
=
0.8s

f
=
x
5
( x )
=
ii
0.75(x
5)
g
iv
( x )
f

=
0.75x
g ( x )
function
of
iii,
the
price
is
$33.75.
From
iv
=
the
0.75x
price
5
is
$32.50.
time.
dierence
is
$1.25.
answers
13
Practice
( x )
From
The
b
=
polygon.
b
v
0.8(1.5w)
function
iii
ofthe
hrs.
time.
12
of
3.8
For
example:
a
(x )
5
5
4
f
=
− x
+ 7; g ( x )
=
x
b
+ 1
f
(x )
=
x ; g (x )
=
4
− 3x
3
1
f
( x )
=
x
;
g ( x )
=
2x
1
3
2
c
f
(x )
=
−5 x
− 8;
g (x )
=
x
f
d
(x )
=
; g (x )
=
x
−
4
3
2
f
( x )
=
3
f
(x )
=
f
(x )
=
5
f
(x )
=
6
f
(x )
=
2x
1;
g ( x )
=
x
x
x
4
x ; g
6x
(x )
− 5;
=
6x
g (x )
e
− 5
=
f
(x )
= 10
; g (x )
=
x
x
Note:
Other
answers
to
Q13
are
possible.
3
x ;
g (x )
=
2x
− 1
Review
in
context
x
2
;
g (x )
=
3x
−
2
1
a
r (t)
=
0.8t
+
70
2
Mixed
b
practice
A  r (t )
=  π (0.8t
+ 70 )
1
a
(
f
+
c
(
f
+
g )( x )
=
c
2
2
1
x
− 9x
+
2
b
( g
d
( g
−
f
)( x )
=
x
+
x
−
135.5
minutes
or
approximately
2
hours
2
4
5
2
−19 x
h )( x )
+
6
4 x
=
−
h )( x )
− 17 x
−
2
2
a
K (C (F ))
( F − 32 ) +
=
4
4
b
373,
273
2
2
x
e
(h +
k )( x )
−
2x
− 8
=
x
≠
0
3
a
r (v (t ))
−
h )( x )
+
40 + 3t
2x
b
− 8
=
;
x
≠
0.202

+ t
0.1
=
500

2
− x
(k
2

;
4 x
f
273
9
=
liters
per
+ 0.2

km
0
4 x
4
a
f
( x )
=
4x
+
6;
g ( x )
=
+ 3
2
2
−5 x
g
(
f
+
k )( x )
+
2x
−
2
=
;
x
≠
4 x
0
g (
x
f
( x ))
+
=
6
+ 3
=2x
+ 3 + 3 =  2 x
+
6
2
1
2
a
Domain
of
f
is
;
domain
of
g
is
x
;
≥
b
Set
the
above
expression
equal
to
the
friend’s
result
2
mentally
solve
the
equation.
1
f

g
(x )
=
2x
− 1 + 1,
domain
of
f

g
is
x
≥
;
2
5
Students’
own
answers
1
g

f
(x )
=
2x
+ 1,
domain
of
g

f
is
x
≥
2
Answers
317
and
1
1
E2.2
5
a
f
(x )
=
;
f
Y
ou
1
a
should
Not
a
already
function,
since
know
b
∈A
c
∈A
does
how
c
Not
It
is
a
a
function,
since
has
not
two
have
its
inverse
is
a
function
with
domain
x
≠
1.
inverse
are
symmetrical
about
the
line
y
=
x
the
line
y
=
x
to:
an
image
in
images
in
B,
p
b
Domain
a
y
of
f
is
x
≠
1.
B
6
b
and
the
1
x
and
=
±
x
+
2
r
f
function.
and
its
inverse
are
symmetrical
about
1
2
a
b
Not
It
a
is
function,
a
since
2
is
mapped
to
1
and
b
3.
Domain
a
It
is
function.
a
It
is
is
function.
not
a
is
x
≥
2,
(x )
f
2 +
=
x ,
x
≥
0,
or
domain
of
x
≤
2,
f
(x )
=
2 −
x ,
x
≥
0.
1
7
b
f
1
f
3
of
a
f
(x )
=
x
+ 5
− 1
function.
f
and
its
inverse
are
symmetrical
about
the
line
y
=
x
2
4
a
f

g (x )
=
2x
+ 3
b
g

f
=
4 x
1
b
Domain
of
f
is
x
≥
−1
f
for
(x )
=
x
+ 5
− 1;
or
domain
of
2
(x )
+
4 x
+
2
1
f
Practice
is
x
1
≤
−1
±
8
a
y
for
2( x
(x )
− 3)
+
=
−
x
+ 5
− 1
2
=
2
1
a
Neither
b
Neither
d
Neither
f
c
Onto,
not
one-to-one
and
its
inverse
are
symmetrical
about
the
2( x
line
− 3)
+
y
=
x
2
1
e
Onto,
g
Both
not
one-to-one
f
Onto,
not
one-to-on
b
Domain
of
f
is
x
≥
1
for
f
(x )
,
=
x
≥
≥
3
3,
or
2
h
One-to-one,
not
onto
−
2( x
− 3)
+
2
1
2
Students’
own
domain
functions
of
f
is
x
≤
1,
f
(x )
=
,
x
2
Practice
2
9
Either
f
( x )
=
x
for
x
≥
0,
or
f
( x )
=
x
for
x
≤
0.
x,
x
1
For
f
( x )
=
x,
x
≥
0,
f
1
( x )
=
x;
for
f
( x )
=
≤
0,
f
( x )
=
x
1
1
a
f
( x )
=
{(1,
3),
(1,
1),
(1,
3),
(1,
1)};
not
a
function
1
b
g
( x )
(
1,
=
3)};
{(4,
is
a
2),
(3,
5),
(
3,
2),
(1,
1),
(
4,
4),
(
2,
1),
10
a
s
≥
b
h
0,
since
h (s)
never
be
less
than
0.
s
198
function.
can
1
(s)
=
1
c
h
( x )
(
1,
=
1)};
{(0,
not
0),
a
(4,
2),
(
4,
1),
(2,
2),
(
3,
3),
(4,
4),
0.9
function.
The
c
2
a
The
inverse
is
also
a
function.
b
The
inverse
is
also
a
function.
c
The
inverse
is
also
a
function.
d
The
inverse
is
also
a
function.
e
The
inverse
is
also
a
function.
inverse
Practice
f
The
inverse
1
3
a
y
is
not
a
2( x
;
also
a
function.
4
1
Yes
2
No
3
Yes
4
No
5
Yes
6
Yes
7
Yes
8
No,
function.
x
=
is
203
function
b
y
=
4)
−
;
4
f
( g ( x ))
=
x
and
g (
f
( x ))
=|
x
function
3
1
c
y
=
−5 − 3 x ;
function
y
d
=
;
−
x
1
x
≠
−1,
function
Mixed
practice
+ 1
5x
1
e
y
=
x
≠
,
function
y
f
=
±
x
− 1;
not
a
function
1
a
i
Yes
ii
R
=
x
b
3
g
y
y
i
Yes
iii
No
ii
S
=
1),
{(0,
(3,
2),
2),
(1,
(4,
3),
3),
(1,
(5,
5),
4)}
(7,
iii
6),
Yes
(0,
0)}
3
=
x ;
3 +
i
{(2,
1
function
h
y
=
x
−
2;
function
1
2x
c
;
=
x
≠ 1,
x
No
iii
2
Practice
i
ii
T
=
{(
1,
0),
(1,
3),
(
1,
4),
(6,
3),
(
1,
3)}
function
1
a
No
Neither
b
Both
3
c
x
+
Both
2
1
1
a
f
(x )
=
.
The
inverse
is
a
function.
3
a
No.
f
is
not
into.
b
Yes.
d
No.
f
is
into
and
onto.
5
f
and
its
inverse
are
symmetrical
about
the
line
y
=
c
x
No.
f
is
not
into
(x )
=
1
2
a
f
(nor
8
(x )
=
2 (7 −
x ) .
The
inverse
is
a
function.
onto)
f
is
not
into.
x
1
4
i
a
f
3
f
and
its
inverse
are
symmetrical
about
the
line
y
=
x
f
1
3
a
f
(x )
=±
x
b
+ 3
and
its
inverse
4( x
f
and
its
inverse
are
symmetrical
about
the
line
y
=
are
symmetrical
about
the
line
y
=
x
symmetrical
about
the
line
y
=
x
function
+ 1)
1
x
ii
a
f
(x )
=
3
1
b
Domain
of
f
is
x
≥
0,
f
(x )
=
x
+ 3 ,
x
≥
−3
or
domain
of
f
1
f
is
x
≤
0,
f
b
(x )
=−
x
+ 3;
x
≥
and
its
inverse
are
function
−3
1
iii
4
a
y
f
=
±
and
1 −
its
a
( x ) =±
f
4
−
x
x
inverse
are
symmetrical
about
the
line
y
=
x
f
and
its
inverse
are
symmetrical
1
b
Domain
of
f
is
x
≥
0,
f
(x )
=
1 −
1
f
is
x
≤
318
0,
f
(x )
=
−
1 −
x ;
x
Answers
≥ 1
x ;
x
≥ 1,
or
domain
of
about
the
line
1
b
For
domain
f
( x )
≥
0
f
(x )
=
4
−
x ,
x
≤
4
y
=
x
x
+
2
Practice
1
iv
( x ) =
f
a
1
±
3

f
and
its
inverse
are
symmetrical
about
the
line
y
=
x
1
a
2
+

2

ii

iii
 4
x
4


i
3

2

2
1
b
For
domain
f
( x )
≥
0
f
(x )
,
=
x
−2
≥
3
2

v
a
and
its
are
symmetrical
about
the
line
y
=
4

1

x.
AD
i


2

=
DA
5

1


=
1
x
5

1
inverse
b
b

vi
4

2x
f
2

v
( x ) =
f

iv
x
1

5

≠
2

1

1


x
1
5
a
f
ii
(x )
=
;
EG
=
=
GE
function
6
6
5
x
2
c
1
f
b
(x )
=
;
The

3
d
1
f
(x )
d
g
=
(x
);
+
2
=
h
(x )
=
3;
(x )
=
(x )
x
+ 3;
−
x
if
x
≥
−3
+ 3 ,
if
the
e
i
2
a
C
F
the
domain
of
f
is
x
≥
domain
of
f
is
x
≤
3
a
(5,
4
a
MN
to
A
but
5

opposite
sign.


iii
0

iv
2

 0
0.
9)
x
2
≠
+
G
b
E
iii
b
(1,
to
C
3
8)
D
c
A
to
c
(0,
D
3)
7


=
2
3x
ii
d
H
to
d
(14,
B
0.

;
=
x
PQ
b
5)

=
 2
3

2
1
g
(x )
;
=
x
1
s
h
x
≠ 1
c
RS
=
x
− 3;
x
≥
For
each
pair
of
Length
8
a
=
39.5
functions
f
and
g,
f
(g ( x ))
=
g (
f
( x ))
=
x
5
DC
After

=
4


2

=
4
cm
3

412;
7
0

7
TU
d

1


=
2
(x )
4

1

6
size,
 5
1
p
same
ii
1
f
the

function
1
h
3
2
5x
1
e

 1
function
1
( x )
have
1

i
1
c
components
function
6
24
months,
a
computer
has
depreciated
from
6

b
1 
$700
b
54.2
to
to
$412.
months;
it
depreciate
takes
from
700
54.2
$700
d
for
a
computer’s
value
7
E.g.
(0,
0)
and
(
5,
3);
(3,
1)
and
(
2,
4);
(6,
11)
and
(1,
14)
$50.
Practice
d (t )
2
1
1
c
months
to
(t )
=
;
d
(t)
is
the
number
of
months
for
a
12
computer
d
depreciate
to
t
=
is
−500,
which
means
1
a
25
b
2
a
5
b
3
a
6.40
b
4
a
6
5
a
9.06
7
Any
Add
8
in
to
subtract
that
after
100
months
the
10
c
2
8.25
c
b
26
b
3.16
10
5
d
5
4.47
d
c
25
c
13.3
2
e
13
f
281
13.2
e
8.94
f
12.5
d
1
e
10
f
5
d
6.80
e
20.8
f
21.0
computer
context
34,
5,
multiply
to
get
the
this
result
answer
by
2,
divide
then
by
6,
2( x
The
result
is
the
original
four
from
(8,
1),
(8,
5),
(9,
0),
(9,
4),
(4,
1),
(4,
5),
9.
(3,
a
c
worthless.
Review
2
5
d (t).
16.7
e
1
to
number,
+
0),
(3,
4)
6 ) − 12
=
since
x
8
x
=
2,
y
=
1
9
x
=
2
2
b
Choose
a
number,
multiply
by
2,
add
12,
divide
the
result
Practice
by
2,
and
then
subtract
6.
The
x
2
3
f
(x )
=
2x
result
is
the

3
for
1


f
(x )
,
=
x
≥ 3;
The
puzzle
would
not
1
real
Students’
 5
numbers
less
than
own
answers
2
a
AB
AB
1
already
know
how
x
=
0,
3
94.1°
y
7.28
=


6

(3
s.f.)
b
e

5
1

 7
5


BA

=
1

1

AB
11

=

BA
11

=
5

2
0
 0
=
7
=

=

to:
c
53

d

BA

a
5
 17 

=

b
should
1
3.
E3.1
1

c


Y
ou

b
a

4
8
1
+ 3;
2
work
3
number.
10

10 

1

d
AB
9
BA

AB
9


=

BA

7

=
 0
e
9


=
0
9


=
 7
Answers
319
4
3
(
4
a
15,
4)

a
=
0,
b
=
−1
b
a
=
1,
b
=
−2
c
a
=
3,
b
=
6
3
2


7
a
=
−1
2
b
=
8
3
OM
10.5
a
87°
b
5.5
6



6
a
=
17
and
CD
=
2
17 .
They
are
dierent
=
DC
= 10u
ABCD
or
12
b
4
37°
5
m
=
6
a
ab
b
It
5
127°
b
67°
c
25°
a
+
(a
is
+
an
=
Any

5
d
2)(2
b)
equation
b
 25

9


15


the
10
a
=
−1
or
odd
11
b
=
−3
13
a
AB
b
OB
=
c
AC
=
a
=
0,
in
so
two
a
b
=
one
out.
=
4
parallel
OC
to
because
OC
=
2 AB
OA +
AB
=
b
+
a
=
a
+
b
unknowns.
AO
+
OC
=
2a
b
=
non-zero
OC
is
−2
multiple
1
d
2
b
X
lies
OX
hence
e

=
Since
on
OB,
OX
is
parallel
to
OB ,

of
Since
X
k OB
=
lies
on
=
k(a
AC,
+
AX
b).
is
parallel
to
AC ,
14.5°
3
hence
AX
=
m AC
8
So
LHS
=
RHS
=
OX
= OB
k (a
+
b)
=
b
6
m
=
,
k
=
a
2
a
b
u
3
a
r
+
v
b
v
w
c
u
+
v
+
b
b
a
c
b
a
d
b
2a
e
a
+
5q
=
3t
m(2a
b)
,
so
ratio
is
1
:
2
in
context
a
76
cm
b
45
cm
c
67
cm
4p
d
s
+
b)
b
2
=
b
w
Review
a
=
AX
m(2a
3
3
1
+
2
1
Practice
+
16
f
4


2

10
15

and
6

35°
 4
9
4

3
,

a

 45 
4

8
36

111°
2
7

 20 
 8
1
c
16
and
18 
is
a

and

3

30 

ABDE
Practice
1
9

+ 5v

7
97°
lengths.

AB
c
and
 20 
AB
84°

8
b

3

=

4
5

c
4

1

 3
 5
Practice
5
b
a
95
cm
e
292
cm
f
447
cm
2u
2
3
a
15
m
3
5
a
+
b
a

2

c
A
rectangle
with
sides
in
the
ratio
2
:

60 
3
c
Mixed
25
b
250
+
600


1
5
+
650
=
1500
657
cm
=
15
m
practice

2


3


11

4
a
AE
b
c

3

,
 55 
d
 4

=

a
36

19 
 4

b

2
a
AB
3


=
BA

3
OF

b
AB


BA
=

3
7
AC
c
AB

,

CF
=
510
cm
45 

,
|
AC |
=
2169
=
3
241
3


3

BA

24
=
=

=
CF
 45 
24

3
12
=

CE


,
 3
c


30 

=
3
7
22
=


=
,
|
CE |
676
=
 20 

=
5

5

AC
is
longer
2

d
AB
13

=
BA

13

d
Perimeter
a
PQ

14


3


9


3
3

b
a


c

area
18.6
m
6
2
32



d
5

24
=

3
m,
=
14 
5

18.2


 15 
PS
60

=
 45 
4
a
c
a
a
=
0,
b
=
1
b
a
=
2,
=
5,
b
=
−4
d
a
=
−2,
b
=
b
−3
=
4

c
SR
a
29
b
5
c
3
41
d
5
30
Answers
hence
not
parallel

17
opposite
320

,

5
25
=
sides
not
parallel.
to
IQ.
Not
rectangular,

6
a
OU
18

=
b
UV
is
25

c
VW
VW |
to WX
so
47

, WX
is
a
µ)
∑
f (x
µ
σ
10
50
20
123
39
20
5
2
41.2
6.15
2.06
197
80.34
5.05
2.06
43
223
102.34
5.186
2.38
23
115
39.56
5
1.72
trapezium.
, VW
=
UV
0, VW
XW
=
0
25 
=
25
2 ,
UV |
=
22
2.
|
XW |
=
47
2
2
Area
e
∑
f x
47 

UVWX
∑
f
3

=
22 

=

d

=

parallel
2
22

so UV
,
 22 
=
17.25
m
45˚
2
Length (cm)
4
mid-value
f
xf
2.5
3
7.5
11.4
389.88
7.5
4
30
6.4
163.84
12.5
6
75
17.5
7
122.5
3.6
90.72
22.5
5
112.5
8.6
369.8
E4.1
0
Y
ou
should
already
know
how
5
to:
<
<
10
1
a
3.9
2
a
A
b
b
population
A
sample
Practice
is
is
the
any
whole
subset
of
x
x
<
≤
≤
x
5
10
≤
15
( x )
µ
x
µ)
f (x
11.76
1.4
80.5
of
all
the
outcomes
in
question.
15
<
x
≤
20
20
<
x
≤
25
population.
1
∑
xf =
2
25
µ)
∑
xf (x
347.5
1
a
µ
5, σ
=
=
2.19.
Mean
is
5,
but
on
average
the
spread
of
the
The
data
b
µ
is
±2.19
= 16.54, σ
This
c
µ
of
each
that
the
= 15 cm, σ
the
data
=
is
mean
±3.07
is
16.54,
on
each
but
on
average
implies
of
13.9
and
the
standard
deviation
is
6.41.
Practice
that
the
3
2.05 cm.
the
mean
is
15
cm,
but
on
average
µ
= 11 kg , σ
data
is
±2.05
cm
on
each
= 1.5, kg.
Fully
grown
mute
swans
will
on
the
average
spread
is
the
side.
1
This
mean
side.
= 3.07.
implies
spread
on
weigh
between
9.5
and
12.5
kg.
side.
2
d
µ
=
kg, σ
40.8
This
implies
spread
of
=
4.75
that
the
kg.
the
data
is
2
mean
±4.75
is
40.8
kg
on
kg,
each
but
on
average
the
a
∑
x
µ
b
side.
c
=
=
27
m,
1.811
There
is
not
standard
2
a
µ
b
This
=
min, σ
25.4
=
2.82
buton
that
average
the
the
mean
spread
length
length
of
of
journey
journey
is
25.4
a
µ
motorway:
=
19.8
min,
σ
=
4.57
µ
lanes:
=
20
±2.82
mins.
3
You
min,
σ
=
1.41
i
the
in
country
the
lanes,
country
m
dramatic
deviation
has
change
in
the
mean,
however
the
changed.
because
lanes
there
the
is
of
the
team
should
lie
within
6.7
cm
of
the
mean.
would
1420
expect
mm,
the
rainfall
to
be
within
the
range
of
so:
1100
two
less
mm
is
exceptional
min
b
b
48.628
0.067
min
a
country
a
=
min,
1109
3
=
σ
min.
68%
implies
∑
x
m,
means
are
equal
deviation
1400
mm
is
not
exceptional
but
4
a
49.55
min
b
48.48
min
2
ii
the
motorway,
because
the
iii
the
motorway,
as
a
iv
the
country
it
has
mean
lower
time
is
faster
minimum
c
4
µ
5
a
µ
=
1.72,
6
a
µ
=
7,
=
8,
σ
=
as
the
maximum
is
µ
=
0.0618
b
µ
=
1.74,
σ
=
0.0906
1
s
3
x
=
4
b
n
σ
=
σ
90.63,
Practice
=
raw
actual
=
c
Student’s
2.45
µ
=
11,
σ
own
=
4.36
2
The
mean
x
=
−7.5
y
=
3.5
is
hours,
s
=
The
data
aected
but
the
standard
deviation
is
to
µ
=
µ
2.94, σ
µ
= 10.46, σ
a
i
63
=
a
distributed
normal
about
the
mean
and
distribution.
x
=
7.84
Katie
people,
could
s
=
2.85
1
assume
that
5
and
11
68%
people
in
of
carriages
would
have
them.
kg
practice
a
=
µ
=
4.4,
σ
=
1.36
b
µ
=
24.5,
σ
=
2.93
d
µ
=
13.3,
σ
=
6.06
6.67
µ
=
4,
σ
=
0.76
5.34
2
kg
53.55
follow
= 1.27
= 15.03, σ
c
b
18.72
2
c
2
=
−1
1
symmetrically
n
1
b
460.16 s
hours
9
Mixed
a
39.53
not.
between
1
is
20
d
Practice
=
n
1550
2.45
or
or
x
−1
=
answer
c
e
∑
x
4
appears
d
or
33.02
n
b
data
lower
4.32
σ
the
value
5
roads,
Either
ii
9.45
With
water:
With
food:
µ
=
13.6,
σ
=
2.2
kg
On
average
the
mean
µ
=
the
was
15.7,
σ
=
2.15
strawberries
greater
and
given
the
food
standard
Answers
gave
a
better
deviation
yield,
was
lower.
3 21
3
µ
a
b
=
an
food.
49.2
c
σ
55.2,
On
About
and
This
b
5.96
is
5
4.25
Practice
day
68%
61.2
g
result,
the
of
of
standard
abnormal
4
=
average
hamster
the
time,
will
the
eat
about
hamster
55.2
will
eat
g
between
1
food.
and
x (1.25)
=
2,
x (1.25)
=
6.5,
x
he
=
a
4
d
deviations
less
should
than
be
normal,
so
it
2
of
is
b
16
an
81
e
25
2
2
a
x
+
3
a
x
x
b
x
4
1.5
x
=
1.5
f
49
3
4
b
5
2
c
x
c
x
13
3
4.25
125
3
3
worried.
c
2
d
x
d
3
7
3
6
2
=
9
4
5
µ
6
a
=
σ
28.15,
=
3.84
15
e
x
=
200.5,
s
=
2.655
b
x
=
200.5,
s
x
7
x
=
9.5,
s
x
=
9.5,
s
=
=
x
x
f
2.49
Practice
x
=
x
16
2.80
−1
3
2.62
−1
1
Yes,
because
mean
is
close
to
10
and
assuming
a
normal
1
34%
of
the
values
will
be
between
9.5
and
b
c
many
will
be
over
8
f
216
in
Patient
x
=
1:
91.3,
s
=
x
5.12
high
blood
pressure
a
5
b
e
81
f
2
c
2401
d
4
c
9
d
2
3
treatment
−1
3
Patient
27
context
2
1
256
10.
e
Review
d
27
8
12.12,
125
so
64
1
a
125
distribution,
Student’s
own
answers.
2:
1
x
=
78.3,
s
=
x
2.62
no
4
treatment
a
b
4
b
2
b
3
b
2
−1
9
Patient
x
=
3:
78.7,
s
=
x
2
x
=
75,
s
=
x
lie
more
tests
needed
Practice
5.53,
s
−1
Provided
68%
15
−1
=
x
the
data
between
is
70
5.6
−1
normally
and
4
7
7
distributed
you
can
1
a
2
2
a
7
3
a
2
80.
8
a
x
=
3.2,
s
=
0.49,
s
x
b
x
=
1.95
=
x
0.5
a
x
=
1
s
=
6.48,
s
x
Provided
25
now
=
x
the
data
and
68%
39,
of
is
the
however
mothers
were
2
d
2
d
2
3
1
4
1
3
1
4
c
3 ×
1
2
2
2
× 5
7.82
normally
ages
30
1
× 7
1
distributed
you
can
1
of
rst
years
ago
time
mothers
68%
of
the
lie
between
22
and
28
years
ages
old.
1
1
assume
e
2
a
2
2
3
2
× 5
f
2
b
3
× 3
between
of
rst
7
5
4
1
3
6
6
time
d
4
3
2
4
that
2
2
1
1
1
× 3
1
b
c
d
1
kg
31.79,
2
2
2
4
2
c
3
15
3
5
7
3
2
assume
1
6
10
c
7
6
3

E5.1
1
4
(
15
Y
ou
should
already
know
how
e
2
g
a
f
1
a
8
x
b
2
a
3
3
a
2
5
(
6
2
c
8x
b
4
c
6
b
2
c
2
n
m
1
)
2
(

1
n
m
× 5
m
n

=
m
n



)

mn

mn
2
× 5
to:
1
6
1

1
+
n
m
)
m

=
+
n

(

mn
a
h
1
1
n
m

)
a
m

=
n
mn


a
5
Practice
5
1
d
8
e
4
f
1
6
0.1
1
a
2
b
3
c
d
5
e
6
7
11
1
1
4
a
c
b
36
6
8
4
f
6
5
3
g
4
5
15
2
h
10
9
7
6
3
i
2
5
j
2
25
3
1
d
2
e
f
3
4
2
2
12
5
a
2
b
3
7
3
×
2
6
4
c
2
×
6
7
3
d
5
2
e
6
7
7
Practice
1
12
1
a
4
2
h
b
2
c
2
d
5
f
a
x
=
3
b
y
3
=
Practice
1
x
b
3x
b
2
5
x
32
3
3
d
e
8
f
6
1
f
4
x
g
0.2
h
x
1
=
8
32 2
b
x
b
y
(32 )
5
=
(2
5
)
3
a
4
x
2.4
b
=
2
1
c
4.6
4
c
=
=
1
3
2
2
1
5
3x
2
1
1000
1
1
x
1000
6
5
a
2
2x
a
c
3
d
1
4
=
10
c
2
x
z
3
1
a
c
10
2
3
2
j
i
3
5
h
g
3
3
3
5
2
1
e
e
5
2
3
a
g
1
2
2
3
5
5
f
1
x
100
Answers
=
1024
5
a
3
b
2
7
Mixed
practice
Practice
1
a
b
3
c
d
3
3
2.32
2
0.661
3
1.7
4
0.232
5
0.58
6
2.58
7
4.2
8
4.3
5
e
f
2
g
4
Practice
2
1
7x
1
c
b
4 x
a
1000
b
100
4
307 850.7
300 000
×
≈
307 851
1.013
t
c
350 000
,
= 300 000 × 1.013
year
=
2012
4.31
=
5
years
25
25
c
b
=
27
2
27
a
P (t)
1
d
c
9
4
and
t
1
a
303 900
2
b
3
3
3
3
a
1
5
3
2
2
1
1
d
3
8
Time
27
16
Time
t
5
5
2
5
a
2
6
a
25
4
3
b
period
Amount
(hours)
caeine
of
(C
mg)
7
c
2
c
7
c
2
6
d
3
1
b
08:00
t
=
0
120
13.00
t
=
1
60
18:00
t
=
2
30
23:00
t =
3
15
2
3
7
7
a
3
2
b
2
6
3
d
1
1
e
2
f
8
g
a
r
=
0.5
h
w
12
12
b
3
12
C
(t)
=
120
×
0.5
7
c
8
Student’s
own
C
(7)
=
120
×
0.5
0.02
=
120
×
0.5
=
0.9375
answers
w
d
,
13
hours
(12.55)
12
3
9
1
7
2
a
b
12
5
c
d
w
4
2
5
4
T
5
a
=
72
×
0.98
,
w
60
=
72
×
0.98
,
w
=
9.02,
thus
after
10
weeks
2
1
10
a
x
=
b
x
=
2
c
x
=
27
3
5
year
9
6
5
11
a
b
12
Mathematics
5
is
c
as
old
9
as
d
1
e
2
population
0
100 000
1
90 000
2
81 000
3
72 900
3
×
3
=
1728
man.
E6.1
t
b
Y
ou
should
already
know
how
P (t)
=
100 000
×
0.9
to:
t
c
25 000
=
100 000
×
0.9
,
t
=
13.15,
thus
after
14
years
7
1
x
2
a
Practice
1
125
b
c
4
4
2
2
1
1
18 242
2
a
years
3
0.866t
3
2
3
a
6
b
=
−0.866
b
A
=
A
2
3
b
c
e
c
0.265
o
5
3
a
1000
4
a
121
5
a
5
6
a
r
b
2013
c
5.58
or
approximately
6
weeks
52
4
2
×
1.18
=
Practice
1
a
c
log
log
(to
the
nearest
whole
number)
1
500
60
10 937
=
=
x
x
=
=
2.70
1.77
b
log
d
log
150
45
=
=
x
x
=
=
2.18
rabbits
b
5
wombats
b
1.73
b
6224
=
0.0199
years
years
or
approximately
c
34
c
log
21
months
years
5.49
2
e
log
6
=
x
=
2.58
Mixed
2
2
practice
3
2
2
a
8
1
=
64
b
8
e
b
=
4
c
10
=
0.1
1
a
log
23
=
x
b
log
95
=
x
7
5
d
4
5
=
x
0
=
b
f
x
d
=
log
y
=
x
e
log
log
9
=
2
b
3
log
0.125
=
c
log
4
10
2
=
a
5
125
b
a
4
e
b
5
f
5
3
c
g
2
d
0
d
5
h
5
3
=
a
7
a
c
b
c
2
d
2401
a
e
4
b
4
a
=
x
b
2.68
e
c
3.13
d
2
=
=
3.36
3.20
(3
(3
s.f.)
s.f.)
b
x
=
2.86
(3
s.f.)
d
x
=
6.49
(3
s.f.)
b
3
2
c
0
1
f
g
a
2.81
1.39
2.73
b
e
b
1.79
2
2.81
1.86
c
f
c
3.57
1.61
3.0
2
d
3
a
d
1
1
a
d
3
0.77
5
8
1
1000
m
1
2
4.09
x
c
=
2
4
7
10
n
=
2
a
c
3
2
6
x
9
3
1
3
3
4
4
=
x
2
=
1000
3
=
1
3
a
1200
12
1
2
3
6
8
47
4
0
d
e
1.46
e
0.91
f
0.65
5
Answers
323
7
1
6
a
1

c
b
8
2
3
6
a
3
1 
i
b
2


1

2
3 
i
2 
2

t
7
a
C ( t )
=
4
×
1.055

26
b
C (26)
=
4
×
1.055
=
16.09
8
a
years,
100°C
or
b
in
the
year
d
2
in
2
54.8°C
c
5
2 
context
4
3
3
a
b
0
c
3
2
a
2

minutes
1
1
1 
3
i

2020
Practice
Review

−
i

30
2 
2
−
c
c
3.4
b
i
7.4
ii
5.27
iii
5.5
iv
5.9
v
1
2
5.9
d
c
91 488 815
d
4280
e
microns
0
f
2
2
microns
g
10
2
a
2.8
b
3.98
3
a
4.2
b
0.005
4
a
110
×
1
h
i
7
10
c
2
2
10
π
2
dB,
1
2
recommended
b
65
db
c
169
dB,
a
For
2π
example
b
For
example
4
necessary
3
π
c
For
3π
example
d
For
example
2
E7.1
2
π
e
Y
ou
should
know
how
For
example
3
to:
2
1
1
already
a
b
3
c
Practice
5
2
2
π
2
a,
b
Suitable
graphs
3
a
Amplitude:
of
sin x
and
cos x
1
2,
frequency:
a
Amplitude
Vertical
dilation
of
scale
factor
3,
vertical
1unit
in
the
positive
y
frequency
=
4;
period
=
0;
vertical
translation
=
a
Circumference
=
24π
=
horizontal
2π
Amplitude
=
1;
frequency
=
3;
period
=
;
horizontal
π;
horizontal
direction
3
π
4
;
0
translation
b
of
3;
2
shift=
b
=
2
75.4
shift=
cm
;
vertical
translation
=
2
3
b
AB
=
4π
=
12.6
cm
c
Amplitude
=
0.5;
frequency
=
2;
period
=
π
shift
Practice
=
;
vertical
translation
=
7
1
8
1
2π
π
π
a
π
b
d
Amplitude
10
5
2π
6;
frequency
=
3;
period
=
;
horizontal
π
shift
2π
e
=
d
3
12
18
π
c
7π
f
=
;
5π
g
vertical
translation
=
−3
3
h
2π
5
10
6
e
9
Amplitude
=
1;
frequency
=
3;
period
=
;
horizontal
3
5π
5π
shift=
4
2
;
vertical
translation
=
2
j
i
3
3
a
30°
b
e
157.5°
f
120°
c
135°
d
108°
h
240°
2
a
y
160°
g
75°
5
i
300°
j
225°
1
2
3
A
=
4
θ
r
π
2
(2π
, 3)
4
3
2
Practice
2
2
1
a
31.4
cm
b
70.7
cm
2
a
39.0
m
b
0.292
c
6
hours
c
1 591 500
π
1
km
y
=
x
3
rotations
2
3
12.00
4
a
m
0
π
π
3π
2π
x
61π
b
143.7
m
2
2
2
Practice
b
3
y
1
2
3
1
a
b
c
3
2
2
0
x
3
d
e
0
f
0
π
π
3π
4
2
4
1
2
3
2
2
h
g
i
1
2
2
y
3
3π
2
a
b
and
4
3π
and
2
324
3
π
2π
and
3
Answers
3
7π
and
c
and
e
2
5π
5π
3
4
π
d
π
7π
4
4
=
cos (4x
−
2π)
−
1
π
b
y
c
2
Amplitude
horizontal
The
1
=
1.3
shift
amplitude
heights
of
m,
=
is
tides.
period
3,
=
vertical
the
12
hours,
shift
maximum
The
period
is
=
0.67.
rise
the
and
time
fall
in
taken
the
for
a
full
0
x
π
2π
3π
4π
5π
cycle
of
high
and
low
tide.
The
vertical
shift
is
the
mean
6π
1
2
c
depth
over
time,
depth
over
time.
1.97
and
the
horizontal
shift
is
the
mean
m
3
d
Important
for
people
whose
livelihood
is
from
the
sea,
e.g.
4
y
=
3 sin(0.5x)
shermen,
2
navigation,
coastal
engineering,
etc.
5
(3π,
3
5)
a
2π
h
6
h(t)
=
t
−
7.5)
+
94
30
250
d
y
(14.7,
200
1
y
=
0.5 sin(2x
−
π)
−
180)
2
150
0
π
π
3π
x
2π
100
1
2
2
(30, 8)
50
2
0
3
t
5
π

3
y
=
2
sin
2
x

−
2
i
y
=
4
π

x
sin
2
Amplitude
30
y
ii
=
4
( x
cos

− π
y
=
π
3 sin
x
6
y
ii
=
3 cos

x
−
=
6
sin
4
x
+ 3
y
ii
=
6
cos
π
4
x
radius
to
make
of
the
one
wheel);
complete
period
=
revolution;
shift
=
7.5
min
(
6
of
the
period);
vertical
shift
137
=
94
m,
the
maximum
height
above
the
ground.
m

−
+ 3


2
(the

4


−

m
time
4
+

π

y
86
(the

c
i
35
1
3π

+
4
=
minutes
3

−

c
30
)
horizontal
i
25
1

+

3

b
20

b
1
a
15
1

4
10
50
4
a

8
π
h
h(t)
=
t
−
+
3π
7
4
5π

d
i
y
=
5 sin 3
x


−
10

6
π
y
ii
=
5 cos 3
x

−
10


20
3

16
(12,
(20,
10)
10)
(28,
10)
12
Practice
6
8
1
a
π
h(t)
h
=
π
t
1.5 sin
+
5
4
2
2
0
t
1
4
8
12
16
20
24
28
32
4
0
t
π
2π
3π
4π
b
1
Amplitude
fromthe
=
3
m
(The
platform
is
6
range
m).
of
the
Period
height
=
8
s,
of
the
the
time
car
it
2
takesto
b
Amplitude
vertical
The
the
=
1.5
m;
translation
amplitude
buoy.
The
is
period
=
the
=
10
s;
horizontal
shift
=
2.5
s;
0.
maximum
period
is
the
rise
time
and
taken
fall
for
in
one
depth
rise
of
of
the
buoy.
The
horizontal
shift
means
that
and
at
0
maximum
height);
of
the
buoy
was
above
its
mean
height.
The
is
the
mean
height
over
the
10
i
0.464
m
ii
1.5
b
Amplitude
a
h
=
shift
22.5;
=
vertical
cycle
12
shift
=
s
on
to
7
the
reach
m
(the
track;
the
height
amplitude
months,
over
1.3 sin
(t
+
3)
+
period
=
12;
horizontal
shift
is
half
the
dierence
temperatures
in
one
of
the
year.
and
the
the
time
lows.
12
The
taken
for
vertical
a
full
shift
is
cycle
the
of
period
(12,
mean
temperature
(Answers
from
GDCs
will
vary):
y
=
−22.5
π
cos
6
1.97)
Mixed
1
b
0
t
20
e
c
5π
3
4
28π
9π
g
f
36
d
5
3
25π
8π
8π
4π
a
15
16
+ 54.5

practice
π
(6,

x
1.97)
1
4
is
months.

1.97)
0;
temperature

c
(24,
(0,
=
maximum
The
0.67
6
2
of
54.5.
minimum
highs
π
=
takes
m
3
h(t)
(it
s.
12
2
complete
the
and
c
s
vertical
The
translation
12
fall
s
vertical
height
one
shift=
theplatform).
5
cycle
complete
horizontal
h
2
15
0.63)
1
4π
i
35π
j
9
9
Answers
325
2
a
15°
e
i
b
75°
138°
72°
c
f
252°
j
51°
270°
g
d
99°
f
96°
h
1
2
1
b
c
3
2
3
3
g
1
b
y
=
−1000
the
number
of
the
tire
of
revolutions
gives
the
total
distance
tire.
π

cos
of
radius

+
4000
x

c

6
4866
d
3500
h
e
3
2
the
3
2
f
e
by
the
or
d
2
measure
by
3
9
a
radian
travelled
1
3
The
multiplied
45°
End
of
April/beginning
of
May,
and
end
of
August/
2
beginning
of
September
1
i
0
j
2
Review
4
5.00
in
context
m
2π
1
5
a
a
=
2;
f
π;
=
p
=
2,
hs
=
0;
vs
=
c
b
15
π
a
=
5;
f
=
4;
p
=
;
hs
=
−
;
2
a
=
0.2;
f
=
6;
p
vs
=
a
i
t ( x )
=
a
12
sin (bx)
0
2
π
c
d
30
3
π
2
b
5π
7π
π
a
0
ii
x
represents
the
time
after
launch
that
the
stone
is
in
theair.
π
=
;
hs
=
−
;
3
vs
=
iii
−1
a
represents
the
maximum
height
of
the
stone
during
3
its
ight.
This
is
inuenced
by
the
catapult’s
size
and
π
d
a
=
3;
f
=
;
p
=
4;
hs
=
3;
vs
=
8
hs
=
2;
vs
=
−1
power.
2
iv
b
represents
the
frequency,
or
number
of
cycles
per
2π
π
e
a
=
1;
f
=
6;
p
=
;
The
more
cycles
there
are,
the
closer
the
point
of
3
impact.
6
a
y
away
y
=
3 cos (2x)
−
The
the
smaller
point
of
the
number
of
cycles,
of
t ( x )
the
further
impact.
3
π
2
v
Domain:
x
is
from
0
to
,
range
is
from
0
to
a
b
0
b
x
π
π
3π
c
represents
the
2
the
horizontal
shift,
which
is
the
position
of
2π
2
catapult
from
the
origin.
The
catapult
can
be
moved
2
toward
the
target
to
hit
it,
for
example.
4
π
c
c
Domain:
x
from
+
,
to
6
b
(stays
the
c
range:
t ( x )
from
0
to
a
b
same).
8
d
Instead
of
moving
backwards
if
the
the
catapult
target
is
too
up,
it
close,
can
for
simply
be
example.
moved
By
y
b
moving
the
catapult
up,
the
sine
curve
does
not
model
1
the
trajectory
of
the
stone
once
the
stone
goes
below
the
0
π
π
3π
catapult’s
x
2π
sine
1
2
height
curve
–
makes
it
it
would
keep
advance
going
down
whereas
forward.
2
e
i
Domain:
x
from
2π
6π,
to
range:
t ( x )
from
0
to
8
π
2
y
=
x−
−
4
π

2
ii
t (x )
=
8 sin
0.25
3
x
−

f
4
i
(2π,
iii
Max
iv
Domain

,
2
0)
2π
5,
at
the
x
(4π,
ii
height
<
<
6π

point
0)
(3π,
5)
5
x
is
from
2π
to
4π,
range
t ( x )
from
0
to
5.
6
c
E8.1
y
1
Y
ou
0
should
already
know
how
to:
x
π
π
3π
2π
1
a
4n
+
1,
81
1
2
2
b
2
c
π
y
=
x
−
+
2.5n
1.2n
+
+
27.5,
41.5,
22.5
65.5
3
4
n
3
2
a
480,
960;
15
×
1
2
25
n
4
b
1.5625,
or
0.390625;
100
×
1
(0.25)
64
5
2
c
6
n
2
,
;
3
162
×
9


1
3
1


1
3
3π

7
a
8
a
y
=
2 sin
x
−

of
The
circumference
is
c
d
e
i
1341
m
=
sin
2x
the
the
the
tire
depends
distance
i
72
cm
2398.9
32 6
m
6,
10,
15
b
n(n
1)
3
4
the
travelled
in
a
2,
5,
8,
11,
14
b
6,
12,
18,
36,
72
diameter.
one
Practice
1
tire.
ii
1424
m
ii
86
ii
2570.2
iii
1676
cm
iii
105
m
iii
3426.9
Answers
1
465
2
5
a
930
b
6
a
2601
b
1131
3
2112
c
51
m
1.2π
i
3,
2
−
on
1,

2
circumference
of
y
b

The
revolution
b
+ 1
4
π


a
900
cm
m
2550
4
18 396
the
7
80
1
1.5
2
4
r
a
=
r
so
=
±
24
8
a
91,
so
correct.
b
385
c
2870
d
a
b
9
4
2485
=
± 96
190
96
If
a
96,
=
S
= 128
=
∞
1
Practice
1
2
4
If
1
a
459
b
198
c
a
513
3
a
a
4
725
=
b
15,
d
=
r
=
−96, S
=
0.9
−76.8
6
x
=
15
7
x
=
−3
10 380
4
b
5
Practice
=
∞
5
2
a
438
47
c
Mixed
5029
practice
120
3
1
a
b
208
c
1365
2
2464
9801
3
210
4
48.2
6
1071
7
159
(3
s.f.)
d
62 499.84
5
394.0625
28 394
1
a
185
b
56.4
c
780
d
351
(3
s.f.)
8
≈
1050
(3
s.f.)
27
2
a
3
720
825
b
441
c
4
308
5
924
d
1904
6
13.2
e
3630
2
9
Practice
1
a
630
3
10
41
2
a
b
11
37 700
12
4
14
96
15
r
=
18
x
=
4
c
333
11
13
a
=
− 56,
16
a
=
36,
19
165
d
=
24
1
413 336
a
=
9
b
23 327.5
c
23.8 576 907
d
1 797 558
b
6075.04
c
d
=
6
a
constant
3
420.04
17
2
d
x
=
5
10
5147.5
2
20
n
3
a
x
=
396
21
1240
cm
(3
s.f.)
1
b
2 × 3
6560
n
Review
Practice
1
1
a
26.622 976
b
39.921 875
c
a
3
8 388 600
6
a
19 530
a
8
b
329 554 400
c
+
(the
1266
b
4
485
(3
s.f.)
5
2n
=
± 2
b
1778
or
8.65
(3
s.f.)
8
n
=
30
4.16
as
3
6
n
b
u
=
3
×
increases
by
amount
180
million.
It
is
only
an
estimate
because
she
has
estimated
602
rate
at
−1
a
attendance
might
inuence
will
fall,
and
other
factors,
such
attendance.
144
=
b
4
which
reviews,
=
1.2.
There
is
a
common
ratio.
120
100
1
capacity
dierence).
c
120
Practice
week
common
28
the
7
each
484 275 610
2
r
context
917 503.125
Since
2
in
5
3960
(3
s.f.)
n
4
c
65 535
d
S
=
a
Because
each
new
machine
connects
to
the
server
and
n
1
1 073 741 823
15
other
2
a
11
b
660
3
a
8
b
32.8
4
a
962
b
1 969 214
b
c
2
2
mm
mm
Adding
1.97
the
d
120 000
1371.6
a
=
mm
(54
=
20 + 5 ×
2
=
u
c
=
20 +
2 (n
− 1)
=
c
7
=
16;
Day
456
the
2
+
3
+
…
16th
(7
=
0.8.
The
sequence
has
a
common
553 392
(to
the
nearest
whole
number
of
followers)
6
a
111 000
b
19 000
c
111 000
7
a
3
b
15
c
20
+
76 000
=
187 000
day
depots
hours
36
minutes)
7
200
1
a
2000
2
54
b
100
3
156.25
4
a
23
b
5
d
1
a
5
2
a
(5,
3
Practice
8)
a
Yes
e
Yes
b
a
f
,
4,
No
c
Yes
g
Yes
d
Yes
No
h
No
1
b
5.98
(3
s.f.)
c
6
2
15
r
=
b
(0,
4
53°
7.5)
57
a
A
b
E
(0,
1
0,
1),
B
(4,
9,
0),
C
(4,
0,
0),
D
(0,
5,
0,
0),
G
(0,
7,
0),
H
c
13
9,
1)
(
5,
7,
0),
F
(
(
5,
7,
2)
a
7
b
6
5
.
=
18
25
to:
9
3
3
b
how
4
4
2
know
8
Practice
1
already
9
27
111
should
5
16
c
99
9
Y
ou
c
35
2
ratio.
750 000
E9.1
Practice
n
120 000
25
minutes
+
50
n
n
+
30
6
b
1
inches)
b
u
gives
96 000
a
150 000
6
together
64
m
5
5
connections
1275
2
≈
machines.
Since
−1 <
r
< 1,
the
sum
is
nite.
6
Answers
327
Practice
1
2
a
25
a
2
Review
cm
b
56.648…
≈
57
9
m
c
cm
27
b
cm
137.2
1
in
a
(112.5,
b
The
context
90,
25.729
≈
26
5
3
4
16.6
cm
were
3
a
7
b
3
c
9
e
37
a
a
3
OA
14
b
to
f
9,
OB
=
a
EF
of
equal
9.
So
OAB
is
isosceles
because
it
has
two
=
room
This
is
is
the
space
greater
hit
be
so
it
is
possible
that
some
than
people
targets.
from
held
at
realistic
ground
around
level,
4
feet
maximum
98.1
feet.
above
the
If
the
laser
ground,
97
feet
range.
≈
193.32
≈
687
m
m
m
≈
8
m
dierence
sides
(214,
29,
18.5)
length.
c
Area
the
feet.
23
b
4
in
193
22.5
7.98
=
possible
length
range,
ring
more
GH
2
has
11
2
d
not
Assume
is
1
maximum
might
c
Practice
which
inches
the
3
distance
mm
diagonal,
c
15)
greatest
36.3
(3
M(118.5,
22.5,
17);
N(120.5,
24.5,
18)
s.f.)
MN
=
3
m
3
3
Practice
a
40
This
b
14°
54.7°
b
54.7°
10.2
b
11.4
1
17°
3
a
4
a
5
a
2
8
m
cm
2
b
4
cm
c
2
c
10°
4
d
6
d
11°
4
is
to
a
d
(99,
an
a
13.6
m
b
Practice
1
a
(2,
d
4,
(4.5,
76
4.8°
7
12.3
estimate
degree
of
29.8,
a
415
b
She
m
c
because
Volume
accuracy;
the
numbers
tunnel
will
≈
687
will
only
m
not
be
have
a
been
perfect
m
will
probably
slope
will
ski
have
left
and
steeper
right,
and
not
in
shallower
a
straight
parts.
m
c
20.7
d
(
s
e
134
f
6.87°
155,
268,
59)
5
8)
b
5,
8)
(13,
e
(
6,
7.5,
4)
c
12.5,
8.5)
f
(
3,
(2,
1,
1.85,
m
11)
−0.15)
E9.2
2
(21,
7,
4
ABCD
5
P (1,
2)
is
3
a
a
=
1,
b
=
3,
c
=
−18
parallelogram.
Y
ou
7,
2),
Q (3,
8,
4),
R (5,
9,
should
already
know
how
to:
6)
2
1
Practice
6
2
a
10.5
cm
b
42
cm
2
a
3.76
cm
b
45.6°
3
36.1
4
a
b
1.75
2
1
a
(10,
15)
2
a
(9,
1,
3
(16,
4
m
b
(
b
(12,
1,
4)
c
(7,
12)
c
(6,
4,
cm
2
2,
2)
4,
1.5)
15)
21.25
cm
n
=
10
:
15
so
let
m
:
n
be
2
:
=
−4,
b
=
1
3.
1
a
radians
28)
Practice
:
∆ABC
∆DEF
B
F
5
e
c
Mixed
d
a
practice
D
E
f
C
A
1
57
cm
2
Yes.
(to
The
the
nearest
space
b
centimeter)
diagonal
is
around
25.7
inches
long,
and
∆LMN
drumsticks
are
usually
quite
∆UVW
L
V
thin.
u
3
a
4
a
(2,
0.5,
6);
7
b
(5.5,
3,
6);
9
c
(11,
0,
7.5);
≈
16.4
m;
Midpoint
=
(7.5,
4.5,
5)
N
M
b
5
a
6
a
It
gains
61
≈
6
7.8
meters
in
m
l
height.
b
22.6°
b
43.11°
Practice
41.77°
2
2
c
No,
since
the
angles
are
not
the
1
a
41.6
2
a
1114
2
cm
b
181
(19,
8
OA
9
z
4,
b
3402
3
=
1,
and
OB
=
189
A
is
15
closer
to
O
34
67
120°
10
Area
=
2
7897
≈
178
P
2
Area
32 8
Answers
c
2
cm
17)
131
=
2
m
same.
2
7
w
9
v
270
=
given
cylinder.
24.4)
55°
the
6
b
4
986
cm
cm
968
km
line;
2
4
a
130
6
cm
a
L
2
b
i
It
estimates
ii
It
is
an
cover
the
number
underestimate,
the
space
of
as
exactly,
tiles
the
and
needed
tiles
will
will
have
to
tile
10
probably
to
be
cut
m
not
leaving
10°
oddly
shaped
sections
so
extras
will
be
25°
needed.
F
5
35°
18
12
Y
5
49°
30°
b
2
Area
106
km
X
L
=
25°,
X
=
30°,
Y
=
125°
cm
c
5.92
km
6
Practice
1
a
a
=
6.3
2
a
91°
3
a
r
4
a
N
=
5
cm
19.0
b
b
b
103°
b
=
E
6.0
=
cm
104°
c
c
=
61.1
c
42°
c
y
=
cm
11.0
d
D
=
73.9°
60°
8
b
2
Area
of
one
face
=
27.7
75°
c
13.4
km
13
m
2
Total
surface
area
=
111
m
60°
Thickness
of
paint
is
1
mm,
so
volume
=
75°
3
111
74
×
0.001
cans
=
0.111
m
=
111
liters.
8
needed.
5
Practice
a
86.4°
b
34.8°
3
2
1
a
Area
=
0.0572
cm
Practice
6
2
b
Area
=
271
cm
2
c
Area
=
4.01
1
2
2
a
2.86
a,
b,
in
the
2
cm
b
0.112
c
49.7
cm
c
17.7
cm
c,
a
4
2
29.1
the
cosine
rule;
three
sides
and
an
angle
are
involved
2
cm
b
90.0
problem.
2
cm
2
3
e:
cm
d,
f:
the
sine
rule;
each
problem
involves
two
sides
and
2
cm
two
angles.
1
2
=
×
49 (θ
θ
=
a
d
0.80
(2
cm
b
74.6°
e
95.0
112
cm
cm
c
132°
f
26.8°
d.p.)
3
a
d
Practice
11.7
sin θ)
2
a
d
=
21.3
=
cm
b
70.6°
e
b
e
=
=
8.27
cm
c
104°
f
c
f
=
24.1°
=
10.1
cm
4
15
3
2
4
1
a
a
=
2
a
A
3
a
4
y
5
a
22.9
cm
b
b
b
A
b
x
=
16.1
cm
c
c
c
B
c
x
=
6.14
Area
≈
=
cm
6.50
units
4
2
5
x
=
=
44°,
B
=
56°
59.5°
=
=
82°,
B
33.5°
=
28°
=
=
57°,
39.3
C
=
a
7.79
cm
12.4
1.413
rad
c
11.8
cm
cm
Practice
=
b
55°
7
cm
1
a
A
R
4.1
72°
25°
Q
B
5.5
26°
2.6
39°
C
P
b
∠
CBA
=
129°,
∠
CAB
=
25°
b
c
c
=
2.7
km,
distance
=
5.3
147°
c
9.21
km
km
Answers
32 9
2
a
14
2
a
146°
b
18.1
3
a
77.6°
b
122°
4
8.8
km
c
241°
50°
km
145°
E10.1
18
Y
ou
1
should
log
y
=
already
know
how
to:
x
5
y
b
95°
c
23.7
2
x
3
a
=
7
x
km
y
2
b
=
y
=
±
x
2
and
+ 1
3
3
a
N
4
Vertical
one
unit
dilation
in
the
scale
factor
negative
a
vertical
translation
direction.
5
5
Algebraically:
f

f
(x )
=
5 ÷
=
x
x
113°
Graphically:
f
( x )
is
a
reection
of
itself
in
the
line
y
=
A
430
m
Practice
210°
310
1
m
P
1
1
a
f
1
(x )
=
log
x
b
f
d
f
x
(x )
=
6
(x )
=
5
4
x
1
c
b
3
1
M
97°
c
560
f
x
(x )
=
e
m
4 x
3
1
4
a
50°
b
039°
c
112
m
d
211
m
e
f
(x )
=

1
e
f
f
(x )
=
11
Mixed
practice
2
2
1
a
10.7
2
cm
b
63.4
cm
2
d
47.6
cm
g
24.7
cm
c
34.6
37.6
y
cm
2
e
a
2
2
cm
f
36.3
25
cm
x
2
f
(x)
=
15
×
1.15
1
20
2
2
a
2.81
2
cm
b
178
cm
b
74.8
(0, 15)
2
3
a
4
a
e
0.109
=
=
5.8
cm,
10.5
b
cm,
15
2
cm
=
f
6.76
=
cm,
14.7
c
=
cm
14.1
cm,
d
=
11.2
10
cm,
cm
5
5
a
=
43.6°,
6
a
=
1.71,
7
20.0°
b
b
=
=
38.6°,
0.547,
c
c
=
=
54.8°,
0.356,
d
d
=
=
101°
0.364
0
x
5
8
42.6
5
10
15
20
25
m
5
9
x
=
62.7°,
perimeter
=
66.1
cm
P
=
15
seagulls
0
Review
1
b
in
∠RST
=
context
85°,
∠STR
=
40°,
∠TRS
=
55°
f
(x )
=
RT
=
14.7
km,
ST
=
12.0
km
d
L
c

log
1.15 
c
x

1
b
15

1.1
y
x
f
(x)
=
15
×
1.15
1
25
20
y
S
=
x
15
100°
9.5
10
86°
x
R
f
54°
5
5
5
T
LS
=
16.8
3 3 0
km,
=
log
LT
=
22.2
1.15
km
Answers
15
x
48°
40°
e
(x)
2
5
12
10
15
20
25
3x

log

2

x
of
x
3
1
d
b
1.1
i
Domain
iii
y
The
is
x
>
graphs
−2;
are
range
is

reections
ii
in
the
f
= 10
( x )
line
y
=
2
x
x
f
(x)
=
15
×
1.15
1
25
1.1
(2.06, 20)
20
y
y
=
x
10
15
8
10
x
(x)
f
=
log
2
6
1.15
15
5
(x) = 3
f
log
1
(x
+
10
5
15
=
x
3
4
x
5
(x)
f
2)
10
(20, 2.06)
20
2
25
5
0
10
2.06
or
approximately
2
8
6
4
x
2
2
4
6
8
10
months.
x
4
3
f
(x)
=
10
2
2
Practice
1
a
B
2
a
a
b
f
2
b
=
6
C
c
10
1
c
8
A
1.2
x
( x )
The
the
=
two
line
+
1.2
Domain
functions
y
=
are
is
,
range
reections
is
of

each
other
in
x
4
x.
1
c
i
Domain
is
x
>
−0.5;
range
is

ii
f
10
(x )
1
=
2
1.1
iii
y
f
(x)
=
log
1
The
graphs
are
reections
in
the
line
y
=
x
(x)
1.2
10
1.1
8
y
6
10
4
8
2
x
f
(x)
=
(1.2)
6
2
f
(x) = 4
log
1
(1
+
2x)
10
x
10
8
6
4
2
2
4
6
8
4
10
2
2
4
f
(x)
=
x
3
6
0
10
8
6
4
x
2
2
4
8
6
10
2
8
x
10
f
(x)
=
x
1
3
f
(x)
4
=
10
2
2
+
3
a
Domain
=
;

range
=

{0};
a
=
3
8
+
b
Domain
=

c
Domain
=

a
i
;
range
=

{0};
a
=
0.5
range
=

{0};
a
=
5
+
;
+
4
Domain
is
;
range
is
10
1

ii
f
( x )
=
ln
=
x
x
2
2
The
graphs
are
reections
in
the
line
y
x
10
2
iii
+
1
d
i
Domain
is
x
>
;
range
is

ii
f
(x )
=
3
3
1.1
iii
The
graphs
are
reections
in
the
line
y
=
x
y
1.1
10
8
y
f
(x)
=
2
log
1
6
f
(x)
=
(3x
2)
10
15
x
3
4
12
2
x
f
(x)
=
+
2
e
1
f
0
10
8
6
4
2
x
2
4
6
8
(x)
=
x
3
9
10
2
f
(x)
=
ln
x
2
2
6
4
2
−
x
+
10
6
(x)
f
2
3
2
=
3
8
10
3
0
3
6
9
12
15
x
3
Answers
3 31
2
x
+
4
e
+
e
i
Domain
is
x
>
0;
range
is
1
1

f
ii
(x )
=
,
x
>
g
0
Horizontal
dilation
scale
,
factor
2
iii
The
graphs
are
reections
in
the
line
y
=
vertical
dilation
scale
5
factor
2,
in
negative
reection
in
the
x-axis,
vertical
translation
3
units
x
the
y-direction.
1
1.1
h
Horizontal
dilation
scale
,
factor
horizontal
translation
2
y
3units
10
in
x
(x)
=
+
x-direction,
vertical
dilation
scale
,
vertical
translation
4
units
in
the
positive
5
8
4
positive
1
factor
1
f
the
e
2
y-direction.
2
6
f
(x)
=
x
3
3
a
Horizontal
translation
of
5
units
in
the
negative
direction;
4
f
(x
+
5).
2
b
Reect
the
original
horizontally
0
10
8
6
4
two
in
the
units
in
y-axis,
the
and
positive
then
translate
x-direction;
f
(
x
+
2)
x
2
2
4
6
8
10
or
2
c
f
(
(x
Reect
2)).
in
the
x-axis
and
translate
vertically
2
units
in
the
4
negative
(x)
f
6
=
ln(2x)
y-direction;
f
(x
2).
4
1
4
1.1
8
y
10
4
1
3
ln x
l
f
(x)
=
x
2
5
y
=
log
x
=
b
=
hence
y
=
b
ln x .
2
a
ln a
2
ln a
x
1
7
f
6
a
y
=
log
x
b
y
=
(x)
=
ln(x)
1
2
5
c
=
log
2
x
0.2
0
x
1
1
2
3
4
1
Practice
1
a
3
Horizontal
translation
of
7
units
in
the
positive
1
x-direction,
Translate
and
vertical
translation1
unit
in
the
positive
f
( x)
horizontally
unit
y-direction.
in
the
positive
x-direction,
2
1
b
Vertical
dilation,
scale
factor
2
and
vertical
translation
of
and
dilate
horizontally
by
scale
factor
2
3
units
in
the
negative
y-direction.
1
c
Horizontal
dilation
scale
5
factor
and
vertical
1.1
translation
6
4units
in
the
negative
y
y-direction.
10
d
Horizontal
translation
2
units
in
the
negative
x-direction,
8
e
vertical
dilation
vertical
translation
Horizontal
vertical
scale
8
factor
units
translation
dilation
scale
5
4,
in
the
units
factor
reection
in
3,
in
positive
the
the
6
y-direction.
positive
vertical
x-axis,
x-direction,
translation
7
4
f
units
(x)
=
log
1
(x)
5
2
(1,
in
the
positive
y-direction.
(
0)
1.8, 0)
1
f
Horizontal
dilation
scale
,
factor
horizontal
0
translation
10
8
6
4
4
units
in
the
negative
x
2
2
4
6
8
10
2
3
x-direction,
vertical
dilation
(0,
scale
1.43)
4
2
factor
,
vertical
translation
10
units
in
the
negative
f
6
3
(x)
=
−log
2
(5
(x
+
2))
5
y-direction.
8
10
2
a
Horizontal
and
translation
vertical
3
translation
units
2
in
units
the
in
positive
the
x-direction,
positive
y-direction.
Translate
b
Vertical
dilation
scale
factor
of
0.5,
and
vertical
f
( x )
horizontally
2
units
in
the
negative
x-direction,
translation
1
1
unit
in
the
negative
dilate
y-direction.
Horizontal
dilation
scale
factor
of
horizontally
by
a
scale
factor
,
of
and
then
reect
in
5
1
c
the
,
and
x-axis.
vertical
2
Mixed
translation
d
Vertical
1
unit
dilation
in
the
scale
negative
factor
2,
and
vertical
translation
1
1
unit
in
the
positive
Horizontal
a
C
b
A
translation
1
unit
in
the
negative
b
a
=
2.2
1
dilation
scale
factor
3,
and
reection
in
the
w
=
Horizontal
2
units
3 32
in
dilation
the
scale
positive
factor
2
y-direction.
Answers
and
a
vertical
f
(s)
s
=
14
2.2
tells
what
x-axis.
the
f
B
x-direction,
c
vertical
c
y-direction.
2
e
practice
y-direction.
translation
speed
you
have
achieved.
week
you
should
be
in
for
1
3
a
i
Domain
iii
The
is
x
graphs
>
3;
are
range
is

reections
ii
in
the
f
x
( x )
line
y
=
=
1
e
+
d
3
x
i
Domain
iii
The
is
x
>−1;
graphs
are
range
y
3
f
(x)
=
x
f
3
(x)
=
3
−
1
f
2
=
3
1
x
=
x
2
10
8
2
x
0
x
4
(x)
3
2
6
=
4
4
8
x
( x )
y
x
+
2
10
f
line
6
x
e
the
8
6
=
ii
in
10
8
(x)

y
10
f
is
reections
8
6
2
4
2
4
6
8
10
2
2
f
(x)
=
ln(x
−
3)
1
4
log
4
(x
+
1)
2
(x)
f
=
1
6
log
(3)
2
6
8
8
10
10
+
e
b
i
Domain
is
;
range
is
y
>
i
Domain
is
;
range
is

2.
1
ii
f
(x )
=
The
2
graphs
are
x
= 1 −
log
x
;
x
>
0
2
>
iii

5
(x )
4
;

f

5 log
7
iii
x

1
ii
reections
in
the
line
y
=
The
graphs
x
are
reections
in
the
line
y
=
x
y
10
y
f
(x)
=
1
−
log
8
8
6
(x)
4
2
10
f
(x)
=
x
3
6
4
f
(x)
=
x
3
4
2
1
f
(x)
=
−
x
4
1
2
0.2 x
f
(x)
=
5
7
+
2
0
1
6
4
x
2
2
4
6
8
10
12
14
2
0
10
8
6
4
x
2
2
4
6
8
10
2
4
4
4
a
Translate
the
graph
2
units
vertically
in
the
positive
graph
2
units
horizontally
in
the
negative
graph
2
units
horizontally
in
the
positive
y-direction.
6
x
f
8
(x)
=
5
−
2
b
Translate
the
log
7
2
5
x-direction.
10
c
Translate
the
x-direction,
then
translate
3
units
vertically
in
the
negative
2 ( x + 5)
2
+ 8
y-direction.
1
c
i
Domain
is
x
>
2;
range
is

ii
f
(x )
=
4
d
Reect
the
vertically
iii
The
graphs
are
reections
in
the
line
y
=
graph
in
the
in
the
y-axis
positive
and
translate
2
units
y-direction.
x
e
Dilate
f
Reect
the
graph
horizontally
by
scale
factor
4.
y
the
graph
in
the
y-axis
and
translate
2
units
10
horizontally
2
(x
+
2
f
(x)
in
the
positive
x-direction.
5)
+
8
8
=
g
2
Reect
the
graph
in
the
y-axis,
dilate
vertically
by
scale
4
factor
6
f
(x)
=
2,
and
reect
in
the
x
axis.
x
3
h
Translate
the
graph
2
units
horizontally
in
the
positive
4
x-direction,
dilate
vertically
by
scale
factor
2
and
translate
2
1
0
10
8
6
4
2
i
x
2
4
6
8
unit
vertically
Reect
in
the
in
the
y-axis,
negative
translate
1
y-direction.
unit
horizontally
in
the
10
negative
2
reect
in
x-direction,
the
x-axis,
dilate
then
vertically
translate
2
by
scale
units
in
factor
the
2,
positive
4
y-direction.
1
6
(x)
f
=
log
1
(4x
−
8)
−
5
2
2
8
10
Answers
3 3 3
y
E10.2
16
5
Y
ou
should
already
know
how
to:
12
+
1
a
Domain
is
,
range
is

b
Domain
is
,
range
is

8
f (x)
=
ln
x
+
c
Domain
is

,
x
≠
1,
range
is

4
(4.31, 1.46)
(1, 0)
1
2
a
y
∝
x
so
y
=
kx
y
b
∝
k
,
so
y
=
x
0
4
x
x
2
2
8
12
14
16
4
3
g (x)
=
−7
ln
(3x
−
12)
+
Horizontal
translation
3
units
in
the
negative
reection
in
translation
x-axis,
vertical
dilation
scale
factor
5
units
in
the
positive
y-direction.
12
4(x
4
a
y
2)
,
=
16
which
is
a
function
3
20
1
b
f
(x )
=
2
±
x ,
which
is
not
a
function
24
1
Dilate
f
( x )
horizontally
by
scale
factor
;
Practice
translate
1
3
horizontally
4
units
vertically
by
scale
vertically
1
in
the
factor
positive
7;
reect
x-direction;
in
the
dilate
x-axis;
1
a
Domain:
x
≠
0,
x
∈ ,
range:
y
≠ 0, y
∈
translate
y
unit
in
the
positive
y-direction.
y
6
4
1
f
(x)
=
1
1
x
g (x)
=
log
(2
−
x)
10
4
0.5
(0, 0.075)
x
=
0
(1, 0)
0
x
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
x
0
y
0.5
1
f (x)
=
log
=
0
x
10
1.5
2
Reect
f
( x )
in
the
y-axis;
translate
horizontally
2
units
in
the
b
Domain:
x
≠ 0,
x
∈ ,
range:
y
≠ 0, y
∈
1
positive
x-direction;
dilate
vertically
by
scale
factor
y
4
7
a
y
=
log
(x
b
Students’
− 3)
4
own
answers
8
f
(x)
=
−
2
x
Review
in
context
x
1
The
2
a
b
standard
The
standard
The
a
is
model
standard
factor
3
model
model
reected
is
in
the
vertically
would
be
=
0
y
x-axis.
dilated
by
vertically
scale
dilated
factor
by
4.
=
0
0
x
scale
2.
i
The
graph
of
f
ii
The
graph
of
f
is
dilated
is
vertically
translated
7.6
by
units
scale
in
the
factor
0.67.
negative
y-direction
11
b
Domain:
2.20
×
10
26
<
E
<
1.86
×
10
c
c
There
is
no
solution
for
r
when
E
=
0.
For
the
model
Domain:
x
≠ 0,
x
∈ ,
range:
y
≠ 0, y
∈
to
y
have
y-intercepts,
translation
4
Horizontal
in
the
dilation
it
would
negative
by
scale
have
to
include
a
horizontal
x-direction.
factor
2120;
vertical
dilation
by
1
f
scale
factor
26 400;
reection
in
the
(x)
=
3
x-axis.
4x
x
=
0
0
3 34
Answers
x-direction,
1
8
y
=
0
x
2,
vertical
d
x
Domain:
≠
0,
x
∈,
range:
y
≠ 0,
y
b
∈
Asymptotes:
y
Range:
x
≠ 0,
=
y
4
and
y
=
0,
Domain:
x
≠
4, x
∈ ,
∈
y
y
2
f
(x)
=
−
1
4
f
3x
(x)
=
2
x
x
=
0
x
0
y
=
=
b
3
a
y
=
200
The
x
0
k
=
=
x
0
is
inversely
proportional
to
the
current.
Asymptotes:
x
=
0
and
y
=
3,
Domain:
x
≠
0,
x
∈ ,
20
Range:
R (I)
=
y
≠ 3,
y
∈
20
20
c
y
hours
resistance
c
b
4
4
0
2
−
or
R ( x )
=
y
x
I
4
a
50
5
a
i
hours
b
10
hours
c
5
hours
1
(x)
f
+
=
3
3
Force
and
acceleration
are
directly
proportional.
x
ii
Acceleration
and
mass
are
inversely
=
x
0
proportional.
y
b
i
Force
=
3
Acceleration
ii
0
Mass
Acceleration
Practice
2
d
1
a
h
=
−3,
x
k
=
Domain:
x
−7,
≠
asymptotes:
−3, x
∈ ,
x
=
Range:
−3,
y
y
≠
=
−7,
−7.
y
Asymptotes:
Range:
y
x
≠ 0,
=
y
2
and
y
=
0,
Domain:
x
≠
2, x
∈ ,
∈
∈
y
b
h
=
−2,
k
Domain:
=
0,
x
≠
asymptotes:
−2, x
∈ ,
x
=
−2,
Range:
y
y
=
≠ 0,
0.
y
∈
1
f
(x)
=
4
c
h
=
5,
k
=
Domain:
d
h
=
0,
k
=
Domain:
8,
x
asymptotes:
≠ 5, x
−11,
x
∈ ,
x
=
∈ ,
x
=
Range:
y
y
Range:
asymptotes:
≠ 0, x
5,
=
≠ 8,
0,
y
8.
y
≠
y
=
x
2
∈
−11.
−11,
y
∈ .
0
e
h
=
1,
k
=
Domain:
π,
x
asymptotes:
≠ 1, x
∈ ,
x
=
1,
Range:
y
y
≠
=
π
y
y
a
Asymptotes:
x
Domain:
0,
x
≠
=
x
0
and
∈ ,
y
=
x
0
∈
x
2
=
π
=
2
−4.
Range:
y
≠
−4,
y
∈
y
e
Asymptotes:
Range:
y
≠ 5,
x
y
=
−2
and
y
=
5,
Domain:
x
≠
−2,
y
∈ ,
∈
y
x
=
1
0
f
(x)
=
−
4
1
x
1
f
(x)
=
+
5
1
x
x
0
=
+
2
−2
x
y
y
=
=
5
−4
0
x
Answers
3 3 5
f
Asymptotes:
x
=
y
∈
3
and
y
=
1,
Domain:
x
Asymptotes:
∈ ,
≠ 3, x
y
=
2,
x
=
4
1
y
Range:
≠ 1,
e
Horizontal
translation
of
unit
in
the
positive
x-direction,
2
y
reection
in
the
y-axis
(or
in
the
x-axis),
vertical
dilation
scale
factor
(or
horizontal
dilation
of
scale
factor
),
5
2
vertical
of
2
5
translation
of
8
units
in
the
negative
y-direction.
1
x
=
3
Asymptotes:
y
=
y
=
−8,
x
=
2
1
f
0
Reection
in
the
y-axis
(or
in
the
x-axis),
vertical
dilation
1
x
of
scale
factor
13
(or
horizontal
dilation
of
scale
factor
),
13
1
f
(x)
vertical
=
+
translation
of
11
units
in
the
negative
y-direction.
1
2
x
−
3
Asymptotes:
Practice
1
3
a
y
1
=
− 5
y
b
b
y
d
+
=
+
x
4
2
3
4
y
y
=
+
4
b
y
=
y
=
+
x
x
x
0
4
3
+
2
4
i
h
>
0
horizontal
translation
in
the
positive
ii
h
<
0
horizontal
translation
in
the
negative
i
k
>
0
vertical
x-direction
translation
in
the
positive
x-direction
d
y-direction
y
ii
k
<
0
vertical
translation
in
the
negative
y
y-direction
1
Practice
y
3
1
a
f
(x )
+
1
1
y
=
x
1
=
− 1 =
=
x
4)
2x
+
1
8
x
0
(x )
y
1
x
0
3
3
1
f
3
1
4
x
4
1
=
2( x
b
x
0
3
c
b
−
+
d
1
=
x
a
=
2
1
5
y
x
3
1
y
2
=
1
x
c
0
7
1
a
=
1
=
x
4
x
y
1
y
−11,
4
x
c
=
a
1
=
y
=
−
2
=
y
2
=
1
x
1
(x
x
3)
3
3
1
1
4
c
f
(x )
=
+ 8
=
e
f
+ 8
−
y
x
3
3
y
f
(x )
2
a
4
+
+
Horizontal
x-direction,
=
2)
+
2x
translation
vertical
+
of
scale
6
+
=
1
of
in
the
scale
vertical
−
1
x
negative
factor
3
(or
translation
of
5
units
x
0
horizontal
=
x
0
3
1
),
1
x
y
units
dilation
factor
−
4
y
of
+
4
1
dilation
3
3
=
−( x
=
3
2
d
y
4 x
1
1
y
3
=
1
in
x
x
3
1
3
the
b
negative
Horizontal
vertical
y-direction.
translation
dilation
of
Asymptotes:
of
scale
2
units
factor
5
in
y
the
(or
=
5,
x
=
− 6
positive
horizontal
x-direction,
dilation
7
5
of
2
a
y
=
−
b
− 1
y
=
1
scale
factor
),
vertical
translation
of
3
units
in
the
positive
5
y-direction.
+ 9
x
x
6
1
Asymptotes:
y
=
3,
x
=
c
2
y
=
6
−
+ 12
d
y
=
−
+ 8
x
+ 5
x
+ 3
1
c
Horizontal
translation
of
units
in
the
negative
10
3
e
x-direction,
vertical
dilation
of
scale
factor
3
(or
y
=
− 3
x
1
1
dilation
of
scale
factor
).
Asymptotes:
y
=
0,
x
=
−
3
d
Horizontal
reection
translation
in
the
y-axis
of
4
(or
units
in
the
in
the
positive
x-axis),
x-direction,
vertical
dilation
5
scale
factor
2
(or
horizontal
dilation
of
scale
2
vertical
3 3 6
factor
),
5
translation
of
2
units
Answers
in
the
positive
3x
−
y-direction.
of
3
8
+ 1
f
=
horizontal
3
x
y
=
4 x
−
+
x
+
2
4
=
x
+
2
3
Mixed
practice
2
a
g (x )
5
x
1
a
Domain:
x
≠ 0, x
∈ ,
range:
y
≠ 0,
y
1
=
x
=
0,
y
=
g (x )
=
+ 1
−
2x
∈ ,
3
asymptotes:
b
5
c
0.
g (x )
1
12
=
d
g (x )
=
1
−
3
x
4
x
y
3
a
Horizontal
translation
in
the
y-axis
of
in
the
positive
9
units
Asymptotes:
x
=
of
reection
x
=
2
2
(or
and
y
units
in
the
in
the
positive
x-axis),
x-direction,
vertical
translation
y-direction.
=
9.
0
y
y
=
0
0
x
5
y
=
x
1
g(x)
=
+
−
x
b
Domain:
x
≠ 0, x
asymptotes:
x
=
∈ ,
0,
y
range:
=
y
≠ 1,
y
∈,
9
2
0
x
1.
y
2x
19
1
g
Inverse:
(x )
=
x
b
x
=
Horizontal
9
translation
vertical
dilation
of
factor
of
of
scale
1
unit
factor
in
3
the
(or
negative
horizontal
x-direction,
dilation
0
1
y
=
1
scale
),
vertical
translation
of
3
units
in
the
3
negative
0
y-direction.
x
Asymptotes:
x
=
−1
1
y
=
and
y
=
−3.
y
+
1
x
3
h(x)
c
Domain:
x
≠
asymptotes:
2, x
x
=
∈ ,
2,
y
=
range:
y
≠
−8,
y
=
∈ ,
3
x
+
1
−8.
0
x
y
x
0
2
y
=
3
8
x
1
Inverse:
2
(x )
h
=
1
x
c
y
=
Horizontal
+ 3
translation
of
5
units
in
the
positive
x-direction,
−8
reection
in
the
y-axis
(or
in
the
x-axis),
vertical
dilation
1
scale
factor
10
(or
horizontal
dilation
of
scale
factor
)
10
Asymptotes:
x
=
x
=
5
and
y
=
0.
2
y
d
Domain:
x
≠
asymptotes:
−3, x
x
=
∈ ,
−3,
y
=
y
range:
≠
2,
y
∈ ,
2.
y
10
k(x)
=
5
x
0
y
=
x
2
0
x
4
x
=
−3
y
=
+
x
+
2
3
10
1
Inverse:
k
(x )
=
−
+ 5
x
Answers
3 37
of
d
Horizontal
translation
of
1
unit
in
the
positive
600
x-direction,
4
reection
of
scale
in
the
factor
y-axis
4
and
(or
in
the
horizontal
x-axis),
vertical
dilation
of
scale
a
translation
of
1
unit
in
the
positive
mg
(3
s.f.)
C
b
=
+ 50
−
factor
+ 12
x
3,
c
vertical
22.7
dilation
Horizontal
translation
of
12
units
in
the
negative
x-direction,
y-direction.
reection
in
the
x-axis
vertical
dilation
of
scale
factor
600,
OR
vertical
Horizontal
reection
translation
in
the
of
y-axis
1
(or
unit
in
in
the
the
positive
x-axis),
horizontal
dilation
of
scale
factor
in
the
positive
y-direction.
context
of
1
unit
in
the
positive
1
854.91
2
a
AUD
=
1
1
x-direction.
slow
pipe
=
p
3 p
−
and
y
=
1
+
b
2
x
in
),
4
3
translation
Asymptotes:
units
3
(or
factor
vertical
50
dilation
Review
scale
of
x-direction,
vertical
4
of
translation
3 p
1.
3
1
1
1
+
c
y
p
=
,
3 p
Time
p
=
10.5
hours.
14
to
ll
with
slow
pipe
=
31.5
hours.
E11.1
0
x
4
m(x)
=
+
3x
+
Y
ou
1
should
already
1
a
b
4
(
)
c
3
3
2
translation
in
300°
the
of
y-axis
2
(or
units
in
the
in
the
positive
x-axis),
1
2
or
a
b
2
reection
d
2
1
Horizontal
30°
=
3x
e
to:
4
3
1
how
3π
2π
Inverse:
know
2
c
1.07
(3
s.f.)
2
2
x-direction,
vertical
3
a
71.3°
(3
s.f.)
4
Horizontal
b
30°
dilation
1
of
scale
factor
5
and
horizontal
dilation
of
scale
factor
4,
dilation
of
scale
factor
,
vertical
dilation
of
2
vertical
translation
of
6
units
in
the
positive
y-direction.
scale
factor
3,
vertical
translation
of
5
units
in
the
positive
OR
ydirection.
Horizontal
translation
of
2
units
in
the
positive
x-direction,
3
reection
in
the
y-axis
(or
in
the
x-axis),
vertical
dilation
of
scale
5
x
=
or
factor
(or
horizontal
dilation
of
scale
factor
x
5
6
a
∠B
=
25.7°
(3
s.f.)
b
4.26
),
Practice
translation
Asymptotes:
=
5
4
vertical
x
2
4
5
=
2
6
units
and
y
in
=
the
positive
1
y-direction.
1
6.
a
30°,
150°
b
104°,
256°
c
45°,
225°
d
225°,
315°
e
143°,
(3
s.f.)
y
5
n(x)
=
+
8
6
323°
f
2π
2
3π
x
3.73
e
π
=
+
4 x
1
p (x )
3π
i
2
2π
4π
5π
3
3
3
2
3
0.464,
2.68,
3.61,
5.82
5
=
horizontal
translation
of
units,
vertical
π
4 x
π
5.64
24
l
f
3
3.79,
k
2
2
f
6
π
j
5
(x )
4.86
7π
6
3π
1
3
,
h
4
5π
,
c
4
1.43,
7π
,
4
4π
,
4
3
2.56,
g
n
210°
3π
b
3
0
π
4π
a
d
Inverse:
150°,
4x
5
5π
4
π
m
n
1.05,
1.91,
4.37,
5.24
p
1.11,
2.16,
4.25,
5.30
1
dilation
of
scale
(or
factor
horizontal
dilation
of
3
scale
4
factor
2π
4).
o
5
Asymptotes:
x
=
and
y
3
=
0,
π ,
, 2π
3
0
π
4π
,
4
3
4
π
y
3
3π
5π
a
7π
, π ,
0,
b
4
4
4
7π
, 2π
4
11π
1
p(x)
4
=
4x
2.30,
, 3.98,
6
x
0
b
5
Practice
6
2
3π
1
a
θ
4π
=
b
4
d
1
1
Inverse:
p
(x )
=
=
c
=
3
2π
4π
3
3
, 2π
0,
+
4 x
3 3 8
θ
5
θ
4
Answers
e
θ
=
0.068, 0.56,1.32,1.82, 2.58,3.07
=
3π
3π
q
4
2
3π
f
5π
11π
g
8
8
π
5π
18
18
13π
=
8
x
13π
17π
Mixed
18
18
1
5π
x
h
7π
13π
15π
π
=
θ
i
8
8
8
=
π
0,
8
a
90°,
d
6
3
16.6°,
π
=
x
k
π
5π
2
6
150°,
e
No
2π
3π
a
3
x
=
0.955,
2.19,
4.10,
π,
0,
2π
5π
2π
4π
π
3
5π
3
3
π
a
0,
180°,
0°,
, 2π
3
360°
75.5°,
b
180°,
d
It
41.4°,
is
not
and
180°,
an
thus
319°
e
identity,
not
210°,
3
b
true
30°,
since
for
Practice
all
150°,
it
is
210°,
true
values
of
π
5π
13π
17π
12
12
12
12
(1
a
225°,
sin x
+
sin
tan x
when
sin x
=
cos x,
3
5
x)
≡
cos
− cos x
∠C
=
>
a
(θ
30°,
c
∠C
∠C
d
2
a
12.8°
=
∠Q
or
∠C
(∠C
(∠C
70.7°
=
∠P
or
53.1°;
=
0°,
∠Q
=
p
=
≡
∠Q
=
=
=
−sin
180°,
π
π
6
2
∠Q
∠P
43.9°,
34.2°,
13.2°,
3
70.0°
5
39
6
a
c
7
m
1
0
=
=
or
or
<
Quadrants
0.
sin θ
If
270°
360°
11π
3π
π
3π
=
I
and
tan θ
0,
b
90°,
d
109°,
5π
b
6
5π
9
IV
=
120°,
0.
If
sin θ
240°,
180°,
<
0,
270°
251°
7π
c
,
2
4
4
4
0,
π,
2π
4
π
π
e
6
x
12
It
13
a
=
167°
=
=
=
isn’t
6
=
isn’t
4π
7π
,
4
3
3
is
an
B
=
identity
36.9°
∠P
=
=
124°,
11.3
∠P
p
=
=
or
because
B
=
it
is
143.1°
true
for
all
values
4
θ
of
b
B
=
60.3°
or
B
=
119.7°
in
b
T
=
65.4°
or
34.6°
possible)
1
a
2
7.70
I
=
3
a
74.6°
p
=
9.93
cm
or
∠Q
=
km
or
or
105.4°
2.08
km
127°;
228
m
or
p
=
31.4
Because
427
m
with
p
=
44.4
the
given
information,
there
are
two
possible
cm
m
or
∠Q
=
for
the
position
of
B
(because
there
are
two
146°,
solutions
for
the
angle
B ).
m
127°,
6.63
context
cm
81.1°,
110.1°
7
2π
f
109°
96.9°;
3.93
∠P
p
35.0°,
17.0°,
2.91, 3.38
2
possible)
c
d
16π
9
120°)
possible
∠P
14π
9
0.
solutions
c
10π
104.6°
139°
∠C
∠P
23.1°;
=
tan θ
a
b
b
b
0,
d
x)
5
41.4°
=
∠C
9

− 1
75.3°
=
>
150°,
Review
b
8π
9
x

(cos
Practice
a
315°
sin θ
If
π
1
4π
9

b

2π
d
x
11
tan x
cos x

306.9°
330°
only

+

or
4
1

53.1°
330°
c
a
6
360°
10
2
6
3
3
90°,
284°,
5π
3
c
5π
0,
3
4π
tan θ
3
7π
f
3
a
c
c
3
π
5π
3
4
2
4π
3
e
, π,
c
3
5π
e
3
π
, π,
b
d
3
π
4
3
3
π
315°
5.33
π
a
338°
225°,
18
18
2
1
158°,
f
5π
4
d
Practice
c
c
3
3π
l
330°
solution
b
,
=
b
163.4°
3
π
θ
270°
2π
,
π
j
practice
=
8
p
=
18.1
km
or
∠Q
4
=
m
145°,
4
km
199
a
16.2°
m
triangle
b
1
triangle
triangles
d
2
triangles
a
B
68.3°
9.65
8
b
Either
235.2
million
km
or
44.8
million
km.
E12.1
63.7°
48°
A
C
10
Y
ou
should
already
know
how
to:
B
8
1
a
2
x
=
0.699
2.73
4
y
=
3x
b
1.10
3
x
c
2.58
3.73
111.7°
=
−1
48°
20.3°
A
4
x
C
10
Inverse
function:
y
+
4
=
3
b
8
a
9.7
2
km
or
3.7
km
b
10.9
m
or
5.4
m
Answers
3 3 9
1.1
x
y
1
y
=
3x
1
( x ) =
h
c
4
4
10
2
15
1
1
y
=
Domain
x
of
h
is
x
>
−
10
;
range
of
h
is
.
Domain
of
h
is
;
2
1
5
x
y
+
1
4
range
of
h
is
y
>
−
=
2
3
0
10
x
5
5
10
1.1
15
5
y
1
h
(x)
10
15
y
=
x
10
The
functions
x + 11

are
reections
in
the
line
y
=
x.
5

h(x)

1
5
y
=

2
5
3
x
0
10
5
5
10
15
5
Practice
1
10
1
a
log
12
b
log
15
c
log
200
1
z
d
2
3
3
9
x
e
x
x
x
log
2 + e
f
log
d
(x ) =
j
;
4
y
+
1
ln
8
Domain
e
96 y
a
ln
a
+
ln
b
b
ln
a
ln
b
c
2
ln
a
+
1
ln
a
e
2
ln
a
f
ln
a
+
2
ln
b
ln
b
Domain
of
j

ln
a
x
e
2y
a
1;
range
of
j
is

1
;
range
of
j
is
y
>
1.
1.1
a
ln
b
h
3
ln
a
2 ln
b
i
y
(ln
a
ln
b)
+
y
b
y
x
c
x
15
y
d
2x
j(x)
y
4
>
2
2
3
x
+
is
1
3
is
1
d
g
j
1
1
2
of
x
2x
f
+
3y
+
3x
g
2
b
y
5x
+
4y
5
2
x
1
j
1
=
10
3x
(x)
1
c
(6x
+
8y
+
3)
d
x
y
1
x
3
5
2
10
15
5
Practice
2
10
1
1
a
f
+
(x )
=
log
x
+ 3;
Domain
of
f
is
;
range
of
f
is

15
2
1
Domain
of
f
1
+
is

;
range
of
f

is
n
2
1.1
g (x )
f
=
( x )
log x
scale
=
n
factor
log
x
g (x )
=
n
log
x
is
a
vertical
dilation
of
n
y
f (x)
15
(x )
3
=
− log
x
+
2.
Reect
the
graph
of
y
=
log
=
x-axis
x
and
x
in
the
3
3
y
translate
it
vertically
2
units
in
the
positive
10
ydirection.
5
1
f
(x)

4
a
i
0

=
2
10
8
log

log
8 −
x
log
2

x
=
3 −
x
log
2
2
x
5
5
10
15
2
5
ii
log
(x
+ 3)
=
2
log
2
(x
+ 3)
2
10
b
i
Reect
y
=
log
x
in
the
x-axis
then
translate
3
units
2
vertically
in
the
positive
y-direction.
2
x
e
1
b
g
1
(x ) =
,
x
>
0;
Domain
of
g
is
x
>
2
+
;
range
of
g
is
ii

Translate
and
+
1
Domain
of
g
3
units
horizontally
in
the
negative
x-direction,
2
is

dilate
vertically
by
scale
factor
2.
1
Range
of
g
is
y
>
0.
Practice
3
1.1
n
3
p
4
1
a
log
p
(x
2)
2
b
q
log
c
y
ln
3
x
q
15
1
g
(x)
x
d
y
=

ln
2
(x
2
5
a
a
b
a
+
x
b
5
10
a)
e
2
x
5
1
(b
d
0
10
c
2
15
5
3
a
b
10
c
34 0
Answers
ln(x
ln(x
ln( x
3)
+
2)
+
+
− 3) −
ln(x
ln(x
ln( x
+
+
+
3)
3)
4)
ln ( x
2
(x

g (x)
1
4
f

2)
b

ln
e
x
10
e
+ b
c
2a
+
c
1)
)
d
ln(x
+
1)
ln(x
+
2)
ln(x
ln(x
+
1)
ln(x
3
2)
x
e
ln(x
6)
+
+
2)
ln(x
2)
1
c
f
2
ln x
+
ln(x
+
2)
+
ln(x
5)
ln(5x
h
2
e
(x )
+ 1
=
20)
3
1
Practice
4
Domain
of
h
Domain
of
h
is
x
>
;
range
of
h
is

1
1
1
a
2.81
b
1.29
c
2.10
1
is
;
range
of
h
is
y
>
3
d
0.732
e
1
f
1.16
x
5
4
+
1
k
d
Practice
(x )
=
;
Domain
of
k

is
;
range
of
k
is
7
4
ln 7
1
2
3
1
a
1
b
ln x
ln x
1
c
d
ln x
5
5
Domain
(ln
10
x
ln x
c
a
y
2
ln x
=
3
+
d
x
=
b

y
=
x
b
x
=
1.91
c
x
=
log
( x );
x
=
0.693
b
x
=
0.79
c
x
Translate
the
positive
a
x
=
0.397
b
x
=
0.417
c
a
c
a
=
1.044
8
4
log
4
(x
9);
x
=
x
0
=
or
1
x
or
=
x
y
Dilate
translate
x
=
e
6
b
x
1.2
=
x
2
=
b
or
−4
10
x
or
=
x
a
log
x
by
3
log
x
vertically
9
units
by
x
the
c
3a
positive
a
+
b
+
c
(3a
+
b)
2c
(a
+
c)
e
2
5)
7)
+
+
ln(x
+
ln(x
5)
+
ln(x
+
1)
ln(x
4)
8)
−7.
=
x
5.
x
c
=
−7
=
−4
ln(x
5)
+
ln(x
+
1)
ln(x
+
4)
ln(x
+
2)
5
=
x
=
e
ln(x
12)
ln(x
12
+
ln(x
+
2)
ln
x
ln(x
+
2)
ln(x
2);
8
is
is
not
not
possible,
hence
possible,
hence
x
=
x
=
ln
x
ln(x
2)
2
9
a
x
=
±
10
a
x
=
15
1
b
x
=
2
b
x
=
112
5
c
x
=
c
4x
3
years
15
is
the
initial
number
of
video
x
=
2
e
x
=
6
sharers.
x
b
21
hours
11
b
a
3x
+
y
+
2y
y
c
d
8
a
c
23
hours
Students’
x
=
x
own
answers
0.746
=
0.333
d
b
x
=
1.35
d
x
=
2
12
a
13
a
y
x
x
=
e
0.24
b
x
y
log
y
c
x
=
1.66
d
x
=
2.05
2 y
c
y
x
practice
14
a
x
−4.64
x
3
1
=
b
x
Mixed
a
+
log
b
+
log
c
b
log
a
+
log
c
log
a
3
ln x
b
ln x
16
b
8
8
1
c
3
log
a
+
2
log
b
+
4
log
c
d
log
a
+
e
log
a
+
log
b
15
a
24
16
a
52
b
5
b
2.8
2
1
log
b
log
c
0.05 x
days
million
17
y
= 15e
2
2
a
log
12.5
b
ln
250
c
log
40
4
d
4
log
Review
6

ln
e
f
a
2

( x
− 1)
in
context

x
log
6)
+
54
4
3
(2x
1

a
8.90
1
b
4 x
g
Scale
factor
of
6.31
times
bigger
3
log
7
c
2
Add
1.3
y
d
3
a
d
y
2x
b
+
2y
e
x
+
2y
3x
c
(x
+
y)
40
times
e
Eastern
a
6.8
bigger
Sichuan
was
40
times
bigger.
2y
2
acidic
b
2.8
c
6.2
1
4
a
5x
+
y
b
1
+
(x
+
3y)
−7
d
Yes
e
1 × 10
−8
− 3.98
× 10
4

c
2(3x
+
4y
2)
d
1
+
1
y

ln
x

7x
2
3
a
f
(x )
137
dB
b
6.3
times
bigger
ln 30
;
=
2
a

1
5
ln
Domain
of
f
is
;
range
of
f
is
+

E13.1
5
+
Domain
of
1
is
f

;
range
of
is
f

Y
ou
3
(x
1
b
g
(x )
+
;
Domain
of
1
Domain
of
g
g
is

;
range
of
g
;
range
of
g
is
already
know
how
to:
;
1
Student
2
a
diagrams
in
DGS
+
1
is
should
+
2)
= 10
is
units
scale
in
1
ln(x
d
7
=
1.29
34
d
=
horizontally
b
ln(x
d
a
y
of
2.58
=
5
5
graph
3
and
b
(2b
a
c
b
the
0.55
b
4

1.30
=
x
is
y-direction.
2
3
k
direction.
d
a
of
3
in
1
2
range
3
6
2.93
;
4
2
9
7
a
≠
ln
factor
1
x

vertically

Practice
,
3
1
x
5
b
is
ln
6
ln x
+
1
k
5

a
of
ln 3
7
2

x
5

b
Reection
Rotation
in
90°
x
=
−1
anticlockwise,
center
(3,
Answers
2)
341
2

c
Translation
2
units
left
and
6
units
down,
Reection
in
x
e
Reection
in
the
f
Rotation
90°
=
0
or
the
line
y
y-axis
=
x
+1
anticlockwise,
center
a
=
110°,
b
=
70°
4
AC
=
1
A
F,
B
4.5
cm,
DE
=
6
cm
3
M
5
∆ ABC
1
E,
C
JK
≅
JF
FK
≅
FK
2
I
K,
4
S
G
(reexive
N
Q,
O
P
V,
L,
T
H
∆ DEF
≅
∠SRX
(alternate
∠PXQ
≅
∠SXR
(vertically
≅
XS
(denition
the
X,
U
≅
=
CD,
AC
=
CA
≅
a
b
No
b
AC
(AAS
≅
CD
or
No
c
BE
=
3
∠BCA
≅
∠ECD
(vertically
Hence
they
ET
AD
≅
DT
is
are
opposite
congruent.
(given)
they
are
BD
is
=
congruent.
(denition
∠BDC
≅
is
a
the
denition
symmetric
of
because
it
=
is
of
a
rhombus.
property.
SSS
of
congruence.
CPCTC.
given
90°
AB
=
=
of
90°
because
because
AC
×
of
it
is
∠CAX
+
≅
∠AXC
⇒
2
∠AXC
⇒
∠AXC
=
=
of
the
≅
given.
AX
∆ ACX
180°
of
an
isosceles
triangle.
congruency.
CPCTC.
triangle
∠AXC
=
RHS
of
because
∆ABX
∠AXB
denition
because
because
≅
the
because
∠CAD
is
is
isosceles.
the
angle
bisector.
(SAS).
(CPCTC).
because
they
lie
on
a
straight
line.
180°
90°
(SSS)
OA
=
OB
because
they
are
both
radii.
bisector)
(denition
of
AM
=
BM
OM
is
because
M
is
the
midpoint
of
AB
perpendicular)
common.
Therefore
they
are
congruent.
rotating
the
pentagon
by
∆OAM
≅
∆OBM
(SSS).
(SAS)
Hence
By
(AAS)
common.
Hence
3
from
it
because
∆DCA
≅
∠AXB
common.
∠ADC
DC
congruent.
because
∠DCA
CA
Hence
(ASA)
4
≅
DA
∠BCA
Therefore
angles)
(given)
ED
=
∠BAX
(given)
AD
=
∆CDA
≅
=
∆BCE
(SSS)
(given)
∠DEC
Hence
c
BC
because
DA
∠CEB
≅
≅
Yes
ASA)
∠BAC
AT
are
2
CE
2
angles)
∆TSU
∠BCE
Yes
triangles
opposite
midpoint)
∆WXV
2
d
of
angles)
∆ PRQ
≅
Practice
a
(SSS)
4
AB
∠DAC
1
congruent.
W
∆ABC
∆ JKL
are
∆ NOM
≅
∆ GHI
property)
triangles
J
1
≅
square)
circle)
∠PQX
Practice
R,
of
of
the
Hence
D
(denition
(denition
Hence
PX
Practice
≅
HF
(0,1)
7
3
HK
6

d
6

or
144°,
∆CDE
will
line
up
∠AMO
≅
∠BMO
with
∠AMO
+
∠BMO
=
180°
because
they
lie
on
a
straight
line.
∆ABC
∆ 2
4
∆ABD,
∆BFE,
∆DCA,
∆EDB,
∆CDF,
∆DEA,
∆FAC,
∆EFC,
∆BAE,
×
∠AMO
180°
∆CBF,
∠AMO
∆BCE,
=
=
90°
∆AFD
5
a
L
1
Practice
1
AB
≅
BM
AC
≅
AM
≅
(denition
AM
(reexive
the
of
midpoint)
property)
triangles
∆ABM
and
≅
CD
BO
≅
DO
(denition
of
circle)
OC
(denition
of
circle)
≅
Hence,
PQ
≅
∆ACM
are
congruent.
(SSS)
Y
AB
AO
3
X
(given)
MC
Hence,
2
3
(given)
the
QR
triangles
are
L
2
congruent.
b
(SSS)
1
OX
2
∠OXZ
≅
OY
=
(they
≅
QX
∠QXR
=
Hence,
∠ABC
the
=
90°
triangles
(denition
are
of
≅
≅
congruent.
∠ACB
AC
(given)
(∆ABC
implies
3
OZ
4
Hence
≅
≅
OA
OC
is
∆ABC
Hence
5
AB
AD
≅
≅
≅
(reexive
AB
AE
Hence
a
circle
in
triangles
(reexive
of
≅
has
a
congruent.
≅
∆OYZ
(RHS)
(CPCTC)
ABCDE
(properties
line
of
a
of
symmetry
regular
perpendicular
pentagon).
Since
K
to
is
DE
through
equidistant
ED,
it
lies
of
triangles
of
of
on
the
KA,
line
they
of
are
symmetry,
equal
and
hence
KC
length.
(SSS)
Mixed
property)
(denition
34 2
YZ
circle)
are
(denition
the
∆OXZ
property)
(denition
the
BD
to
isosceles
Practice
circle)
1
BE
tangent
radius)
common
XZ
reection
Hence
the
isosceles)
to
OB
to
(RHS)
B
AO
radii)
(the
altitude)
10
AB
both
90°
(reexive)
∠QXP
c
4
=
(given)
perpendicular
QX
are
∠OYZ
a
No
c
Yes
(only
SA
given)
b
No
(only
SSA
given)
d
No
(only
SSA
given)
circle)
are
congruent.
Answers
(SSS)
(AAS)
is
a
c
Review
in
context
1.1
y
1
a
Since
folding
rectangle
they
b
EG
are
is
is
is
a
equivalent
mirror
to
image
reection,
of
the
other
one
and
hence
congruent.
the
full
width
of
the
page.
EX
is
half
the
width
of
thepage.
y
=
4 log
(2 x)
10
c
30°
d
Since
(0.5, 0)
GDZ
is
the
empty
space
previously
occupied
by
the
x
0
triangle
∠EGD
e
30°
f
Since
GDE,
≅
the
∠DGZ
GF
is
half
=
two
triangles
are
congruent.
30°
the
width
of
the
paper
(the
edge
GE )
1.1
d
and
EY
is
half
the
width
of
the
paper,
they
are
the
y
same
length.
2
g
GF
≅
EY
( by
f,
above)
y
=
x
∠FGH
h
=
∠YED
=
30°
(by
∠GFH
=
90°
(since
ABFG
∠EYD
=
90°
(since
the
Hence
∠EYD
≅
∠GFH
Hence
∠GFH
=
∠EYD
FH
≅
DY
d
and
is
fold
a
is
e,
−
4
above)
(0, 0.5)
rectangle)
parallel
to
ABC )
x
0
(ASA)
(CPCTC)
E14.1
e
Y
ou
1
x
<
should
already
know
how
1.1
to:
7
y
2
a
2x
−
3
3x
+
6
=
1.1
(1.5, 0)
y
x
0
(
4, 0)
(1, 0)
0
x
2
y
=
x
+
3x
−
4
f
1.1
y
(
1.5,
6.25)
y
b
=
√x
−
5
1.1
0
x
(5, 0)
y
3
x
=
−3
6
±
(0.405, 0)
Practice
0
1
x
(0,
1)
1
a
c
x
x
<
≤
−2
or
−0.4
x
or
>
x
6
≥
5
b
1.5
≤
d
4
x
<
x
<
≤
2
1.5
x
y
=
2e
−
3
2
3
e
<
x
<
3
2
2
a
b
h (t)
>
9600
About
3
x
(x
+
4
x (12
6)
23
>
x)
m
seconds
216;
≤
24;
x
0
>
12
m
<
cm;
x
≤
(x
+
2.54
6)
m
>
or
18
cm
9.46
m
≤
x
<
12
m
2
5
y
=
0.005x
0.23x
+
22
>
25;
x
>
56
years
Answers
34 3
6
a
h (t)
>
0
b
3.56
d
seconds
y
6
7
x
>
2.73
8
4.47
cm
5
cm
×
4.47
cm
×
2
cm
4
Practice
x
=
2
2
3
1
1
a
x
>
1
or
1
<
x
<
0
b
0
<
x
≤
2
4
y
c
1
<
x
e
1.5
g
3
<
<
4
<
x
x
<
<
d
−1
or
x
R
f
>
4
<
<
x
x
>
2)
+
3
<
<
−1
−2
or
or
1
x
<
<
x
(2.05, 0)
−5
<
0
6
x
2
3
5
6
2R
1
=
⇒
> 1,
R
>
2
1
2 +
ln(x −
1
5.09
1.5
h
1
a
1.57
0
2R
2
>
x
R
2 +
y
e
R
1
1
3
b
R
>
7.48
2
2
x +
y
≤
1
e
−
3
1
Practice
3
(0.099, 0)
1
y
a
x
4
3
2
1
8
1
2
3
4
5
6
1
6
2
y
4
2
y
>
x
+
3x −
=
−3
3
1
2
(
3.3, 0)
(0.303, 0)
f
y
x
6
2
4
4
2
8
6
2
(0,
1)
6
(
1.5,
y
3.25)
x +
3
x −
3
<
4
6
2
(
x
=
3
3, 0)
8
x
8
6
2
4
4
y
b
2
(0,
6
10
12
1)
1.0
4
(1.5, 0.25)
0.5
(1, 0)
6
(2, 0)
0
x
1.0
2
1.0
2
a
y
<
− x
2
−
4 x
+ 5
b
y
≥
0.5 x
+
2x
y
≤
− 1
0.5
x
2x
1.0
c
y
a
i
≤
+ 1
e
−
d
4
>
y
x
2
y
≤
−x
+
3x −
2
2
2
3
1.5
f
(x )
=
x
−
4
ii
y
>
f
( x )
ii
y
>
f
( x )
ii
y
<
f
( x )
1
2.0
(0,
2
2)
b
i
f
(x )
=
(x
− 1)
+ 3
3
2.5
2
c
i
f
( x )
=
x
6x
+
5
(x
−
45)
3
2
4
a
f
(x )
=
−
+ 3
2025
b
c
Domain
0
≤
y
≤
{x
f
0
≤
x
≤
90},
range
{y
0
≤
3}
(x )
y
c
6
(
Practice
0.75, 5.13)
4
5
1
a
1
≤
x
≤
3
b
x
<
2
or
x
>
≤
4
8
(0, 4)
4
c
3
≤
x
≤
4
d
10
≤
x
2
y
≥
4
−
3x
−
2x
1
1
3
e
3
<
x
<
f
2
2
1
g
<
1
(
0
3
2
0
<
x
≤
1.70
3
Student’s
cm
x
1.0
1
2
3
34 4
<
(0.851, 0)
2.35, 0)
2.0
x
2
Answers
own
x
<
or
3
2
answers
x
>
1
Practice
c
5
y
2
1
a
x
<
0.209;
x
>
(
4.79
b
6.32
<
x
<
0.317
c
2.19
≤
x
≤
0.687
x
<
0.333
or
x
>
(4.35, 0)
0
8
d
0.345, 0)
6
4
x
2
2
2
4
e
2.79
≤
x
≤
1.79
2
y
<
2x
−
8x −
3
6
f
x
<
−0.414
g
0.290
h
9.12
<
≤
or
x
<
x
>
2.41
0.690
8
x
or
x
≥
−0.877
−1.18
or
x
>
0.425
10
2
i
x
j
0.209
a
<
∆
=
0,
<
x
<
4.79
hence
only
one
root,
x
=
−1.
Since
the
graph
above
the
x-axis
except
for
this
one
value,
11)
lies
d
entirely
(2,
12
there
y
is
1.5
only
b
∆
=
solution,
0,
hence
f
( x )
only
=
0
one
for
x
root,
=
x
−1.
=
2.
Since
the
graph
1.0
lies
y
entirely
above
or
on
the
x-axis,
the
solution
is
x
∈
<
log
(x
+
1)
10

0.5
c
Since ∆ < 0, the quadratic has no roots. Since a > 0, its graph
(0, 0)
is always above the x-axis for all values of x. Hence x∈ 
x
2.0
d
1.0
1.0
Since ∆ < 0, the quadratic has no roots. Since a < 0, its graph
2.0
3.0
0.5
is always below the x-axis for all values of x. Hence x ∈ 
1.0
Practice
1
6
1.5
a
x
<
−1
or
x
>
1
b
c
x
<
−1
or
x
>
4
d
1
<
x
≤
4
y
e
x
≥
3
or
x
<
2
10
2
1
e
0
<
x
<
f
x
>
5
or
x
≤
−
2
3
<
x
<
8
3
4
−1
6
3
About
17
(0, 5.14)
4
x −
y
Mixed
1
≥
2
e
+
5
practice
a
x
<
−3
c
x
<
−3.59
or
e
0
x
>
2
−1
b
x
≤
−8.58
or
d
x
>
0
<
f
0.333
x
≥
0.583
0
10
<
x
<
or
x
>
2.09
0.286
or
<
x
x
<
8
6
4
x
2
−3
2
8
10
y
y
a
6
5
f
2
4
2
8
6
6
2
y
>
x
−
4x +
2
4
4
y
x
2
=
x −
2
x +
4
>
2
−4
(3.41, 0)
(0.586, 0)
x
12
0
6
4
10
8
6
4
2
2
4
6
8
x
2
2
4
2
6
(0,
1)
2
(2,
2)
4
4
6
6
3
y
b
a
6
c
7
<
<
x
<
x
<
1
b
5
x
d
<
8
−2
<
x
and
<
x
>
12
2
6
(1.5, 4.5)
3
e
0
<
x
<
5.5
f
x
≤
4
or
x
≥
2
5
2
y
≥
−2x
+
6x
g
x
>
3
or
x
<
−5
h
1
<
x
<
1
2
(3, 0)
(0, 0)
2
4
36.2
meters
5
Between
11
6
4.6
<
radius
7
1.66
<
x
x
0
4
2
4
6
and
30
tourists
2
cm
<
9.7
cm;
6.8
cm
<
height
<
29.8
cm
4
in
<
6.16
in
6
Review
1,
2
in
context
Student’s
own
answers.
Answers
34 5
d
E15.1
Y
ou
1
should
already
know
how
a
Song
to:
A
B
2
9
1980
17
Song
1
10
1980
18
e
8
1990
17
A
B
10
1980
17
8
1990
18
7
1990
17
Practice
5
1
b
17
1
1
1
P (B
|
A)
=
B
36
32
2
3
a
4
b
52
36
c
169
1
221
P (B
A
1
|
A
)
=
36
2
35
B
1
15
3
a
P ( A)
b
=
P(A ∪
B )
P (B )
36
= 1
1
=
1
+
=
72
27
72
36
1
B
4
36
U
1
A
2
A
B
35
26
7
21
B
36
P (B
∴
A)
=
P (B
independent
A
)
=
P(B )
events
5
5
2
5
P (B
a
A )
=
B
9
9
3
5
P (B
A
A
)
=
5
9
4
B
P (B )
=
P (A
∩
B )
+
P (A′
∩
9
3
=
5
B
=
+
9
5
2
9
9
9
2
A
b
5
4
B
A
9
B
P (B
∴
A)
=
P (B
independent
A
)
=
P (B )
events
3
4
4
B
9
P (B
A)
=
c
9
5
5
A
P (B
A
)
=
10
9
A
B
5
B
2
9
P (B )
B
9
5
A
10
4
B
9
P (B
The
34 6
Answers
A)
≠
P (B
events
are
A
)
not
≠
P (B )
independent.
=
5
+
9
5
1
=
18
2
B )
4
6
17
B
5
10
a
ii
35
6
P( B
7
A)
b
iii
35
35
7
13
9
A
1
17
19
i
i
ii
iii
=
4
B
18
17
19
10
10
7
c
28
1
B
2
6
1
A
7
P( B
A
)
Practice
3
=
2
1
a
1
Male
(M )
Female
(F )
Total
B
2
Professional
The
events
are
not
P (B
A)
5
12
17
5
6
11
10
18
28
Amateur(A)
5
5
(P )
independent.
=
9
Total
6
P (B
A
)
=
,
therefore
not
independent
events.
9
5
b
i
P (M
P )
10
5
=
ii
P (M
A)
=
iii
P (M
)
=
5
6
P (B
A)
17
=
28
11
11
c
No
they
are
not
independent
events
6
P (B
A
)
=
,
therefore
not
independent
events.
2
11
7
8
=
soccer
P (B
A)
P (B
A
)
=
65%,
P (Y
X )
=
0.8
P (Y
X
)
No
soccer
Total
90%
=
therefore
not
independent
0.8
∴
P (Y
∴
independent
X
)
=
P (Y
X )
=
P (Y
Left
10
30
40
Right
15
45
60
Total
25
75
100
events.
)
10
P (L
S )
=
events
25
Practice
30
2
P ( L
11
1
=
75
11
a
b
26
2
NS )
2
32
P ( L )
=
They
are
5
a
K
independent
events.
C
3
20
12
Yes,
because
P (F
and
S )
=
P (F )
×
P (S )
14
Practice
4
1
1
a
P( LS
operation)
=
P ( LS
No
operation)
=
P ( LS )
=
4
9
The
i
P( K )
are
independent
events
the
claim
is
not
valid.
12
32
b
results
=
ii
P (C
K
)
=
b
1
32
50
LS
9
12
26
iii
P (C )
=
iv
P (K
C )
=
26
50
1
O
4
3
a
8
LS
9
A
D
1
7
5
10
LS
9
3
O
4
8
8
LS
9
5
12
b
i
P ( A)
=
ii
P (D
A)
=
It
12
30
is
evident
P ( LS
P (D
∪
A)
=
iv
P (D
9
15
a
b
25
the
tree
diagram
( LS
operation)
=
∩
A
D
∪
A)
c
=
Student’s
own
answer
with
justication.
22
30
4
from
operation)
5
22
iii
No
4
c
25
9
Answers
34 7
2
a
T
Mixed
Total
T
1
G
10
15
practice
a
44
25
O
99
40
G
30
10
40
40
25
65
99
O
A
45
Total
100
15
B
99
b
The
events
are
not
independent.
45
c
Student’s
own
O
answer.
99
40
3
a
i
P (E )
iii
P (L
v
P (E
vii
P (E′
=
0.4
ii
P (L
iv
P (E
E )
vi
P (E′
∩
viii
P (L)
=
=
0.8
39
100
99
A
E
∩
)
=
L
)
0.4
=
0.08
∩
L)
L)
=
A
0.32
=
0.24
15
B
b
No,
∩
L
)
because
=
0.36
the
conditional
0.56
probabilities
are
99
dierent.
45
O
99
15
40
Practice
5
100
99
B
1
A
a
14
B
A
W
23
10
99
8
3750
25
9900
55
40
=
b
c
d
66
99
99
6
12
1
2
10
2
a
b
c
13
i
P (A )
=
ii
P (W
)
=
10
P (A
W
)
d
3
5
1
e
3
g
f
13
51
51
iii
5
18
33
b
13
2
4
10
=
iv
P (W
A)
18
=
3
2
a
33
I
3
8
v
P (W
A
)
=
S
4
18
5
1
I
A
c
A
3
W
10
8
18
W
23
10
33
33
18
51
1
I
3
1
S
5
2
I
3
12
12
2
4
b
a
c
2
8
×
b
=
70
30
5
P (M )
does
therefore
not
not
equal
P (M
FT )
does
not
equal
P (M
FT
),
c
independent.
3
P( I
∴
15
S )
not
≠
P (I
S
)
independent
4
3
a
88%
4
b
a
S
0.2
12
c
No,
the
from
the
tree
diagram
the
second
branches
are
not
R
0.5
same.
0.8
4
S
14
14
a
b
70
30
S
c
Yes
they
are
independent
as
0.8
0.5
16
P (M
L)
=
64
=
30
P (M
L
)
R
=
120
0.2
S
b
Th
Th
34 8
Answers
2:
3:
=
0.1
P (R )
P (R
∩
S )
=
0.5
P (S )
=
0.1
=
0.5
P (S
R)
=
0.2
P (S
R
=
0.8
)
P (R
+
∩
S )
≠
P (R )
×
P (S )
0.4
∴
Not
independent
∴
Not
independent
5
a
Review
T
in
context
T
45
B
19
29
1
48
a
i
P (D )
200
11
B
11
1
9
=
=
ii
P (N )
=
2
40
1
22
iii
P (D
N )
iv
P (D
N
)
10
30
40
=
=
20
100
70
b
They
are
not
independent
from
Theorem
3.
29
30
b
c
d
Not
independent,
by
Th
3
2
a
T
48
70
6
7
35
=
0.98
a
B
0.01
H
0.02
B
4
4
T
2
T
0.05
0.99
4
8
b
B
10
6
i
ii
iii
14
0.95
T
14
14
b
4
6
iv
14
3
14
4
12
The
events
14
0.9503
Smoker
(A)
Non-smoker
(A
)
14
Cancer
B )
=
independent.
ix
(B )
1763
981
2744
73 237
124 019
197 256
75 000
125 000
200 000
4
4
P (H
0.9405
are
4
4
viii
14
c
+
vi
14
vii
0.0098
8
v
=
P (H
B
)
=
8
6
(B
)
4
P (B
H
)
=
P (B
H
8
7
)
=
6
a
1763
Should
have
Should not have
a
Total
P (B
A)
=
75000
dress
code
dress code
981
b
Middle
school
15
30
P (B
A
)
=
45
124
High
school
Total
40
80
120
55
110
165
c
d
They
are
not
Students’
019
independent
own
from
Theorem
2
answer
10
5
b
Theorem
a
15
and
0.69
b
P(A)
=
0.01
c
P(B )
=
code
)
=
P (MS )
×
P (dress
The
events
are
not
independent.
code)
55
×
165
Theorem
P (MS
=
330
Dress
45
=
165
A)
2
b
P (MS
P(B
165
3
DC )
=
P (MS
DC
)
30
15
=
55
110
Therefore
they
both
work.
Answers
34 9
Index
cosine
A
r ule
183–4,
application
A4
paper
additive
203–4
systems
ambiguous
amplitude
case
of
a
formula
3
for
areas
the
sine
sinusoidal
Angle-Side-Angle
angular
CPCTC
r ule
Congr uence
displacement
243–5
function
192,
cuboids
168–176
135–7
(ASA/AAS)
269,
277
D
127
data
of
a
parallelogram
76–7
188
dispersion
area
of
a
segment
area
of
a
triangle
area
of
a
triangle
82
184–9
normal
formula
185,
sine
of
an
and
sine
equilateral
cosine
and
cosine
r ules
triangle
189–94,
r ules,
distribution
problems,
of
a
applying
logarithm
series
146,
decimal
exponents
decimal
number
218,
226,
170,
229,
211,
cur ve
binar y
111
3–17
86
180
221–2,
229
16–18
functions
functions
space
formula
dividing
space
higher
213
229
the
168–75
172,
180
175–9
diagonals
nding
211,
221–2,
167–8
distance
face
numbers
9–10
231
describing
number
171,
213,
dimensions
bases,
5,
32
164
B
logarithm
105–6
111
151,
rational
a
98,
system
functions
logarithmic
of
94
114–116
195–9
dilation
asymptotes
and
188
diagonals
arithmetic
growth
199
decomposing
argument
86–7,
199
decay
area
77–87
188
mean
bell
199
273
183–4
area
base
189–94
195–9
170
midpoint
dimensional
planes
169,
section
formula
176,
space
180
179–80
180
C
change
of
circles
base
formula
111
space
124–5
area
of
radian
a
dispersion
segment
measure
circle
column
diagonals
188
and
125–34,
vectors
distance
the
unit
dot
178,
180
170–1,
180
77–87
formula
product
68,
172,
180
73
142
56–7,
73
E
common
ratio,
components,
composition
of
of
compounded
conditional
153–4
vectors
probability
decisions
representing
with
notation
1
Theorem
2
302,
312
Theorem
3
302,
312
is
equations
exponential
functions
from
302,
probability
301,
312
309–12
decimal
probabilities
264–5,
for
273,
305–9
exponents
98,
exponents
probability?
98–102,
301–4
face
diagonals
four
dimensions
170
277
congr uence
269,
179–80
277
163
277
congr uent
triangles
exponents
97–102,
106
265–73
of
a
sinusoidal
266
function
135–7
270
function
using
congr uence
to
prove
operations
results
21–2,
35–6
273–6
decomposing
similarity
to
prove
other
results
Index
functions
32–3
276–7
doing
3 5 0
22
other
functions
using
functions
105–6
97–8,
F
frequency
sign
logarithmic
205
fractional
fractional
establishing
shapes
to
121
203
exponents
312
fractals
CPCTC
exponential
inverses
109,
202,
302
conditional
conditions
axiom
273
exponential
299–300
conditional
Theorem
congr uence
Euclid
29–32
117
probabilities
mathematical
what
59
functions
interest
conditional
making
r
and
undoing
50–1
106
203–5
exponential
function
inverse
functions
logarithmic
modelling
and
sinusoidal
of
logarithmic
dilation
44–50,
51
202–12,
compositions
38–44,
37–8,
functions
reversible
types
42,
functions
functions
rational
202–5
22–9
functions
using
one-to-one
onto
functions
operations
from
213
of
29–32
51
51
of
reection
211,
134–41,
44–50
translation
142
logarithms
37–43
base
111,
change
of
211,
111,
functions
206–7,
functions
203–5
209,
213
205–10
logarithmic
210–12,
213
213
108–9,
argument
functions
213
functions
changes
213
logarithmic
logarithmic
transformations
irreversible
to
logarithmic
proper ties
216–31
functions
of
202–3,
213
exponential
graphs
functions
functions
211,
109–14
121
121
of
base
formula
111,
121,
257,
261
G
creating
Gallivan,
Gauss’s
Britney
method
generalizations
proving
series
162,
grad/gradian
of
of
164
of
logarithms
249–54,
logarithms
117–21,
r ule
proving
generalizations
using
125
zero
206–7,
209,
251,
r ule
r ule
261
251,
251,
251,
255–8
261
261
logarithms
r ule
261
121
261
r ule
unitar y
164
251,
product
quotient
logarithmic
functions
graphs
154,
geometric
series
graphs
146,
255–8
250–4
natural
power
250–4
generalizations
geometric
innite
laws
204
149
generalizations
to
solve
equations
114–17,
258–61
261
213
sinusoidal
M
functions
growth
and
134–41
decay
problems
magnitude,
114–116
mapping
Mayan
vectors
diagram
number
60–61
49
system
11–12
H
mean
hexadecimal
hieroglyphic
midpoint,
17
compositions
3–14
asymptote
equation
226,
218,
176,
180
multiple
horizontal
dilations
horizontal
line
horizontal
shift
horizontal
translation
horizontal
translations
136,
of
multiplication
51
29–32
transformations,
order
254
37,
of
functions
231
229
test
nding
modelling
number
system,
horizontal
82
229
tables,
dierent
bases
16
142
of
a
sinusoidal
function
135–7
N
hyperbolas
254
Napier,
219
John
non-linear
118,
251,
inequalities
algebraic
solutions
256
282–3,
to
297
non-linear
inequalities
291–6
I
indices
innite
97–8,
inter net
inverse
102–5,
geometric
search
ranking
functions
exponential
inverse
isosceles
of
a
106
series
42,
rational
164
255
44–50,
functions
triangle
162,
51
non-linear
inequalities
in
one
variable
non-linear
inequalities
in
two
variables
function
274
a
quadratic
inequality
algebraically
solving
a
quadratic
inequality
graphically
solving
non-linear
graphically
230–1
288–91
solving
205
theorem
283–8
normal
number
distribution
systems
binar y
in
in
two
variables
297
86–7,
94
16–18
dierent
bases
3–12
K
performing
Koch
snowake
operations
12–16
163
O
L
one-to-one
onto
Leibniz,
Gottfried
of
symmetr y
37–8,
38–44,
51
51
6
ordered
line
functions
functions
pairs
41
42
Index
297
297
2–3
numbers
numbers
equalities
289,
292,
284,
3 51
scalars
P
61,
scalar
paper
folding
formula
parallelograms,
area
of
per pendicular
height
per pendicular
vectors
phase
place
shift
135,
value
planes
169,
versus
57,
numbers
positive
real
product
121
111,
r ule
101,
logarithms
106
in
real-life
250–1
order
graphs
of
logarithmic
functions
157
206
theorem
170,
Pythagorean
identity
Pythagorean
triples
Pythagorean
quadr uples
192
(SAS)
(SSS)
269,
269,
277
277
199
shift
134–41,
136,
transformations
135,
136,
142
142
141,
142
142
132
167–8
face
171
164
159–63
195–9
shift
distance
171
164
162,
163
functions
describing
239
73
189–94
SOHCAHTOA
space
164
Congr uence
189,
of
phase
65,
155–9
Triangle
sinusoidal
151,
Congr uence
183–4,
formula
250–1
logarithms
scheme
Pythagoras’
series
r ule
vector
154,
series
fractions
horizontal
of
of
and
application
99,
255
proper ties
series
sine
146,
geometric
Sier pinski
121
numbers
of
91–4
146,
series
Side-Side-Side
73
a
164
series
Side-Angle-Side
255
proving
pyramid
innite
population
positive
proving
geometric
4
of
177–8
146
145–55,
arithmetic
142
117,
vectors
r ule
73
growth
equation
power
series
69,
formula
sequences
79–80
population
sample
188
180
populations
position
section
185
136,
system
204
73
multiplication
space
diagonals
nding
higher
168–75
formula
the
172,
180
170
midpoint
dimensional
planes
169,
section
formula
176,
180
space
179–80
180
Q
quadrants
quotient
128–9
r ule
proving
of
space
logarithms
250–1
diagonals
special
255
triangles
standard
178,
170,
dierent
180
132
deviation
calculating
180
171,
from
76–7,
94–5
summar y
formulae
for
statistics
dierent
89–90
pur poses
R
sample
radians
125–34,
angles
ultimate
142
sum
127
radian
radicals
measures
97–8,
versus
of
a
population
measure
sequence
of
91–4
dispersion
77–87
147
126
102–5,
106
T
radicands
Ratio
104,
identity
rational
106
239
functions
dilation
inverse
221–2,
of
a
transforming
translation
exponential
logarithmic
230–1
rational
domain
rational
and
range
functions
226
224–9
229
217,
hyperbola
219
219
vectors
tree
logarithmic
reversible
Right
and
functions
211,
irreversible
213
changes
Angle-Hypotenuse-Side
(RHS)
269,
trial
44
Congr uence
277
211–12
210–13
224–9
functions
141,
142
translation
rational
reection
functions
functions
functions
sinusoidal
logarithmic
functions
rectangular
198
transformation
229
function,
reciprocal
231
rational
function
parent
theodolite
216–24,
63,
diagrams
and
211,
213
229
73
301–11
improvement
triangles
area
functions
functions
110–111
183–4
184–9
area
of
a
area
of
an
triangle
formula
185,
199
equilateral
triangle
188
S
CPCTC
samples
establishing
79
sample
versus
3 5 2
273,
population
Index
91–4
277
congr uent
triangles
265–73
87–91
triangular
numbers
trigonometr y
ambiguous
cosine
case
r ule
sine
239,
183–4,
with
per pendicular
position
189–99
identity
identity
r ule
operations
242–6
183–4,
Pythagorean
ratio
147
233–4
239,
vectors
resultants
246
scalar
246
63,
189–99
equations
trigonometric
identities
234–8
235,
238–42,
246
63,
addition
vector
geometr y
zero
vectors
diagrams
ver tical
62–70
69,
73
73
73
vector
Venn
57,
multiplication
translations
trigonometric
vectors
vectors
law
57,
of
a
vector
65,
73
73
63–64
70–2
73
305–310
asymptote
218,
231
U
equation
unbiased
unit
estimate
circle
unitar y
volume
92
226,
229
diagonal
171
126–133
r ule
of
logarithms
250–1
X
x
y
plane
169
V
vectors
54–60
Z
column
dot
vectors
product
equal
68,
56–7,
73
components
magnitude
73
60–2,
59,
73
zero
r ule
zero
vectors
of
logarithms
57,
250–1
73
73
Index
3 53
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