# Math Problems ```ID: 8768cbea
Name: Jed Albarico
1. What is the graph of 𝒙𝟐 − 𝒚𝟐 = 𝟐𝟑? What are its lines of symmetry? What
are the domain and range?
Given:
𝑥 2 − 𝑦 2 = 23?
Required:
a. Graph
b. Lines of symmetry
c. Domain and range
Solution:
First, identify what conic section is expressed by the equation by checking the
standard equations of each. Hint: inspect the x and y values

Circle:
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2
sum of squares

Ellipse:
(𝑥−ℎ)2
sum of squares

Parabola:
(𝑦 − 𝑘)2 = 4𝑝(𝑥 − ℎ)
a square and a linear

Hyperbola:
(𝑥−ℎ)2
difference of squares
𝑎2
𝑎2
+
−
(𝑦−𝑘)2
𝑏2
(𝑦−𝑘)2
𝑏2
=1
=1
Since the given equation is a difference of squares, ∴ it is a hyperbola.
Next, transform the given equation to the standard equation.
𝑥 2 − 𝑦 2 = 23
Given
𝑥 2 −𝑦 2
Divide both sides by 23
23
𝑥2
23
= 23
𝑦2
Simplify, separate terms
− 23 = 1
23
(𝑥−0)2
(√23)
2
−
(𝑦−0)2
(√23)
2
=1
Rewrite based on the standard equation
Having the original equation transformed into the standard equation of a
hyperbola, we can now identify the values of the variables which will be needed
for further calculations.
ℎ = 0;
𝑘 = 0;
𝑎 = √23 = 4.80;
𝑏 = √23 = 4.80
a. Graph
First, locate the center of the hyperbola through (ℎ, 𝑘)
Center = (0,0)
Second, make a rectangle using the center point, and the values of 𝑎 = 4.80 and
𝑏 = 4.80 as guide to plotting in the graph later. Move 𝑎 and 𝑏 units depending on
what variable is found at its numerator. (In this problem, it will not matter since a
and b are equal).
First, from the center, move 𝑎 units (4.80) to the right, then move 𝑏 units
(4.80) up and down. Mark these two points {(4.80,4.80),(4.80,-4.80)}.
Then, from the center, move 𝑎 units (4.80) to the left, then move 𝑏 units
(4.80) up and down again. Mark these two points {(-4.80,4.80),(-4.80,4.80)}.as well.
Connect these four points to make a rectangle. Use broken lines or pencil
since this rectangle is not really part of the graph.
Third, connect the diagonals of this rectangle and extend it as a line. These two
lines are the asymptotes of the hyperbola where the graph will try to approach but
will not touch. Draw the asymptotes using a pencil or broken line as well.
Fourth, identify where will the hyperbola will face through the transverse and
conjugate axes. The transverse and conjugate axes are lines parallel to the x- and
y- axes and will pass through the center point. Tip: comparing the x and y
variables, the positive value is the transverse axis and that is where the hyperbola
will face.
The transverse axis is parallel to x-axis since x is positive, while the
conjugate axis is parallel to y-axis.
Passing through the center point (0,0) are the x-axis and y-axis themselves.
Fifth. Locate the vertices. Vertices are two points where the transverse axis and
the rectangle will meet.
Sixth, construct the graph using the vertex, transverse axis, and the asymptotes.
Knowing where the hyperbola faces, start at the vertex, then make a curve line
approaching the asymptote but not touch it. Draw arrowheads at the end to
indicate the graph of hyperbola extends.
b. Lines of symmetry
In the lines of symmetry, the graphs on both sides are mirror images of each other.
Hyperbolas are symmetric in their transverse and conjugate axes.
The lines of symmetry are the x- and y- axes.
c. Domain and Range
Determine the Domain and Range through the graph.
The domain of the equation is the set of all possible values of x. Looking at the
graph, there are no x-values within the rectangle, and that is within (-4.80,4.80).
So the domain is the set of values of x such that x is greater than or equal to 4.80
and less than or equal to -4.80, or:
𝔻 = {𝑥|𝑥 ≥ 4.80 ∪ 𝑥 ≤ −4.80}
The range of the equation is the set of all possible values of y. Looking at the
graph, all y-values will have values. So the range is the set of values of y such that
y is a real number, or
ℝ = {𝑦|𝑦 ∈ ℝ}
2. Divide the polynomials 𝒙𝟐 + 𝟐𝒙𝟐 𝒚 − 𝟐𝒙𝒚 + 𝟐𝒙𝒚𝟐 − 𝟑𝒚𝟐 and 𝒙 + 𝒚 and explain
Given:
𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2 divided by 𝑥 + 𝑦
Required:
Quotient
Solution:
Dividing polynomials is like dividing numbers. Usually, the long method is
employed to get the quotient of the two polynomials.
𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
Write the expression.
𝒙 + 𝑦√𝒙𝟐 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
Look at the first expression in both
dividend and divisor. Divide them
first in order to get the multiplier.
𝑥2
=𝒙
𝑥
Multiply this multiplier to the
divisor to get the product.
(𝑥 + 𝑦) ∗ 𝒙 = 𝒙𝟐 + 𝒙𝒚
𝒙
𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
𝒙𝟐
+ 𝒙𝒚
−3𝑥𝑦
Going back to the long method,
align the product under the same
term of the dividend, then do
subtraction. Write the multiplier at
the quotient.
𝑥
𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
𝑥2
+ 𝑥𝑦
2
2𝑥 𝑦 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
𝒙
𝒙 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
𝑥2
+ 𝑥𝑦
𝟐
𝟐𝒙 𝒚 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
Bring down or copy other terms of
the dividend.
Look at the first expression in both
the difference and divisor. Divide
them first in order to get the
multiplier.
2𝑥 2 𝑦
= 𝟐𝒙𝒚
𝑥
Multiply this multiplier to the
divisor to get the product.
(𝑥 + 𝑦) ∗ 𝟐𝒙𝒚 = 𝟐𝒙𝟐 𝒚 + 𝟐𝒙𝒚𝟐
𝒙 + 𝟐𝒙𝒚
𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
𝒙𝟐
+ 𝒙𝒚
2
2𝑥 𝑦 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
𝟐𝒙𝟐 𝒚
+ 𝟐𝒙𝒚𝟐
𝟎
𝟎
𝑥 + 2𝑥𝑦
𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
𝑥2
+ 𝑥𝑦
2
2𝑥 𝑦 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
2𝑥 2 𝑦
+ 2𝑥𝑦 2
−3𝑥𝑦
− 3𝑦 2
Going back to the long method,
align the product under the same
term of the dividend, then do
the quotient.
Bring down or copy other terms of
the dividend.
𝑥 + 2𝑥𝑦
𝒙 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
𝑥2
+ 𝑥𝑦
2
2𝑥 𝑦 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
2𝑥 2 𝑦
+ 2𝑥𝑦 2
−𝟑𝒙𝒚
− 3𝑦 2
Look at the first expression in both
the difference and divisor. Divide
them first in order to get the
multiplier.
−3𝑥𝑦
= −3𝑦
𝑥
Multiply this multiplier to the
divisor to get the product.
(𝑥 + 𝑦) ∗ −𝟑𝒚 = −𝟑𝒙𝒚 − 𝟑𝒚𝟐
𝑥 + 2𝑥𝑦 − 𝟑𝒚
𝑥 + 𝑦√𝑥 2 + 2𝑥 2 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
𝑥2
+ 𝑥𝑦
2
2𝑥 𝑦 − 3𝑥𝑦 + 2𝑥𝑦 2 − 3𝑦 2
2𝑥 2 𝑦
+ 2𝑥𝑦 2
−3𝑥𝑦
− 3𝑦 2
−𝟑𝒙𝒚
− 𝟑𝒚𝟐
𝟎
𝟎
2
2
2
2
𝑥 + 2𝑥 𝑦 − 2𝑥𝑦 + 2𝑥𝑦 − 3𝑦
𝑥+𝑦
= 𝒙 + 𝟐𝒙𝒚 − 𝟑𝒚
Going back to the long method,
align the product under the same
term of the dividend, then do
the quotient.
If no other term remains and the
difference is zero, the quotient is
3. Suppose a number from 1 to 100 is selected at random. What is the probability
that a multiple of 4 or 5 is chosen? Explain each step.
Given:
Population: 1 to 100
Sample: multiple of 4 OR 5
Required:
Probability of picking multiple of 4 OR 5
Solution:
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑠
The formula of the probability is: 𝑃 = 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠
Based on the problem, the total number of all values in the population is
100 (total numbers from 1-100).
For the number of possible values, consider this Venn Diagram:
Multiples
of 4 and 5
Multiples
of 4 only
Multiples
of 5 only
Numbers 1-100
Picking multiples of 4 OR 5 means getting the multiples of 4 and multiples
of 5, which could be obtained by dividing those from 100.
To get total multiples of 4 from 1 to 100:
100
= 25
4
To get total multiples of 5 from 1 to 100:
100
= 20
5
Total multiples of 4 + Total multiples of 5:
25 + 20 = 𝟒𝟓
However, there are also multiples of both 4 and 5 which will be recounted
if not being careful. To account for the double count, we will find the
multiple of the least common multiple of 4 and 5.
Least common multiple:
4 ∗ 5 = 20
To get total multiples of 20 from 1 to 100:
100
=𝟓
20
There are 5 numbers which have overlapped since they are multiples of 4
and 5. We need to account for this by subtracting it to the total multiples
of 4 and multiples of 5. So, to get the multiples of 4 OR 5:
25 + 20 − 5 = 𝟒𝟎
Substituting to the probability formula:
𝑃=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑠
40
=
= 0.40 ≈ 40%
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 100
There is 40% chance of picking a multiple of 4 or 5 from 1 to 100.
4. Prove that 𝟏𝟐 + 𝟐𝟐 + 𝟑𝟐 + ⋯ + 𝒏𝟐 =
Given:
12 + 22 + 32 + ⋯ + 𝑛2 =
Required:
Proof
𝒏(𝒏+𝟏)(𝟐𝒏+𝟏)
𝟔
. Explain each step.
𝑛(𝑛+1)(2𝑛+1)
6
Solution:
Eq 1
Eq 2
Eq 3
Eq 4
Eq 5
Eq 6
Eq 7
𝑎3 − 𝑏 3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 )
Cubic difference identity
𝑛3 − (𝑛 − 1)3 = [𝑛 − (𝑛 − 1)][𝑛2 + 𝑛 ∗ (𝑛 − 1)
+ (𝑛 − 1)2 ]
Substituting 𝑎 → 𝑛, 𝑏 → 𝑛 − 1
where 𝑛 is any natural number to
Eqn 1
𝑛3 − (𝑛 − 1)3 = [𝑛 − 𝑛 + 1][𝑛2 + 𝑛2 − 𝑛 + 𝑛2 − 2𝑛 + 1]
Applying distributive properties
𝑛3 − (𝑛 − 1)3 = [3𝑛2 − 3𝑛 + 1]
Combining like terms
𝑛3 − (𝑛 − 1)3 = 3𝑛2 − 3𝑛 + 1
Simplifying right side of the
equation
(𝑛 − 1)3 − [(𝑛 − 1) − 1]3 = 3(𝑛 − 1)2 − 3(𝑛 − 1) + 1
Substituting 𝑛 → 𝑛 − 1 to Eqn 2
(𝑛 − 1)3 − [𝑛 − 2]3 = 3(𝑛 − 1)2 − 3(𝑛 − 1) + 1
Combining like terms
(𝑛 − 2)3 − [(𝑛 − 2) − 1]3 = 3(𝑛 − 2)2 − 3(𝑛 − 2) + 1
Substituting 𝑛 → 𝑛 − 2 to Eqn 2
(𝑛 − 2)3 − [𝑛 − 3]3 = 3(𝑛 − 2)2 − 3(𝑛 − 2) + 1
Combining like terms
23 − (2 − 1)3 = 3(2)2 − 3(2) + 1
Substituting 𝑛 → 2 to Eq 2
23 − 13 = 3(2)2 − 3(2) + 1
Combining like terms
13 − (1 − 1)3 = 3(1)2 − 3(1) + 1
Substituting 𝑛 → 1 to Eq 2
13 − 03 = 3(1)2 − 3(1) + 1
Combining like terms
𝑛3 − (𝑛 − 1)3 = 3𝑛2 − 3𝑛 + 1
+
(𝑛 − 1)3 − [𝑛 − 2]3 = 3(𝑛 − 1)2 − 3(𝑛 − 1) + 1
+
(𝑛 − 2)3 − [𝑛 − 3]3 = 3(𝑛 − 2)2 − 3(𝑛 − 2) + 1
+
…
Adding Eq 2 to 6 for all possible
values of 𝑛
+
3
3
2 − 1 = 3(2)2 − 3(2) + 1
+
13 − 03 = 3(1)2 − 3(1) + 1
Eq 7.1
Eq 7.2
[𝑛3 − (𝑛 − 1)3 ]+[ (𝑛 − 1)3 − [𝑛 − 2]3 ]+[ (𝑛 − 2)3 −
[𝑛 − 3]3 ]+…+[ 23 − 13 ]+[ 13 − 03 ]
Getting the sum of the left side of
the equation, combining like
terms
𝑛3 − 0 ⇒ 𝑛3
Simplifying left side of the
equation
3𝑛2 + 3(𝑛 − 1)2 + 3(𝑛 − 2)2 + ⋯ + 3(2)2 + 3(1)2
Getting the sum of the first terms
of the right side of the equation
3[𝑛2 + (𝑛 − 1)2 + (𝑛 − 2)2 + ⋯ + (2)2 + (1)2 ]
Factor out common multiple
𝑛
3 ∑ 𝑛2
𝑛=1
Rewriting to summation notation
−3𝑛 − 3(𝑛 − 1) − 3(𝑛 − 2) − ⋯ − 3(2) − 3(1)
Getting the sum of the second
terms of the right side of the
equation
−3[𝑛 + (𝑛 + 1) + (𝑛 + 2) + ⋯ + 2 + 1]
Factor out common multiple
𝑛
Eq 7.3
Rewriting to summation notation
−3 ∑ 𝑛
𝑛=1
1 + 1 + 1 + ⋯+ 1 + 1 ⟹ 𝑛
Getting the sum of the third terms
of the right side of the equation
Rewriting
𝑛
𝑛
3
2
𝑛 = 3∑𝑛 −3∑𝑛+𝑛
𝑛=1
𝑛
𝑛=1
𝑛3 = 3 ∑ 𝑛2 − 3
𝑛=1
𝑛
3 ∑ 𝑛2 = 𝑛3 +
𝑛=1
𝑛
3 ∑ 𝑛2 =
𝑛=1
𝑛
3 ∑ 𝑛2 =
𝑛=1
𝑛(𝑛 + 1)
+𝑛
2
3𝑛(𝑛 + 1)
−𝑛
2
Simplifying Eq 7 with Eq 7.1, 7.2,
and 7.3
Summation of natural numbers
identity
Transpose
2𝑛3 + 3𝑛(𝑛 + 1) − 2𝑛
2
Simplifying right side of the
equation to a single fraction
2𝑛3 + 3𝑛2 + 3𝑛 − 2𝑛
2
Distributive property
𝑛
2𝑛3 + 3𝑛2 + 𝑛
3∑𝑛 =
2
2
𝑛=1
𝑛
3 ∑ 𝑛2 =
𝑛=1
𝑛
3 ∑ 𝑛2 =
𝑛=1
𝑛
∑ 𝑛2 =
𝑛=1
Combine like terms
𝑛(2𝑛2 + 3𝑛 + 1)
2
Factor out common multiple
𝑛(𝑛 + 1)(2𝑛 + 1)
2
Factor binomials
𝑛(𝑛 + 1)(2𝑛 + 1)
6
Dividing both sides by 3 ∎
5. Are the matrices A and B inverses? Explain each step.
𝟏
𝑨 = [−𝟐
−𝟏
𝟎
𝟏
𝟏
𝟐
𝟏]
𝟐
−𝟏
𝑩 = [−𝟑
𝟏
Given:
1
𝐴 = [−2
−1
Required:
Check if inverses
−𝟐 𝟐
−𝟒 𝟓 ]
𝟏 −𝟏
0 2
−1 −2
1 1], 𝐵 = [−3 −4
1 2
1
1
2
5]
−1
Solution:
To check that matrices A and B are inverses, their product must be equal to the
identity matrix.
1 0
𝐴 &times; 𝐵 = 𝐼 = [0 1
0 0
0
0]
1
Write side by side the two matrices. A 3x3 matrix multiplied to another 3x3 matrix
will yield a product which is a 3x3 matrix also.
𝑎
1 0 2 −1 −2 2
𝐴 &times; 𝐵 = [−2 1 1] [−3 −4 5 ] = [𝑑
𝑔
−1 1 2 1
1 −1
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
Next, get the dot product of the first row of A and the first column in B by
multiplying each corresponding (i.e., first element multiplied to the first element)
and so on, then get the sum.
1 0 2 −1 −2 2
[−2 1 1] [−3 −4 5 ]
−1 1 2 1
1 −1
= (1)(−1) + (0)(−3) + (2)(1)
= −1 + 0 − 2 = 1
Fill the answer in the element of the product matrix where the factors meet.
𝑎
[𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
(First row and first column  𝑎)
1
[𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
Then, get the dot product of the second row of A and the first column in B, then
fill the answer where these two meet.
1 0 2 −1 −2 2
[−2 1 1] [−3 −4 5 ]
−1 1 2 1
1 −1
= (−2)(−1) + (1)(−3) + (1)(1)
=2−3+1=0
𝑎
[𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
(Second row and first column  𝑑)
1
[0
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
Repeat this for the third row of A and the first column of B.
1 0 2 −1 −2 2
[−2 1 1] [−3 −4 5 ]
−1 1 2 1
1 −1
= (−1)(−1) + (1)(−3) + (2)(1)
=1−3+2=0
𝑎
[𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
(Third row and first column  𝑔)
1 𝑏
[0 𝑒
0 ℎ
Moving on the second column of B and back at the first row of A,
𝑐
𝑓]
𝑖
1 0 2 −1 −2 2
[−2 1 1] [−3 −4 5 ]
−1 1 2 1
1 −1
= (1)(−2) + (0)(−4) + (2)(1)
= −2 + 0 + 2 = 0
𝑎
[𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
(First row and second column  𝑏)
1 0
[0 𝑒
0 ℎ
𝑐
𝑓]
𝑖
The second row of A again,
1 0 2 −1 −2 2
[−2 1 1] [−3 −4 5 ]
−1 1 2 1
1 −1
= (−2)(−2) + (1)(−4) + (1)(1)
=4−4+1=1
𝑎
[𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
(Second row and second column  𝑒)
1 0
[0 1
0 ℎ
𝑐
𝑓]
𝑖
And the third row of A.
1 0 2 −1 −2 2
[−2 1 1] [−3 −4 5 ]
−1 1 2 1
1 −1
= (−1)(−2) + (1)(−4) + (2)(1)
=2−4+2=0
𝑎
[𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
(Third row and second column  𝑏)
1 0
[0 1
0 0
𝑐
𝑓]
𝑖
Moving to the third column of B and back at the first row of A,
1 0 2 −1 −2 2
[−2 1 1] [−3 −4 5 ]
−1 1 2 1
1 −1
= (1)(2) + (0)(5) + (2)(−1)
=2+0−2=0
𝑎
[𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
(First row and third column  𝑐)
1 0
[0 1
0 0
0
𝑓]
𝑖
The second row of A again,
1 0 2 −1 −2 2
[−2 1 1] [−3 −4 5 ]
−1 1 2 1
1 −1
= (−2)(2) + (1)(5) + (1)(1)
= −4 + 5 + 1 = 0
𝑎
[𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
(Second row and third column  𝑓)
1 0
[0 1
0 0
0
0]
𝑖
And finally, the third row of A.
1 0 2 −1 −2 2
[−2 1 1] [−3 −4 5 ]
−1 1 2 1
1 −1
= (−1)(2) + (1)(5) + (2)(−1)
= −2 + 5 − 2 = 1
𝑎
[𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
(Third row and second column  𝑖)
1 0
[0 1
0 0
0
0]
1
Since the product of A and B is the identity matrix, then A and B are inverses of
each other.
1 0 2 −1 −2 2
1
𝐴 &times; 𝐵 = [−2 1 1] [−3 −4 5 ] = [0
−1 1 2 1
1 −1
0
0 0
1 0] = 𝐼
0 1
```