MATH 150
Quiz Key #1
9/8-10/2015
(1) In the blank after each equation write the mathematical property from the
list that was used.
Properties:
Commutative: (a) a + b = b + a. (b) ab = ba.
Associative: (a) ( a + b) + c = a + (b + c) . (b) ( ab) c = a (bc) .
Distributive: a (b + c) = ab + ac.
Additive Identity: a + 0 = 0 + a = a.
Multiplicative Identity: a · 1 = 1 · a = a.
Additive Inverse: a + (− a) = 0.
Multiplicative Inverse: If a 6= 0, then a
1
a
= 1.
Subtraction: a − b = a + (−b) . Division: If b 6= 0, thena ÷ b = a 1b .
(a) 7 − 13 = 7 + (−
13) . Subtraction
−1 = 1. Multiplicative Inverse
(b) −127 127
√
√
√
(c) 13 y 2x3 − 7x = 26x3 y − 91x y. Distributive
√
√
√
√ (d) 9x2 − 5 x + 7 x = 9x2 + −5 x + 7 x . Associative
√
√
(e) 10 x 8x2 − 5x + 3 = 8x2 − 5x + 3 10 x. Commutative
(2) Circle all the numbers below that are rational numbers
√
√
9π −13 −715 −13i 1.279 5 7 −11.529
36 − 6
Solution: Rational numbers are numbers that can be written as fractions. This
includes all integers, fractions, terminating decimals, and repeating decimals.
Therefore, the following numbers from the above list are rational:
√
−13 −715 1.279 −11.529
36 − 6
The others are not rational numbers:
9π
√
−13i 5 7
1
(3) Simplify the following expression:
−1245 · 5 + 100 · 444 · 347
−44 .
1
30 1611 (−36)22 + 12
Solution: Before beginning, recall that one should break things down to prime
factorizations before simplifying power expressions. Also, following order of
operations, simplify the numerator as much as possible and the denominator
as much as possible before attempting to cancel things out. Finally, remember
that when raising a negative number to an even power, the result is always
positive.
45
44
− 22 · 3 · 5 + 22 · 52 · 22 · 347
=
−44 11
22
2 · 3 · 5 (24 ) (−22 · 33 ) + 221·3
=
−290 · 345 · 5 + 22 · 52 · 288 · 347
h
i
44
2 · 3 · 5 244 · 244 · 344 + (22 · 3)
−290 · 345 · 5 + 290 · 347 · 52
2 · 3 · 5 (288 · 344 + 288 · 344 )
290 · 345 · 5 −1 + 32 · 5
=
289 · 345 · 5 (1 + 1)
=
=
290 · 345 · 5 (44)
= 44.
290 · 345 · 5
2
(4) Fully simplify the following expression and rationalize the denominator
if necessary.
√
13 x6 z4
p
.
z−2 3 −8y3 ( x + 3)
Solution:
q
2
( x3 )
q
2
· ( z2 )
13
q
=
p
√
3
z−2 (−2)3 · 3 y3 · 3 x + 3
13 x3 · z2 −13 x3 z2 · z2
√
√
= −2
=
.
z (−2y) 3 x + 3
2y 3 x + 3
Now,
that no matter what real number z is, z2 is not negative. Therefore,
2 note
z = z2 . Therefore, we can continue to simplify as follows:
−13 x3 z2 · z2
−13 x3 z4 ( x + 3)2/3
√
=
·
2y 3 x + 3
2y ( x + 3)1/3 ( x + 3)2/3
3 4 q
3 4
2/3
x z 3 ( x + 3)2
−
13
−13 x z ( x + 3)
=
.
=
2y ( x + 3)
2y ( x + 3)3/3
3