Unit Test — Cumulative exam Answers
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21
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Two numbers are 10 units away in different directions from their midpoint, m, on a number line. The product of the numbers is –99.
A
(m – 5)(m + 5) = 99B
(m – 10)(m + 10) = 99C
m2 – 25 = –99D
m2 – 100 = –9922
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Marisol grouped the terms and factored the GCF out of the groups of the polynomial 6x3 – 22x2 – 9x + 33. Her work is shown.Step 1: (6x3 – 22x2) – (9x + 33)Step 2: 2x2(3x – 11) – 3(3x + 11)
A
Marisol should realize that her work shows that the polynomial is prime.B
Marisol should go back and group the terms in Step 1 as (6x3 – 22x2) – (9x – 33).C
Marisol should go back and group the terms in Step 1 as (6x3 – 22x2) + (9x – 33).D
Marisol should refactor the expression in Step 2 as 2x2(3x + 11) – 3(3x + 11).23
What values of b satisfy 4(3b + 2)2 = 64?
A
b = and b = –2
B
b = 2 and b =
C
b = and b = 3D
b = 2 and b =
24
Which diagram represents the factors of m2 – 10m + 16 ?
A
A 2-column table with 2 rows. First column is labeled m with entries m squared, 4 m. Second column is labeled negative 6 with entries negative 6 m, 16. First row is labeled m with entries m squared, negative 6 m. Second row is labeled 4 with entries 4 m, 16.
B
A 2-column table with 2 rows. First column is labeled m with entries m squared, 2 m. Second column is labeled negative 8 with entries negative 8 m, 16. First row is labeled m with entries m squared, negative 8 m. Second row is labeled 2 with entries 2 m, 16.
C
A 2-column table with 2 rows. First column is labeled m with entries m squared, negative 6 m. Second column is labeled negative 4 with entries negative 4 m, 16. First row is labeled m with entries m squared, negative 4 m. Second row is labeled negative 6 with entries negative 6 m, 16.
D
A 2-column table with 2 rows. First column is labeled m with entries m squared, negative 8 m. Second column is labeled negative 2 with entries negative 2 m, 16. First row is labeled m with entries m squared, negative 2 m. Second row is labeled negative 8 with entries negative 8 m, 16.
25
On a coordinate plane, an absolute value graph has a vertex at (negative 4, negative 2).
A
g(x) = |x + 4| – 2B
g(x) = |x – 4| – 2C
g(x) = |x – 2| – 4D
g(x) = |x – 2| + 4Did you find these answers helpful?
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