ASVAB Math Knowledge Practice Test 176377 Results

Your Results Global Average
Questions 5 5
Correct 0 3.44
Score 0% 69%

Review

1

Solve for y:
y2 + 9y + 39 = -4y - 3

48% Answer Correctly
6 or 4
8 or -3
-6 or -7
6 or -2

Solution

The first step to solve a quadratic expression that's not set to zero is to solve the equation so that it is set to zero:

y2 + 9y + 39 = -4y - 3
y2 + 9y + 39 + 3 = -4y
y2 + 9y + 4y + 42 = 0
y2 + 13y + 42 = 0

Next, factor the quadratic equation:

y2 + 13y + 42 = 0
(y + 6)(y + 7) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (y + 6) or (y + 7) must equal zero:

If (y + 6) = 0, y must equal -6
If (y + 7) = 0, y must equal -7

So the solution is that y = -6 or -7


2

If c = 2 and z = 6, what is the value of 4c(c - z)?

68% Answer Correctly
96
49
-32
-60

Solution

To solve this equation, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.)

4c(c - z)
4(2)(2 - 6)
4(2)(-4)
(8)(-4)
-32


3

A(n) __________ is two expressions separated by an equal sign.

77% Answer Correctly

formula

problem

equation

expression


Solution

An equation is two expressions separated by an equal sign. The key to solving equations is to repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the equal sign and the answer on the other.


4

If the base of this triangle is 7 and the height is 9, what is the area?

59% Answer Correctly
30
21
24\(\frac{1}{2}\)
31\(\frac{1}{2}\)

Solution

The area of a triangle is equal to ½ base x height:

a = ½bh
a = ½ x 7 x 9 = \( \frac{63}{2} \) = 31\(\frac{1}{2}\)


5

A right angle measures:

91% Answer Correctly

180°

45°

90°

360°


Solution

A right angle measures 90 degrees and is the intersection of two perpendicular lines. In diagrams, a right angle is indicated by a small box completing a square with the perpendicular lines.