ASVAB Math Knowledge Practice Test 109499 Results

Your Results Global Average
Questions 5 5
Correct 0 3.22
Score 0% 64%

Review

1

Solve for z:
-6z + 2 = \( \frac{z}{-3} \)

46% Answer Correctly
\(\frac{5}{44}\)
-\(\frac{12}{13}\)
\(\frac{6}{17}\)
1

Solution

To solve this equation, 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.

-6z + 2 = \( \frac{z}{-3} \)
-3 x (-6z + 2) = z
(-3 x -6z) + (-3 x 2) = z
18z - 6 = z
18z - 6 - z = 0
18z - z = 6
17z = 6
z = \( \frac{6}{17} \)
z = \(\frac{6}{17}\)


2

A coordinate grid is composed of which of the following?

88% Answer Correctly

y-axis

origin

all of these

x-axis


Solution

The coordinate grid is composed of a horizontal x-axis and a vertical y-axis. The center of the grid, where the x-axis and y-axis meet, is called the origin.


3

Solve for z:
z2 - 11z + 18 = 0

58% Answer Correctly
8 or -2
2 or 9
5 or 2
5 or -3

Solution

The first step to solve a quadratic equation that's set to zero is to factor the quadratic equation:

z2 - 11z + 18 = 0
(z - 2)(z - 9) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (z - 2) or (z - 9) must equal zero:

If (z - 2) = 0, z must equal 2
If (z - 9) = 0, z must equal 9

So the solution is that z = 2 or 9


4

If side x = 7cm, side y = 7cm, and side z = 10cm what is the perimeter of this triangle?

84% Answer Correctly
24cm
30cm
28cm
32cm

Solution

The perimeter of a triangle is the sum of the lengths of its sides:

p = x + y + z
p = 7cm + 7cm + 10cm = 24cm


5

Solve for b:
7b - 3 > \( \frac{b}{1} \)

44% Answer Correctly
b > \(\frac{8}{9}\)
b > -\(\frac{6}{19}\)
b > \(\frac{1}{2}\)
b > -1\(\frac{1}{5}\)

Solution

To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the > sign and the answer on the other.

7b - 3 > \( \frac{b}{1} \)
1 x (7b - 3) > b
(1 x 7b) + (1 x -3) > b
7b - 3 > b
7b - 3 - b > 0
7b - b > 3
6b > 3
b > \( \frac{3}{6} \)
b > \(\frac{1}{2}\)