ASVAB Math Knowledge Practice Test 585626 Results

Your Results Global Average
Questions 5 5
Correct 0 3.19
Score 0% 64%

Review

1

Solve for a:
-2a + 1 > -8 - 3a

54% Answer Correctly
a > -\(\frac{4}{7}\)
a > -1
a > \(\frac{1}{3}\)
a > -9

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.

-2a + 1 > -8 - 3a
-2a > -8 - 3a - 1
-2a + 3a > -8 - 1
a > -9


2

Solve for x:
x2 + 5x - 6 = 0

57% Answer Correctly
4 or 2
5 or -9
1 or -6
-6 or -9

Solution

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

x2 + 5x - 6 = 0
(x - 1)(x + 6) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (x - 1) or (x + 6) must equal zero:

If (x - 1) = 0, x must equal 1
If (x + 6) = 0, x must equal -6

So the solution is that x = 1 or -6


3

Breaking apart a quadratic expression into a pair of binomials is called:

74% Answer Correctly

normalizing

deconstructing

squaring

factoring


Solution

To factor a quadratic expression, apply the FOIL (First, Outside, Inside, Last) method in reverse.


4

If a = c = 6, b = d = 1, and the blue angle = 55°, what is the area of this parallelogram?

65% Answer Correctly
6
36
3
21

Solution

The area of a parallelogram is equal to its length x width:

a = l x w
a = a x b
a = 6 x 1
a = 6


5

If the area of this square is 49, what is the length of one of the diagonals?

68% Answer Correctly
7\( \sqrt{2} \)
8\( \sqrt{2} \)
\( \sqrt{2} \)
2\( \sqrt{2} \)

Solution

To find the diagonal we need to know the length of one of the square's sides. We know the area and the area of a square is the length of one side squared:

a = s2

so the length of one side of the square is:

s = \( \sqrt{a} \) = \( \sqrt{49} \) = 7

The Pythagorean theorem defines the square of the hypotenuse (diagonal) of a triangle with a right angle as the sum of the squares of the other two sides:

c2 = a2 + b2
c2 = 72 + 72
c2 = 98
c = \( \sqrt{98} \) = \( \sqrt{49 x 2} \) = \( \sqrt{49} \) \( \sqrt{2} \)
c = 7\( \sqrt{2} \)