ASVAB Math Knowledge Practice Test 382472 Results

Your Results Global Average
Questions 5 5
Correct 0 2.64
Score 0% 53%

Review

1

If side a = 3, side b = 9, what is the length of the hypotenuse of this right triangle?

63% Answer Correctly
\( \sqrt{73} \)
\( \sqrt{90} \)
\( \sqrt{82} \)
\( \sqrt{40} \)

Solution

According to the Pythagorean theorem, the hypotenuse squared is equal to the sum of the two perpendicular sides squared:

c2 = a2 + b2
c2 = 32 + 92
c2 = 9 + 81
c2 = 90
c = \( \sqrt{90} \)


2

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

74% Answer Correctly

factoring

squaring

deconstructing

normalizing


Solution

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


3

Solve 3b + 8b = -4b + 8x + 3 for b in terms of x.

34% Answer Correctly
4\(\frac{1}{2}\)x + 1
6x + 8
x + \(\frac{3}{7}\)
3\(\frac{1}{2}\)x + \(\frac{1}{2}\)

Solution

To solve this equation, isolate the variable for which you are solving (b) on one side of the equation and put everything else on the other side.

3b + 8x = -4b + 8x + 3
3b = -4b + 8x + 3 - 8x
3b + 4b = 8x + 3 - 8x
7b = + 3
b = \( \frac{ + 3}{7} \)
b = \( \frac{}{7} \) + \( \frac{3}{7} \)
b = x + \(\frac{3}{7}\)


4

Solve for b:
b2 + 2b - 9 = 5b - 5

48% Answer Correctly
-5 or -8
3 or 2
-7 or -7
-1 or 4

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:

b2 + 2b - 9 = 5b - 5
b2 + 2b - 9 + 5 = 5b
b2 + 2b - 5b - 4 = 0
b2 - 3b - 4 = 0

Next, factor the quadratic equation:

b2 - 3b - 4 = 0
(b + 1)(b - 4) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (b + 1) or (b - 4) must equal zero:

If (b + 1) = 0, b must equal -1
If (b - 4) = 0, b must equal 4

So the solution is that b = -1 or 4


5

Find the value of b:
8b + z = 8
-6b + 8z = 5

42% Answer Correctly
\(\frac{59}{70}\)
\(\frac{5}{18}\)
-\(\frac{1}{2}\)

Solution

You need to find the value of b so solve the first equation in terms of z:

8b + z = 8
z = 8 - 8b

then substitute the result (8 - 8b) into the second equation:

-6b + 8(8 - 8b) = 5
-6b + (8 x 8) + (8 x -8b) = 5
-6b + 64 - 64b = 5
-6b - 64b = 5 - 64
-70b = -59
b = \( \frac{-59}{-70} \)
b = \(\frac{59}{70}\)