ASVAB Math Knowledge Practice Test 12235 Results

Your Results Global Average
Questions 5 5
Correct 0 3.07
Score 0% 61%

Review

1

If side a = 6, side b = 4, what is the length of the hypotenuse of this right triangle?

64% Answer Correctly
\( \sqrt{45} \)
\( \sqrt{58} \)
\( \sqrt{52} \)
5

Solution

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

c2 = a2 + b2
c2 = 62 + 42
c2 = 36 + 16
c2 = 52
c = \( \sqrt{52} \)


2

Simplify (y + 6)(y - 2)

63% Answer Correctly
y2 - 8y + 12
y2 - 4y - 12
y2 + 4y - 12
y2 + 8y + 12

Solution

To multiply binomials, use the FOIL method. FOIL stands for First, Outside, Inside, Last and refers to the position of each term in the parentheses:

(y + 6)(y - 2)
(y x y) + (y x -2) + (6 x y) + (6 x -2)
y2 - 2y + 6y - 12
y2 + 4y - 12


3

If b = 6 and x = -5, what is the value of -2b(b - x)?

68% Answer Correctly
-140
-132
24
-630

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.)

-2b(b - x)
-2(6)(6 + 5)
-2(6)(11)
(-12)(11)
-132


4

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

76% Answer Correctly

formula

equation

problem

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.


5

Solve -7a + 6a = 8a + 7z + 1 for a in terms of z.

34% Answer Correctly
-2z + 3\(\frac{1}{2}\)
4z + 6
-\(\frac{3}{4}\)z + 1\(\frac{1}{8}\)
-\(\frac{1}{15}\)z - \(\frac{1}{15}\)

Solution

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

-7a + 6z = 8a + 7z + 1
-7a = 8a + 7z + 1 - 6z
-7a - 8a = 7z + 1 - 6z
-15a = z + 1
a = \( \frac{z + 1}{-15} \)
a = \( \frac{z}{-15} \) + \( \frac{1}{-15} \)
a = -\(\frac{1}{15}\)z - \(\frac{1}{15}\)