ASVAB Math Knowledge Practice Test 738878 Results

Your Results Global Average
Questions 5 5
Correct 0 3.76
Score 0% 75%

Review

1

What is 6a9 - 8a9?

73% Answer Correctly
-2a9
-2a18
48a18
-2

Solution

To combine like terms, add or subtract the coefficients (the numbers that come before the variables) of terms that have the same variable raised to the same exponent.

6a9 - 8a9 = -2a9


2

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

76% Answer Correctly

equation

expression

formula

problem


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.


3

If the area of this square is 64, what is the length of one of the diagonals?

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

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 = \( \sqrt{a} \) = \( \sqrt{64} \) = 8

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 = 82 + 82
c2 = 128
c = \( \sqrt{128} \) = \( \sqrt{64 x 2} \) = \( \sqrt{64} \) \( \sqrt{2} \)
c = 8\( \sqrt{2} \)


4

What is the area of a circle with a radius of 5?

69% Answer Correctly
25π

Solution

The formula for area is πr2:

a = πr2
a = π(52)
a = 25π


5

Which of the following is not a part of PEMDAS, the acronym for math order of operations?

88% Answer Correctly

addition

exponents

pairs

division


Solution

When solving an equation with two variables, 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.)