ASVAB Math Knowledge Practice Test 606659 Results

Your Results Global Average
Questions 5 5
Correct 0 3.55
Score 0% 71%

Review

1

What is 4a + 4a?

81% Answer Correctly
0
8a
a2
16a2

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.

4a + 4a = 8a


2

If a = c = 2, b = d = 7, and the blue angle = 77°, what is the area of this parallelogram?

65% Answer Correctly
36
14
3
24

Solution

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

a = l x w
a = a x b
a = 2 x 7
a = 14


3

Simplify 7a x 6b.

86% Answer Correctly
42a2b2
42\( \frac{a}{b} \)
42\( \frac{b}{a} \)
42ab

Solution

To multiply monomials, multiply the coefficients (the numbers that come before the variables) of each term, add the exponents of like variables, and multiply the different variables together.

7a x 6b = (7 x 6) (a x b) = 42ab


4

If the area of this square is 25, what is the length of one of the diagonals?

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

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 = \( \sqrt{a} \) = \( \sqrt{25} \) = 5

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 = 52 + 52
c2 = 50
c = \( \sqrt{50} \) = \( \sqrt{25 x 2} \) = \( \sqrt{25} \) \( \sqrt{2} \)
c = 5\( \sqrt{2} \)


5

Which types of triangles will always have at least two sides of equal length?

54% Answer Correctly

equilateral and right

equilateral, isosceles and right

equilateral and isosceles

isosceles and right


Solution

An isosceles triangle has two sides of equal length. An equilateral triangle has three sides of equal length. In a right triangle, two sides meet at a right angle.