ASVAB Math Knowledge Practice Test 421707 Results

Your Results Global Average
Questions 5 5
Correct 0 2.87
Score 0% 57%

Review

1

What is 3a8 + 9a8?

75% Answer Correctly
-6
12a8
27a16
12a16

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.

3a8 + 9a8 = 12a8


2

Find the value of b:
5b + y = -4
5b + 2y = 1

42% Answer Correctly
3\(\frac{3}{4}\)
-\(\frac{28}{45}\)
-3\(\frac{1}{5}\)
-1\(\frac{4}{5}\)

Solution

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

5b + y = -4
y = -4 - 5b

then substitute the result (-4 - 5b) into the second equation:

5b + 2(-4 - 5b) = 1
5b + (2 x -4) + (2 x -5b) = 1
5b - 8 - 10b = 1
5b - 10b = 1 + 8
-5b = 9
b = \( \frac{9}{-5} \)
b = -1\(\frac{4}{5}\)


3

What is 6a3 - 7a3?

74% Answer Correctly
42a3
42a6
-1a3
13a6

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.

6a3 - 7a3 = -1a3


4

For this diagram, the Pythagorean theorem states that b2 = ?

47% Answer Correctly

c2 - a2

c2 + a2

a2 - c2

c - a


Solution

The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2 + b2): c2 = a2 + b2 or, solved for c, \(c = \sqrt{a + b}\)


5

On this circle, a line segment connecting point A to point D is called:

46% Answer Correctly

circumference

diameter

chord

radius


Solution

A circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter. A chord is a line segment that connects any two points along its perimeter. The diameter of a circle is the length of a chord that passes through the center of the circle and equals twice the circle's radius (2r).