ASVAB Math Knowledge Practice Test 605699 Results

Your Results Global Average
Questions 5 5
Correct 0 2.65
Score 0% 53%

Review

1

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

47% Answer Correctly

c2 + a2

c2 - a2

c - a

a2 - c2


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, \(c = \sqrt{a + b}\)


2

A cylinder with a radius (r) and a height (h) has a surface area of:

54% Answer Correctly

π r2h2

4π r2

2(π r2) + 2π rh

π r2h


Solution

A cylinder is a solid figure with straight parallel sides and a circular or oval cross section with a radius (r) and a height (h). The volume of a cylinder is π r2h and the surface area is 2(π r2) + 2π rh.


3

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

54% Answer Correctly

equilateral and isosceles

isosceles and right

equilateral and right

equilateral, 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.


4

Order the following types of angle from least number of degrees to most number of degrees.

75% Answer Correctly

right, acute, obtuse

right, obtuse, acute

acute, right, obtuse

acute, obtuse, right


Solution

An acute angle measures less than 90°, a right angle measures 90°, and an obtuse angle measures more than 90°.


5

Solve b + 7b = -2b + 2x + 4 for b in terms of x.

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

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.

b + 7x = -2b + 2x + 4
b = -2b + 2x + 4 - 7x
b + 2b = 2x + 4 - 7x
3b = -5x + 4
b = \( \frac{-5x + 4}{3} \)
b = \( \frac{-5x}{3} \) + \( \frac{4}{3} \)
b = -1\(\frac{2}{3}\)x + 1\(\frac{1}{3}\)