ASVAB Math Knowledge Practice Test 566830 Results

Your Results Global Average
Questions 5 5
Correct 0 3.18
Score 0% 64%

Review

1

Solve for a:
a2 + 3a - 40 = 0

58% Answer Correctly
4 or -2
4 or 3
5 or -8
7 or -4

Solution

The first step to solve a quadratic equation that's set to zero is to factor the quadratic equation:

a2 + 3a - 40 = 0
(a - 5)(a + 8) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (a - 5) or (a + 8) must equal zero:

If (a - 5) = 0, a must equal 5
If (a + 8) = 0, a must equal -8

So the solution is that a = 5 or -8


2

Which of the following is not true about both rectangles and squares?

63% Answer Correctly

the lengths of all sides are equal

the area is length x width

all interior angles are right angles

the perimeter is the sum of the lengths of all four sides


Solution

A rectangle is a parallelogram containing four right angles. Opposite sides (a = c, b = d) are equal and the perimeter is the sum of the lengths of all sides (a + b + c + d) or, comonly, 2 x length x width. The area of a rectangle is length x width. A square is a rectangle with four equal length sides. The perimeter of a square is 4 x length of one side (4s) and the area is the length of one side squared (s2).


3

Solve for y:
2y - 7 < -1 + 6y

55% Answer Correctly
y < \(\frac{1}{2}\)
y < 1\(\frac{1}{2}\)
y < -\(\frac{1}{3}\)
y < -1\(\frac{1}{2}\)

Solution

To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the < sign and the answer on the other.

2y - 7 < -1 + 6y
2y < -1 + 6y + 7
2y - 6y < -1 + 7
-4y < 6
y < \( \frac{6}{-4} \)
y < -1\(\frac{1}{2}\)


4

What is 7a9 + 2a9?

75% Answer Correctly
9a9
5
9a18
5a18

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.

7a9 + 2a9 = 9a9


5

Simplify (9a)(7ab) + (9a2)(7b).

65% Answer Correctly
126ab2
2b
126a2b
256ab2

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.

(9a)(7ab) + (9a2)(7b)
(9 x 7)(a x a x b) + (9 x 7)(a2 x b)
(63)(a1+1 x b) + (63)(a2b)
63a2b + 63a2b
126a2b