| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.04 |
| Score | 0% | 61% |
The formula for volume of a cube in terms of height (h), length (l), and width (w) is which of the following?
h x l x w |
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lw x wh + lh |
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2lw x 2wh + 2lh |
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h2 x l2 x w2 |
A cube is a rectangular solid box with a height (h), length (l), and width (w). The volume is h x l x w and the surface area is 2lw x 2wh + 2lh.
Which of the following is not a part of PEMDAS, the acronym for math order of operations?
exponents |
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division |
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addition |
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pairs |
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.)
Solve for y:
-8y - 1 = \( \frac{y}{3} \)
| -\(\frac{4}{5}\) | |
| 1\(\frac{5}{13}\) | |
| -1\(\frac{7}{41}\) | |
| -\(\frac{3}{25}\) |
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 equal sign and the answer on the other.
-8y - 1 = \( \frac{y}{3} \)
3 x (-8y - 1) = y
(3 x -8y) + (3 x -1) = y
-24y - 3 = y
-24y - 3 - y = 0
-24y - y = 3
-25y = 3
y = \( \frac{3}{-25} \)
y = -\(\frac{3}{25}\)
Find the value of c:
-6c + z = 7
-9c - 8z = 8
| -1\(\frac{7}{57}\) | |
| -\(\frac{1}{5}\) | |
| 1\(\frac{1}{9}\) | |
| \(\frac{4}{15}\) |
You need to find the value of c so solve the first equation in terms of z:
-6c + z = 7
z = 7 + 6c
then substitute the result (7 - -6c) into the second equation:
-9c - 8(7 + 6c) = 8
-9c + (-8 x 7) + (-8 x 6c) = 8
-9c - 56 - 48c = 8
-9c - 48c = 8 + 56
-57c = 64
c = \( \frac{64}{-57} \)
c = -1\(\frac{7}{57}\)
A cylinder with a radius (r) and a height (h) has a surface area of:
4π r2 |
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2(π r2) + 2π rh |
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π r2h |
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π r2h2 |
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.