| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.97 |
| Score | 0% | 59% |
Factor y2 - 3y - 40
| (y + 8)(y - 5) | |
| (y - 8)(y + 5) | |
| (y - 8)(y - 5) | |
| (y + 8)(y + 5) |
To factor a quadratic expression, apply the FOIL method (First, Outside, Inside, Last) in reverse. First, find the two Last terms that will multiply to produce -40 as well and sum (Inside, Outside) to equal -3. For this problem, those two numbers are -8 and 5. Then, plug these into a set of binomials using the square root of the First variable (y2):
y2 - 3y - 40
y2 + (-8 + 5)y + (-8 x 5)
(y - 8)(y + 5)
Solve for y:
-4y - 5 = \( \frac{y}{-3} \)
| 2\(\frac{7}{19}\) | |
| -1\(\frac{1}{20}\) | |
| \(\frac{32}{73}\) | |
| -1\(\frac{4}{11}\) |
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.
-4y - 5 = \( \frac{y}{-3} \)
-3 x (-4y - 5) = y
(-3 x -4y) + (-3 x -5) = y
12y + 15 = y
12y + 15 - y = 0
12y - y = -15
11y = -15
y = \( \frac{-15}{11} \)
y = -1\(\frac{4}{11}\)
If AD = 25 and BD = 19, AB = ?
| 9 | |
| 4 | |
| 18 | |
| 6 |
The entire length of this line is represented by AD which is AB + BD:
AD = AB + BD
Solving for AB:AB = AD - BDOn this circle, line segment AB is the:
circumference |
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chord |
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diameter |
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radius |
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).
A cylinder with a radius (r) and a height (h) has a surface area of:
4π r2 |
|
2(π r2) + 2π rh |
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π r2h2 |
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π r2h |
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.