| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.84 |
| Score | 0% | 57% |
Simplify (y + 1)(y - 2)
| y2 - 3y + 2 | |
| y2 + 3y + 2 | |
| y2 - y - 2 | |
| y2 + y - 2 |
To multiply binomials, use the FOIL method. FOIL stands for First, Outside, Inside, Last and refers to the position of each term in the parentheses:
(y + 1)(y - 2)
(y x y) + (y x -2) + (1 x y) + (1 x -2)
y2 - 2y + y - 2
y2 - y - 2
The endpoints of this line segment are at (-2, -3) and (2, 3). What is the slope-intercept equation for this line?
| y = -3x - 2 | |
| y = 1\(\frac{1}{2}\)x + 2 | |
| y = x - 3 | |
| y = 1\(\frac{1}{2}\)x + 0 |
The slope-intercept equation for a line is y = mx + b where m is the slope and b is the y-intercept of the line. From the graph, you can see that the y-intercept (the y-value from the point where the line crosses the y-axis) is 0. The slope of this line is the change in y divided by the change in x. The endpoints of this line segment are at (-2, -3) and (2, 3) so the slope becomes:
m = \( \frac{\Delta y}{\Delta x} \) = \( \frac{(3.0) - (-3.0)}{(2) - (-2)} \) = \( \frac{6}{4} \)Plugging these values into the slope-intercept equation:
y = 1\(\frac{1}{2}\)x + 0
Solve for a:
8a - 9 < \( \frac{a}{6} \)
| a < 1\(\frac{3}{17}\) | |
| a < \(\frac{8}{15}\) | |
| a < -3\(\frac{3}{13}\) | |
| a < 1\(\frac{7}{47}\) |
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.
8a - 9 < \( \frac{a}{6} \)
6 x (8a - 9) < a
(6 x 8a) + (6 x -9) < a
48a - 54 < a
48a - 54 - a < 0
48a - a < 54
47a < 54
a < \( \frac{54}{47} \)
a < 1\(\frac{7}{47}\)
Which of the following expressions contains exactly two terms?
binomial |
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polynomial |
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quadratic |
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monomial |
A monomial contains one term, a binomial contains two terms, and a polynomial contains more than two terms.
For this diagram, the Pythagorean theorem states that b2 = ?
c2 + a2 |
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c - a |
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c2 - a2 |
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a2 - c2 |
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, \(c = \sqrt{a + b}\)