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Solve for b:
-2b - 6 < -1 - 4b
| b < 1\(\frac{3}{5}\) | |
| b < -\(\frac{3}{4}\) | |
| b < 3\(\frac{1}{2}\) | |
| b < 2\(\frac{1}{2}\) |
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.
-2b - 6 < -1 - 4b
-2b < -1 - 4b + 6
-2b + 4b < -1 + 6
2b < 5
b < \( \frac{5}{2} \)
b < 2\(\frac{1}{2}\)
Solve for b:
b2 + 6b + 5 = 0
| 6 or 1 | |
| 4 or 1 | |
| 9 or -8 | |
| -1 or -5 |
The first step to solve a quadratic equation that's set to zero is to factor the quadratic equation:
b2 + 6b + 5 = 0
(b + 1)(b + 5) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (b + 1) or (b + 5) must equal zero:
If (b + 1) = 0, b must equal -1
If (b + 5) = 0, b must equal -5
So the solution is that b = -1 or -5
The endpoints of this line segment are at (-2, 0) and (2, 2). What is the slope of this line?
| 2\(\frac{1}{2}\) | |
| \(\frac{1}{2}\) | |
| -2\(\frac{1}{2}\) | |
| -3 |
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, 0) and (2, 2) so the slope becomes:
m = \( \frac{\Delta y}{\Delta x} \) = \( \frac{(2.0) - (0.0)}{(2) - (-2)} \) = \( \frac{2}{4} \)A cylinder with a radius (r) and a height (h) has a surface area of:
2(π r2) + 2π rh |
|
π r2h2 |
|
4π r2 |
|
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.
This diagram represents two parallel lines with a transversal. If b° = 140, what is the value of y°?
| 34 | |
| 140 | |
| 23 | |
| 33 |
For parallel lines with a transversal, the following relationships apply:
Applying these relationships starting with b° = 140, the value of y° is 140.