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|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.55 |
| Score | 0% | 51% |
Solve for x:
-x + 3 > \( \frac{x}{4} \)
| x > -\(\frac{6}{7}\) | |
| x > 2\(\frac{2}{5}\) | |
| x > -8\(\frac{1}{10}\) | |
| x > -\(\frac{9}{20}\) |
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.
-x + 3 > \( \frac{x}{4} \)
4 x (-x + 3) > x
(4 x -x) + (4 x 3) > x
-4x + 12 > x
-4x + 12 - x > 0
-4x - x > -12
-5x > -12
x > \( \frac{-12}{-5} \)
x > 2\(\frac{2}{5}\)
The endpoints of this line segment are at (-2, 1) and (2, 5). What is the slope of this line?
| \(\frac{1}{2}\) | |
| -\(\frac{1}{2}\) | |
| 1 | |
| -1 |
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, 1) and (2, 5) so the slope becomes:
m = \( \frac{\Delta y}{\Delta x} \) = \( \frac{(5.0) - (1.0)}{(2) - (-2)} \) = \( \frac{4}{4} \)The dimensions of this trapezoid are a = 5, b = 7, c = 8, d = 8, and h = 4. What is the area?
| 15 | |
| 17\(\frac{1}{2}\) | |
| 30 | |
| 37\(\frac{1}{2}\) |
The area of a trapezoid is one-half the sum of the lengths of the parallel sides multiplied by the height:
a = ½(b + d)(h)
a = ½(7 + 8)(4)
a = ½(15)(4)
a = ½(60) = \( \frac{60}{2} \)
a = 30
Solve for y:
y2 - y + 7 = 5y - 1
| 5 or -8 | |
| 9 or -2 | |
| 2 or 4 | |
| 6 or 2 |
The first step to solve a quadratic expression that's not set to zero is to solve the equation so that it is set to zero:
y2 - y + 7 = 5y - 1
y2 - y + 7 + 1 = 5y
y2 - y - 5y + 8 = 0
y2 - 6y + 8 = 0
Next, factor the quadratic equation:
y2 - 6y + 8 = 0
(y - 2)(y - 4) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (y - 2) or (y - 4) must equal zero:
If (y - 2) = 0, y must equal 2
If (y - 4) = 0, y must equal 4
So the solution is that y = 2 or 4
If the area of this square is 49, what is the length of one of the diagonals?
| 7\( \sqrt{2} \) | |
| \( \sqrt{2} \) | |
| 8\( \sqrt{2} \) | |
| 5\( \sqrt{2} \) |
To find the diagonal we need to know the length of one of the square's sides. We know the area and the area of a square is the length of one side squared:
a = s2
so the length of one side of the square is:
s = \( \sqrt{a} \) = \( \sqrt{49} \) = 7
The Pythagorean theorem defines the square of the hypotenuse (diagonal) of a triangle with a right angle as the sum of the squares of the other two sides:
c2 = a2 + b2
c2 = 72 + 72
c2 = 98
c = \( \sqrt{98} \) = \( \sqrt{49 x 2} \) = \( \sqrt{49} \) \( \sqrt{2} \)
c = 7\( \sqrt{2} \)