| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.08 |
| Score | 0% | 62% |
This diagram represents two parallel lines with a transversal. If d° = 160, what is the value of y°?
| 155 | |
| 40 | |
| 167 | |
| 160 |
For parallel lines with a transversal, the following relationships apply:
Applying these relationships starting with d° = 160, the value of y° is 160.
Simplify 4a x 2b.
| 8ab | |
| 8a2b2 | |
| 8\( \frac{b}{a} \) | |
| 8\( \frac{a}{b} \) |
To multiply monomials, multiply the coefficients (the numbers that come before the variables) of each term, add the exponents of like variables, and multiply the different variables together.
4a x 2b = (4 x 2) (a x b) = 8ab
If the area of this square is 25, what is the length of one of the diagonals?
| 5\( \sqrt{2} \) | |
| 7\( \sqrt{2} \) | |
| 6\( \sqrt{2} \) | |
| 9\( \sqrt{2} \) |
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 = \( \sqrt{a} \) = \( \sqrt{25} \) = 5
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 = 52 + 52
c2 = 50
c = \( \sqrt{50} \) = \( \sqrt{25 x 2} \) = \( \sqrt{25} \) \( \sqrt{2} \)
c = 5\( \sqrt{2} \)
On this circle, line segment CD is the:
diameter |
|
circumference |
|
radius |
|
chord |
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).
Solve -8c - 8c = 3c + 3z - 8 for c in terms of z.
| -z + \(\frac{8}{11}\) | |
| -z - 3 | |
| 1\(\frac{2}{5}\)z - \(\frac{3}{5}\) | |
| -1\(\frac{1}{3}\)z - 2\(\frac{1}{3}\) |
To solve this equation, isolate the variable for which you are solving (c) on one side of the equation and put everything else on the other side.
-8c - 8z = 3c + 3z - 8
-8c = 3c + 3z - 8 + 8z
-8c - 3c = 3z - 8 + 8z
-11c = 11z - 8
c = \( \frac{11z - 8}{-11} \)
c = \( \frac{11z}{-11} \) + \( \frac{-8}{-11} \)
c = -z + \(\frac{8}{11}\)