| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.17 |
| Score | 0% | 63% |
Solve for y:
-y + 2 < \( \frac{y}{-9} \)
| y < 4\(\frac{2}{7}\) | |
| y < 2\(\frac{1}{4}\) | |
| y < -\(\frac{16}{19}\) | |
| y < -1\(\frac{5}{9}\) |
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.
-y + 2 < \( \frac{y}{-9} \)
-9 x (-y + 2) < y
(-9 x -y) + (-9 x 2) < y
9y - 18 < y
9y - 18 - y < 0
9y - y < 18
8y < 18
y < \( \frac{18}{8} \)
y < 2\(\frac{1}{4}\)
If BD = 26 and AD = 30, AB = ?
| 8 | |
| 3 | |
| 17 | |
| 4 |
The entire length of this line is represented by AD which is AB + BD:
AD = AB + BD
Solving for AB:AB = AD - BDIf a = c = 9, b = d = 10, what is the area of this rectangle?
| 56 | |
| 90 | |
| 35 | |
| 3 |
The area of a rectangle is equal to its length x width:
a = l x w
a = a x b
a = 9 x 10
a = 90
If side a = 6, side b = 1, what is the length of the hypotenuse of this right triangle?
| \( \sqrt{50} \) | |
| \( \sqrt{37} \) | |
| \( \sqrt{13} \) | |
| \( \sqrt{41} \) |
According to the Pythagorean theorem, the hypotenuse squared is equal to the sum of the two perpendicular sides squared:
c2 = a2 + b2
c2 = 62 + 12
c2 = 36 + 1
c2 = 37
c = \( \sqrt{37} \)
Solve for c:
-8c - 7 > -8 + 9c
| c > -9 | |
| c > -1 | |
| c > \(\frac{7}{8}\) | |
| c > \(\frac{1}{17}\) |
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.
-8c - 7 > -8 + 9c
-8c > -8 + 9c + 7
-8c - 9c > -8 + 7
-17c > -1
c > \( \frac{-1}{-17} \)
c > \(\frac{1}{17}\)