| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.10 |
| Score | 0% | 62% |
What is the circumference of a circle with a radius of 19?
| 38π | |
| 10π | |
| 9π | |
| 18π |
The formula for circumference is circle diameter x π. Circle diameter is 2 x radius:
c = πd
c = π(2 * r)
c = π(2 * 19)
c = 38π
Solve for a:
3a + 3 = -5 + 6a
| 1\(\frac{2}{7}\) | |
| 1 | |
| 2\(\frac{2}{3}\) | |
| -1\(\frac{1}{7}\) |
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 equal sign and the answer on the other.
3a + 3 = -5 + 6a
3a = -5 + 6a - 3
3a - 6a = -5 - 3
-3a = -8
a = \( \frac{-8}{-3} \)
a = 2\(\frac{2}{3}\)
If the area of this square is 1, what is the length of one of the diagonals?
| 4\( \sqrt{2} \) | |
| 5\( \sqrt{2} \) | |
| 9\( \sqrt{2} \) | |
| \( \sqrt{2} \) |
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 = \( \sqrt{a} \) = \( \sqrt{1} \) = 1
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 = 12 + 12
c2 = 2
c = \( \sqrt{2} \)
Find the value of c:
-6c + z = 3
-9c + 8z = -6
| 1\(\frac{1}{33}\) | |
| 2 | |
| -\(\frac{10}{13}\) | |
| -2\(\frac{9}{10}\) |
You need to find the value of c so solve the first equation in terms of z:
-6c + z = 3
z = 3 + 6c
then substitute the result (3 - -6c) into the second equation:
-9c + 8(3 + 6c) = -6
-9c + (8 x 3) + (8 x 6c) = -6
-9c + 24 + 48c = -6
-9c + 48c = -6 - 24
39c = -30
c = \( \frac{-30}{39} \)
c = -\(\frac{10}{13}\)
What is the area of a circle with a radius of 2?
| 8π | |
| 7π | |
| 4π | |
| 81π |
The formula for area is πr2:
a = πr2
a = π(22)
a = 4π