| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.12 |
| Score | 0% | 62% |
Simplify (4a)(6ab) - (4a2)(3b).
| 12a2b | |
| -12ab2 | |
| 70ab2 | |
| 36ab2 |
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)(6ab) - (4a2)(3b)
(4 x 6)(a x a x b) - (4 x 3)(a2 x b)
(24)(a1+1 x b) - (12)(a2b)
24a2b - 12a2b
12a2b
If side a = 3, side b = 6, what is the length of the hypotenuse of this right triangle?
| \( \sqrt{97} \) | |
| \( \sqrt{45} \) | |
| \( \sqrt{89} \) | |
| \( \sqrt{50} \) |
According to the Pythagorean theorem, the hypotenuse squared is equal to the sum of the two perpendicular sides squared:
c2 = a2 + b2
c2 = 32 + 62
c2 = 9 + 36
c2 = 45
c = \( \sqrt{45} \)
If the area of this square is 4, what is the length of one of the diagonals?
| 9\( \sqrt{2} \) | |
| 4\( \sqrt{2} \) | |
| 2\( \sqrt{2} \) | |
| 3\( \sqrt{2} \) |
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 = \( \sqrt{a} \) = \( \sqrt{4} \) = 2
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 = 22 + 22
c2 = 8
c = \( \sqrt{8} \) = \( \sqrt{4 x 2} \) = \( \sqrt{4} \) \( \sqrt{2} \)
c = 2\( \sqrt{2} \)
The dimensions of this cylinder are height (h) = 3 and radius (r) = 2. What is the surface area?
| 90π | |
| 36π | |
| 20π | |
| 182π |
The surface area of a cylinder is 2πr2 + 2πrh:
sa = 2πr2 + 2πrh
sa = 2π(22) + 2π(2 x 3)
sa = 2π(4) + 2π(6)
sa = (2 x 4)π + (2 x 6)π
sa = 8π + 12π
sa = 20π
What is the area of a circle with a diameter of 8?
| 2π | |
| 36π | |
| 9π | |
| 16π |
The formula for area is πr2. Radius is circle \( \frac{diameter}{2} \):
r = \( \frac{d}{2} \)
r = \( \frac{8}{2} \)
r = 4
a = πr2
a = π(42)
a = 16π