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If side a = 3, side b = 9, what is the length of the hypotenuse of this right triangle?
| \( \sqrt{73} \) | |
| \( \sqrt{90} \) | |
| \( \sqrt{82} \) | |
| \( \sqrt{40} \) |
According to the Pythagorean theorem, the hypotenuse squared is equal to the sum of the two perpendicular sides squared:
c2 = a2 + b2
c2 = 32 + 92
c2 = 9 + 81
c2 = 90
c = \( \sqrt{90} \)
Breaking apart a quadratic expression into a pair of binomials is called:
factoring |
|
squaring |
|
deconstructing |
|
normalizing |
To factor a quadratic expression, apply the FOIL (First, Outside, Inside, Last) method in reverse.
Solve 3b + 8b = -4b + 8x + 3 for b in terms of x.
| 4\(\frac{1}{2}\)x + 1 | |
| 6x + 8 | |
| x + \(\frac{3}{7}\) | |
| 3\(\frac{1}{2}\)x + \(\frac{1}{2}\) |
To solve this equation, isolate the variable for which you are solving (b) on one side of the equation and put everything else on the other side.
3b + 8x = -4b + 8x + 3
3b = -4b + 8x + 3 - 8x
3b + 4b = 8x + 3 - 8x
7b = + 3
b = \( \frac{ + 3}{7} \)
b = \( \frac{}{7} \) + \( \frac{3}{7} \)
b = x + \(\frac{3}{7}\)
Solve for b:
b2 + 2b - 9 = 5b - 5
| -5 or -8 | |
| 3 or 2 | |
| -7 or -7 | |
| -1 or 4 |
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:
b2 + 2b - 9 = 5b - 5
b2 + 2b - 9 + 5 = 5b
b2 + 2b - 5b - 4 = 0
b2 - 3b - 4 = 0
Next, factor the quadratic equation:
b2 - 3b - 4 = 0
(b + 1)(b - 4) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (b + 1) or (b - 4) must equal zero:
If (b + 1) = 0, b must equal -1
If (b - 4) = 0, b must equal 4
So the solution is that b = -1 or 4
Find the value of b:
8b + z = 8
-6b + 8z = 5
| \(\frac{59}{70}\) | |
| \(\frac{5}{18}\) | |
| -\(\frac{1}{2}\) |
You need to find the value of b so solve the first equation in terms of z:
8b + z = 8
z = 8 - 8b
then substitute the result (8 - 8b) into the second equation:
-6b + 8(8 - 8b) = 5
-6b + (8 x 8) + (8 x -8b) = 5
-6b + 64 - 64b = 5
-6b - 64b = 5 - 64
-70b = -59
b = \( \frac{-59}{-70} \)
b = \(\frac{59}{70}\)