| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.62 |
| Score | 0% | 52% |
If b = 4 and y = -6, what is the value of 5b(b - y)?
| 200 | |
| -78 | |
| 264 | |
| -48 |
To solve this equation, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.)
5b(b - y)
5(4)(4 + 6)
5(4)(10)
(20)(10)
200
Solve for x:
-7x + 1 < 3 - 8x
| x < \(\frac{2}{3}\) | |
| x < -\(\frac{1}{3}\) | |
| x < -6 | |
| x < 2 |
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.
-7x + 1 < 3 - 8x
-7x < 3 - 8x - 1
-7x + 8x < 3 - 1
x < 2
Solve -7b - 6b = 8b + 3y - 9 for b in terms of y.
| \(\frac{7}{11}\)y + \(\frac{2}{11}\) | |
| -8\(\frac{1}{2}\)y + 1\(\frac{1}{2}\) | |
| 1\(\frac{1}{2}\)y - \(\frac{5}{6}\) | |
| -\(\frac{3}{5}\)y + \(\frac{3}{5}\) |
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.
-7b - 6y = 8b + 3y - 9
-7b = 8b + 3y - 9 + 6y
-7b - 8b = 3y - 9 + 6y
-15b = 9y - 9
b = \( \frac{9y - 9}{-15} \)
b = \( \frac{9y}{-15} \) + \( \frac{-9}{-15} \)
b = -\(\frac{3}{5}\)y + \(\frac{3}{5}\)
Solve for c:
c2 - 12c + 36 = 0
| 2 or -9 | |
| 4 or 3 | |
| -3 or -3 | |
| 6 |
The first step to solve a quadratic equation that's set to zero is to factor the quadratic equation:
c2 - 12c + 36 = 0
(c - 6)(c - 6) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, (c - 6) must equal zero:
If (c - 6) = 0, c must equal 6
So the solution is that c = 6
For this diagram, the Pythagorean theorem states that b2 = ?
a2 - c2 |
|
c - a |
|
c2 - a2 |
|
c2 + a2 |
The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2 + b2): c2 = a2 + b2 or, solved for c, \(c = \sqrt{a + b}\)