| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.34 |
| Score | 0% | 67% |
If b = -3 and y = 1, what is the value of 3b(b - y)?
| -64 | |
| 36 | |
| -351 | |
| -768 |
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.)
3b(b - y)
3(-3)(-3 - 1)
3(-3)(-4)
(-9)(-4)
36
Which of the following is not true about both rectangles and squares?
the lengths of all sides are equal |
|
all interior angles are right angles |
|
the perimeter is the sum of the lengths of all four sides |
|
the area is length x width |
A rectangle is a parallelogram containing four right angles. Opposite sides (a = c, b = d) are equal and the perimeter is the sum of the lengths of all sides (a + b + c + d) or, comonly, 2 x length x width. The area of a rectangle is length x width. A square is a rectangle with four equal length sides. The perimeter of a square is 4 x length of one side (4s) and the area is the length of one side squared (s2).
Which of the following statements about math operations is incorrect?
you can subtract monomials that have the same variable and the same exponent |
|
all of these statements are correct |
|
you can multiply monomials that have different variables and different exponents |
|
you can add monomials that have the same variable and the same exponent |
You can only add or subtract monomials that have the same variable and the same exponent. For example, 2a + 4a = 6a and 4a2 - a2 = 3a2 but 2a + 4b and 7a - 3b cannot be combined. However, you can multiply and divide monomials with unlike terms. For example, 2a x 6b = 12ab.
The dimensions of this cube are height (h) = 4, length (l) = 8, and width (w) = 2. What is the volume?
| 64 | |
| 84 | |
| 192 | |
| 56 |
The volume of a cube is height x length x width:
v = h x l x w
v = 4 x 8 x 2
v = 64
Solve for x:
x2 + 6x - 2 = 3x - 4
| 9 or 1 | |
| -1 or -2 | |
| 8 or 7 | |
| 5 or 1 |
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:
x2 + 6x - 2 = 3x - 4
x2 + 6x - 2 + 4 = 3x
x2 + 6x - 3x + 2 = 0
x2 + 3x + 2 = 0
Next, factor the quadratic equation:
x2 + 3x + 2 = 0
(x + 1)(x + 2) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (x + 1) or (x + 2) must equal zero:
If (x + 1) = 0, x must equal -1
If (x + 2) = 0, x must equal -2
So the solution is that x = -1 or -2