| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.98 |
| Score | 0% | 60% |
Simplify (7a)(5ab) + (4a2)(4b).
| -19a2b | |
| 96ab2 | |
| 19ab2 | |
| 51a2b |
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.
(7a)(5ab) + (4a2)(4b)
(7 x 5)(a x a x b) + (4 x 4)(a2 x b)
(35)(a1+1 x b) + (16)(a2b)
35a2b + 16a2b
51a2b
Which of the following statements about math operations is incorrect?
you can multiply monomials that have different variables and different exponents |
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all of these statements are correct |
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you can subtract monomials that have the same variable and the same exponent |
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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.
Which of the following statements about a parallelogram is not true?
a parallelogram is a quadrilateral |
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the perimeter of a parallelogram is the sum of the lengths of all sides |
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the area of a parallelogram is base x height |
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opposite sides and adjacent angles are equal |
A parallelogram is a quadrilateral with two sets of parallel sides. Opposite sides (a = c, b = d) and angles (red = red, blue = blue) are equal. The area of a parallelogram is base x height and the perimeter is the sum of the lengths of all sides (a + b + c + d).
When two lines intersect, adjacent angles are __________ (they add up to 180°) and angles across from either other are __________ (they're equal).
supplementary, vertical |
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acute, obtuse |
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obtuse, acute |
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vertical, supplementary |
Angles around a line add up to 180°. Angles around a point add up to 360°. When two lines intersect, adjacent angles are supplementary (they add up to 180°) and angles across from either other are vertical (they're equal).
Solve for y:
y2 + 3y + 0 = y + 3
| 9 or 5 | |
| 1 or -3 | |
| 8 or 1 | |
| -2 or -8 |
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:
y2 + 3y + 0 = y + 3
y2 + 3y + 0 - 3 = y
y2 + 3y - y - 3 = 0
y2 + 2y - 3 = 0
Next, factor the quadratic equation:
y2 + 2y - 3 = 0
(y - 1)(y + 3) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (y - 1) or (y + 3) must equal zero:
If (y - 1) = 0, y must equal 1
If (y + 3) = 0, y must equal -3
So the solution is that y = 1 or -3