| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.21 |
| Score | 0% | 64% |
A trapezoid is a quadrilateral with one set of __________ sides.
right angle |
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equal angle |
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parallel |
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equal length |
A trapezoid is a quadrilateral with one set of parallel sides.
Which of the following expressions contains exactly two terms?
quadratic |
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polynomial |
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binomial |
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monomial |
A monomial contains one term, a binomial contains two terms, and a polynomial contains more than two terms.
On this circle, line segment CD is the:
diameter |
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chord |
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radius |
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circumference |
A circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter. A chord is a line segment that connects any two points along its perimeter. The diameter of a circle is the length of a chord that passes through the center of the circle and equals twice the circle's radius (2r).
If the area of this square is 36, what is the length of one of the diagonals?
| 6\( \sqrt{2} \) | |
| 7\( \sqrt{2} \) | |
| 3\( \sqrt{2} \) | |
| 9\( \sqrt{2} \) |
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 = \( \sqrt{a} \) = \( \sqrt{36} \) = 6
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 = 62 + 62
c2 = 72
c = \( \sqrt{72} \) = \( \sqrt{36 x 2} \) = \( \sqrt{36} \) \( \sqrt{2} \)
c = 6\( \sqrt{2} \)
Factor y2 - 4
| (y - 2)(y - 2) | |
| (y + 2)(y + 2) | |
| (y - 2)(y + 2) | |
| (y + 2)(y - 2) |
To factor a quadratic expression, apply the FOIL method (First, Outside, Inside, Last) in reverse. First, find the two Last terms that will multiply to produce -4 as well and sum (Inside, Outside) to equal 0. For this problem, those two numbers are -2 and 2. Then, plug these into a set of binomials using the square root of the First variable (y2):
y2 - 4
y2 + (-2 + 2)y + (-2 x 2)
(y - 2)(y + 2)