| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.43 |
| Score | 0% | 69% |
If BD = 16 and AD = 20, AB = ?
| 1 | |
| 8 | |
| 2 | |
| 4 |
The entire length of this line is represented by AD which is AB + BD:
AD = AB + BD
Solving for AB:AB = AD - BDOn this circle, a line segment connecting point A to point D is called:
circumference |
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diameter |
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radius |
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chord |
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).
A right angle measures:
360° |
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45° |
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180° |
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90° |
A right angle measures 90 degrees and is the intersection of two perpendicular lines. In diagrams, a right angle is indicated by a small box completing a square with the perpendicular lines.
Simplify (3a)(6ab) - (3a2)(7b).
| -3a2b | |
| 90ab2 | |
| 3ab2 | |
| 90a2b |
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.
(3a)(6ab) - (3a2)(7b)
(3 x 6)(a x a x b) - (3 x 7)(a2 x b)
(18)(a1+1 x b) - (21)(a2b)
18a2b - 21a2b
-3a2b
If the area of this square is 9, what is the length of one of the diagonals?
| 9\( \sqrt{2} \) | |
| 3\( \sqrt{2} \) | |
| 7\( \sqrt{2} \) | |
| 5\( \sqrt{2} \) |
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 = \( \sqrt{a} \) = \( \sqrt{9} \) = 3
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 = 32 + 32
c2 = 18
c = \( \sqrt{18} \) = \( \sqrt{9 x 2} \) = \( \sqrt{9} \) \( \sqrt{2} \)
c = 3\( \sqrt{2} \)