| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.17 |
| Score | 0% | 63% |
If the area of this square is 25, what is the length of one of the diagonals?
| 2\( \sqrt{2} \) | |
| 5\( \sqrt{2} \) | |
| 8\( \sqrt{2} \) | |
| 6\( \sqrt{2} \) |
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 = \( \sqrt{a} \) = \( \sqrt{25} \) = 5
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 = 52 + 52
c2 = 50
c = \( \sqrt{50} \) = \( \sqrt{25 x 2} \) = \( \sqrt{25} \) \( \sqrt{2} \)
c = 5\( \sqrt{2} \)
On this circle, a line segment connecting point A to point D is called:
chord |
|
circumference |
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radius |
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diameter |
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 c = -5 and z = 6, what is the value of -8c(c - z)?
| -64 | |
| -440 | |
| -108 | |
| 0 |
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.)
-8c(c - z)
-8(-5)(-5 - 6)
-8(-5)(-11)
(40)(-11)
-440
A right angle measures:
360° |
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180° |
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45° |
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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.
Solve for a:
-4a - 5 > \( \frac{a}{-6} \)
| a > 3\(\frac{3}{7}\) | |
| a > -1\(\frac{7}{23}\) | |
| a > -\(\frac{8}{11}\) | |
| a > \(\frac{6}{53}\) |
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.
-4a - 5 > \( \frac{a}{-6} \)
-6 x (-4a - 5) > a
(-6 x -4a) + (-6 x -5) > a
24a + 30 > a
24a + 30 - a > 0
24a - a > -30
23a > -30
a > \( \frac{-30}{23} \)
a > -1\(\frac{7}{23}\)