| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.41 |
| Score | 0% | 48% |
Which of the following statements about parallel lines with a transversal is not correct?
angles in the same position on different parallel lines are called corresponding angles |
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same-side interior angles are complementary and equal each other |
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all acute angles equal each other |
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all of the angles formed by a transversal are called interior angles |
Parallel lines are lines that share the same slope (steepness) and therefore never intersect. A transversal occurs when a set of parallel lines are crossed by another line. All of the angles formed by a transversal are called interior angles and angles in the same position on different parallel lines equal each other (a° = w°, b° = x°, c° = z°, d° = y°) and are called corresponding angles. Alternate interior angles are equal (a° = z°, b° = y°, c° = w°, d° = x°) and all acute angles (a° = c° = w° = z°) and all obtuse angles (b° = d° = x° = y°) equal each other. Same-side interior angles are supplementary and add up to 180° (e.g. a° + d° = 180°, d° + c° = 180°).
Which of the following statements about a parallelogram is not true?
opposite sides and adjacent angles are equal |
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the perimeter of a parallelogram is the sum of the lengths of all sides |
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a parallelogram is a quadrilateral |
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the area of a parallelogram is base x height |
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).
The formula for the area of a circle is which of the following?
a = π r |
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a = π r2 |
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a = π d2 |
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a = π d |
The circumference of a circle is the distance around its perimeter and equals π (approx. 3.14159) x diameter: c = π d. The area of a circle is π x (radius)2 : a = π r2.
Solve for a:
5a + 1 > \( \frac{a}{1} \)
| a > 4\(\frac{1}{2}\) | |
| a > \(\frac{16}{19}\) | |
| a > -\(\frac{1}{4}\) | |
| a > -2 |
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.
5a + 1 > \( \frac{a}{1} \)
1 x (5a + 1) > a
(1 x 5a) + (1 x 1) > a
5a + 1 > a
5a + 1 - a > 0
5a - a > -1
4a > -1
a > \( \frac{-1}{4} \)
a > -\(\frac{1}{4}\)
Solve -6a - 4a = 9a + z - 6 for a in terms of z.
| 10z + 5 | |
| z + \(\frac{1}{5}\) | |
| \(\frac{5}{12}\)z + \(\frac{2}{3}\) | |
| -\(\frac{1}{3}\)z + \(\frac{2}{5}\) |
To solve this equation, isolate the variable for which you are solving (a) on one side of the equation and put everything else on the other side.
-6a - 4z = 9a + z - 6
-6a = 9a + z - 6 + 4z
-6a - 9a = z - 6 + 4z
-15a = 5z - 6
a = \( \frac{5z - 6}{-15} \)
a = \( \frac{5z}{-15} \) + \( \frac{-6}{-15} \)
a = -\(\frac{1}{3}\)z + \(\frac{2}{5}\)