| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.60 |
| Score | 0% | 52% |
Which types of triangles will always have at least two sides of equal length?
equilateral, isosceles and right |
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equilateral and isosceles |
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isosceles and right |
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equilateral and right |
An isosceles triangle has two sides of equal length. An equilateral triangle has three sides of equal length. In a right triangle, two sides meet at a right angle.
Find the value of a:
2a + z = -2
-7a + 2z = 1
| 4\(\frac{9}{11}\) | |
| -1\(\frac{4}{15}\) | |
| -7\(\frac{7}{8}\) | |
| -\(\frac{5}{11}\) |
You need to find the value of a so solve the first equation in terms of z:
2a + z = -2
z = -2 - 2a
then substitute the result (-2 - 2a) into the second equation:
-7a + 2(-2 - 2a) = 1
-7a + (2 x -2) + (2 x -2a) = 1
-7a - 4 - 4a = 1
-7a - 4a = 1 + 4
-11a = 5
a = \( \frac{5}{-11} \)
a = -\(\frac{5}{11}\)
If the base of this triangle is 3 and the height is 1, what is the area?
| 1\(\frac{1}{2}\) | |
| 65 | |
| 39 | |
| 15 |
The area of a triangle is equal to ½ base x height:
a = ½bh
a = ½ x 3 x 1 = \( \frac{3}{2} \) = 1\(\frac{1}{2}\)
Solve for z:
z2 - 5z - 6 = 0
| 2 or -5 | |
| 8 or 8 | |
| -1 or 6 | |
| 1 or -5 |
The first step to solve a quadratic equation that's set to zero is to factor the quadratic equation:
z2 - 5z - 6 = 0
(z + 1)(z - 6) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (z + 1) or (z - 6) must equal zero:
If (z + 1) = 0, z must equal -1
If (z - 6) = 0, z must equal 6
So the solution is that z = -1 or 6
On this circle, line segment CD is the:
radius |
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circumference |
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diameter |
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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).