| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.60 |
| Score | 0% | 52% |
On this circle, a line segment connecting point A to point D is called:
circumference |
|
radius |
|
diameter |
|
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).
Solve for z:
z2 + 4z - 27 = 2z - 3
| -7 or -9 | |
| 4 or -6 | |
| -3 or -6 | |
| 1 or -5 |
The first step to solve a quadratic expression that's not set to zero is to solve the equation so that it is set to zero:
z2 + 4z - 27 = 2z - 3
z2 + 4z - 27 + 3 = 2z
z2 + 4z - 2z - 24 = 0
z2 + 2z - 24 = 0
Next, factor the quadratic equation:
z2 + 2z - 24 = 0
(z - 4)(z + 6) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (z - 4) or (z + 6) must equal zero:
If (z - 4) = 0, z must equal 4
If (z + 6) = 0, z must equal -6
So the solution is that z = 4 or -6
If BD = 9 and AD = 14, AB = ?
| 10 | |
| 18 | |
| 5 | |
| 8 |
The entire length of this line is represented by AD which is AB + BD:
AD = AB + BD
Solving for AB:AB = AD - BDFor this diagram, the Pythagorean theorem states that b2 = ?
a2 - c2 |
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c2 - a2 |
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c2 + a2 |
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c - a |
The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2 + b2): c2 = a2 + b2 or, solved for c, \(c = \sqrt{a + b}\)
Solve for c:
6c + 3 > \( \frac{c}{1} \)
| c > -\(\frac{48}{53}\) | |
| c > \(\frac{4}{27}\) | |
| c > -\(\frac{4}{17}\) | |
| c > -\(\frac{3}{5}\) |
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.
6c + 3 > \( \frac{c}{1} \)
1 x (6c + 3) > c
(1 x 6c) + (1 x 3) > c
6c + 3 > c
6c + 3 - c > 0
6c - c > -3
5c > -3
c > \( \frac{-3}{5} \)
c > -\(\frac{3}{5}\)