+9491
| 5x2 | - | 13x | + | 6 | Let's split the middle term. | ||||
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Since 5 * 6 = 30 , we need two numbers that add to -13 and multiply to 30 . What two numbers add to -13 and multiply to 30 ? → -10 and -3
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| = | 5x2 | - | 10x | - | 3x | + | 6 | Notice if we combine like terms, we get the original expression. | |
| Factor 5x out of the first two terms. | |||||||||
| = | 5x(x - 2) | - | 3x | + | 6 | ||||
| Factor -3 out of the last two terms. | |||||||||
| = | 5x(x - 2) | - | 3(x - 2) | ||||||
| Factor (x - 2) out of both terms. | |||||||||
| = | (x - 2)(5x - 3) | ||||||||
+9491 | gray circle area | = | π | * | (radius)2 | ||
| radius of the smaller circle = \(\frac{OD}{2}\) | ||||||
| gray circle area | = | π | * | (\(\frac{OD}{2}\))2 | ||
| gray circle area | = | π | * | \(\frac{OD^2}{4}\) | ||
| gray circle area | = | \(\frac{πOD^2}{4}\) | ||||
| larger circle area | = | π | * | (radius)2 | ||
| radius of the larger circle = OD | ||||||
| larger circle area | = | π | * | OD2 |
What, percent, is the gray circle's area out of the larger circle's area?
\(\frac{\text{gray circle area}}{\text{larger circle area}}\,=\,\frac{\frac{\pi OD^2}{4}}{\pi OD^2}\,=\,\frac{\pi OD^2}{4}\,*\,\frac{1}{\pi OD^2}\,=\,\frac14\,=\,\frac{25}{100}\,=\,25\%\)
.+9491 | f(x) | = | a * xr | ||
| f(2) | = | a * 2r | ||
| Since f(2) = 1 , we can replace f(2) with 1 . | ||||
| 1 | = | a * 2r | ||
| Divide both sides of the equation by 2r . | ||||
| \(\frac{1}{2^r}\) | = | a |
| f(32) | = | a * 32r | ||
| Since f(32) = 4 , we can replace f(32) with 4 . | ||||
| 4 | = | a * 32r | ||
| Divide both sides of the equation by 32r . | ||||
| \(\frac{4}{32^r}\) | = | a |
And since a = a .....
| \( \frac1{2^r}\) | = | \(\frac4{32^r}\) | ||
| Since 20 = 1 , 22 = 4 , and 25 = 32 , we can say... | ||||
| \( \frac{2^0}{2^r}\) | = | \( \frac{2^2}{(2^5)^r}\) | ||
| And (xa)b = xab | ||||
| \( \frac{2^0}{2^r}\) | = | \( \frac{2^2}{2^{5r}}\) | ||
| And \(\frac{x^a}{x^b}=x^{a-b}\) | ||||
| 20 - r | = | 22 - 5r | ||
| Now the bases are equal, so the exponents are equal. | ||||
| 0 - r | = | 2 - 5r | ||
| 0 | = | 2 - 4r | ||
| -2 | = | -4r | ||
| \(\frac12\) | = | r |
+9491 I think the points on the graph are:
(0, negative 4) and (pi over 2, 0) and (pi, 4) and (3 pi over 2, 0 ) and (2 pi, negative 4)
There is the point (0, -4) . So, when x = 0 , y = -4 .
There is the point (\(\frac{\pi}{2}\), 0) . So, when x = \(\frac{\pi}{2}\) , y = 0 .
average rate of change = \(\frac{\text{change in y}}{\text{change in x}}=\frac{-4 - 0}{0 - \frac{\pi}{2}}=\frac{4}{\frac{\pi}{2}}=4\cdot\frac2{\pi}=\frac8{\pi}\)
Here's a graph.
