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Whitepunk [10]
2 years ago
15

The area of sector AOB is 20.25 pi ft ^2 . Find the exact area of the shaded region.

Mathematics
1 answer:
pishuonlain [190]2 years ago
4 0

Remark

If you have the area of the sector, then what you need is the area of the triangle in order to find the area of the shaded area. Just subtract the two.

Area of the triangle

The right angle makes finding the base and height very easy since they are equal

Area_Triangle = 1/2 * b * h

b = 9

h = 9

Area = 1/2 * 9 * 9

Area = 1/2 * 81

Area = 40.5

Area of the Shaded Area

Area_shaded = area of the sector - area of the triangle.

Area_shaded = 20.25 * pi - 40.5

Area_shaded = 20.25(pi - 2) <<<< Answer

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3 years ago
0.5 of what number is 15
Alex

Answer:

30

Step-by-step explanation:

0.5 *  unknown number = 15

but  

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6 0
3 years ago
Please help, this chapter was on derivatives...
weqwewe [10]

(3) Differentiating both sides of

2x^{3/2} + y^{3/2} = 29

with respect to <em>x</em> gives

3x^{1/2} + \dfrac32 y^{1/2} \dfrac{\mathrm dy}{\mathrm dx} = 0

Solve for d<em>y</em>/d<em>x</em> :

\dfrac32 y^{1/2} \dfrac{\mathrm dy}{\mathrm dx} = -3x^{1/2} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{-3x^{1/2}}{\frac32y^{1/2}} = \dfrac{-2x^{1/2}}{y^{1/2}} = -2\sqrt{\dfrac xy}

Then the slope of the tangent line to the curve at (1, 9) is

\dfrac{\mathrm dy}{\mathrm dx} = -2\sqrt{\dfrac19} = -\dfrac23

The equation of the tangent line would then be

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(4) The slope of the tangent line to

y=\dfrac{ax+1}{x-2}

at a point <em>(x, y)</em> on the curve is

\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{a(x-2)-(ax+1)}{(x-2)^2} = -\dfrac{2a+1}{(x-2)^2}

When <em>x</em> = -1, we have a slope of 2/3, so

-(2<em>a</em> + 1)/(-1 - 2)² = 2/3

Solve for <em>a</em> :

-(2<em>a</em> + 1)/9 = 2/3

2<em>a</em> + 1 = -18/3 = -6

2<em>a</em> = -7

<em>a</em> = -7/2

7 0
2 years ago
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Answer:

the scale factor is 2.

Step-by-step explanation:

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8 0
3 years ago
Tell whether the lines that pass through the following points are parallel,
wel

Answer:

These Lines will be parallel.

Step-by-step explanation:

You have to find the slope of each line. For (1,1) and (3,3) the slope is 1 and for Line 2 the slope is also 1. (let me know if you want me to show the work for slope). To have parallel lines the slopes have to be the same and y-ints have to be different. Here the slopes are the same and in Line 1 the yint is (0,0) which is different than Line 2. Therefor the Lines are parallel.  

Sorry if im wrong!. Hope this helps

4 0
3 years ago
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