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STatiana [176]
2 years ago
4

Ashley has a sprinkler that has several varieties of coverage. The quarter circle sprinkler head sprays water a distance of up t

o 20 feet from the lead. What area will be covered by the spray of the quartercircle sprinkler head to the nearest square foot?
Mathematics
1 answer:
Vladimir [108]2 years ago
8 0

The area the quarter-circle sprinkler sprays with water is 314 square feet.

Step-by-step explanation:

Step 1; It is given that the quarter-circle sprinkler can reach a distance of 20 feet, the radius of the circle will be equal to 20 feet.

Step 2; The area of any given circle is π times the square of the radius. The radius of this circle is equal to 20 feet.

Area of the circle = π × r² = 3.1415 × 20 × 20 = 1,256.6370 square feet.

Step 3; 1,256.6370 square feet is the area of a full circle with a radius of 20 feet. The sprinkler only covers a quarter of the circle so we divide the area by 4 to convert it into a quarter-circle.

Area of the quarter circle = 1,256.6370 square feet / 4 = 314.1592 square feet.

So the quarter sprinkler covers an area of 314.1592 square feet.

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Clara lends half her collection of formal dresses, d, to her sister, Susan. Clara then buys four more dresses. If she has 12 now
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Answer:

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2 years ago
Aisha owns an apparel store. She bought some shirts and jeans from a wholesaler for $3,900. Each shirt cost $12 and each pair of
Lilit [14]
For this case we can define the following variables:
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5 0
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Consider the curve y=4x-x^3 and the chord AB joining points A(-3, 15) and B(2, -15) on the curve. Find the x- and y-coordinates
Elis [28]
We are asked to find the coordinates of the points where the slope of the curve is equal to the slope of the chord AB. We are given the coordinates of A and B. We can use these and the slope formula to find the slope of the chord.

Recall the slope formula is: m= \frac{ y_{2} - y_{1} }{ x_{2} - x_{1} }
We label the points as follows (note that you could change the way you label the points as you will get the same slope regardless of what point is designated with the 1s and which is designated with the 2s).
A=( x_{1} , y_{1)}=(-3,15)
B=( x_{2} , y_{2}=(2,-15)
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Next consider the curve y=4x- x^{3}. We can find the slope of the curve by taking the first derivative. Thus, the slope of the curve is given by: y^{'} =4-3 x^{2}

Let us set the slope of the curve equal to -6 to find the points (x,y) where the chord and the curve have the same slope.

That is, -6=4-3 x^{2}
3 x^{2} =10
x^{2} = \sqrt{ \frac{10}{3} }
x= \sqrt{ \frac{10}{3} },x= -\sqrt{ \frac{10}{3} }

We now have the x-coordinates of the points at which the slope of the chord equals that of the curve. To find the y-coordinates we substitute the values we found for x in the original equation as follows:

y=4( \sqrt{ \frac{10}{3} }) -( \sqrt{ \frac{10}{3} }) ^{3} =4(  \frac{10}{3}) ^{ \frac{1}{2} }-( \frac{10}{3}) ^{ \frac{3}{2} }
and
y=4( -\sqrt{ \frac{10}{3} }) -(- \sqrt{ \frac{10}{3} }) ^{3} =-4( \frac{10}{3}) ^{ \frac{1}{2} }+( \frac{10}{3}) ^{ \frac{3}{2} }

Thus the two points we seek (x,y) are:
( \sqrt{ \frac{10}{3} }, 4( \frac{10}{3}) ^{ \frac{1}{2} }-( \frac{10}{3}) ^{ \frac{3}{2} } )
and
( -\sqrt{ \frac{10}{3} }, -4( \frac{10}{3}) ^{ \frac{1}{2} }+( \frac{10}{3}) ^{ \frac{3}{2} } )


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