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Alex_Xolod [135]
2 years ago
10

The seat of a bench is the shape of a Trapezoid with bases of 8 feet and 7 feet and a height on 3.5 feet whats the are of the se

at
Mathematics
1 answer:
Pepsi [2]2 years ago
6 0

Answer:

26.25 square feet

Step-by-step explanation:

Calculation to determine the area of the seat

Using this formula

Area=1/2*(b1+b2)*h

Where,

b1=8 feet

b2=7 feet

h=3.5 feet

Let plug in the formula

Area=1/2*(8+7)*3.5

Area=1/2*15*3.5

Area=26.25

Therefore the area of the seat is 26.25 square feet

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Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )
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(a) First find the intersections of y=e^{2x-x^2} and y=2:

2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}

So the area of R is given by

\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx

If you're not familiar with the error function \mathrm{erf}(x), then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line y=1 with y=e^{2x-x^2}.

1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2

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\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx
\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx

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(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve y=e^{2x-x^2} and the line y=1, or e^{2x-x^2}-1. The area of any such circle is \pi times the square of its radius. Since the curve intersects the axis of revolution at x=0 and x=2, the volume would be given by

\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx
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-10q+10q=-10q+10q-7

0=-7 ( zero will never equal -7 hence there are no solutions to this equation)

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