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1 year ago

Help with math Plsss I don't understand how to do this (I got this wrong but I have another try)

Mathematics

2 answers:

Anuta_ua [19.1K]1 year ago

6 0

<h3>Answer: 920</h3>

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Explanation:

The front face is a triangle with area of base*height/2 = 15*8/2 = 60 square feet.

The back face is identical to the front face, so we have another 60 square feet.

The left rectangular wall is 20 ft by 8 ft tall. Its area is 20*8 = 160 ft^2

The right slanted rectangular face is 20 ft by 17 ft. Its area is 20*17 = 340 ft^2

Lastly, the bottom rectangle floor is 15 ft by 20 ft to give an area of 15*20 = 300 ft^2

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To summarize so far

Front face = 60 ft^2
Back face = 60 ft^2
Left face = 160 ft^2
Right slanted face = 340 ft^2
Bottom floor = 300 ft^2

Add up those areas to get the overall surface area.

60+60+160+340+300 = 920

8090 [49]1 year ago

5 0

I may be wrong but I think the answer should be 920ft squared. Maybe wait for someone else to confirm this. Good luck!

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3 years ago

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</span></span>

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2 years ago

How do you do this question?

Ksivusya [100]

Answer:

V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

$\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}$

The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.

$V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\$

$\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)$

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.

5 0

3 years ago

Help with math Plsss I don't understand how to do this (I got this wrong but I have another try)​

Help with math Plsss I don't understand how to do this (I got this wrong but I have another try)