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3 years ago

Can you check my paper? If I got any wrong please tell me the correct answers.

Mathematics

1 answer:

djyliett [7]3 years ago

3 0

There is a lot to go over here. Unfortunately it looks like you got a lot incorrect. I'll focus on two problems. Hopefully these examples below will help correct the other mistakes.

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Problem 7)

We have the starting value be 20 and the ending value be 11. Subtract the values: (end)-(start) = 11 - 20 = -9. The negative indicates we have a drop or decrease.

We'll focus on the positive version of this number, so 9. Divide this value over the starting amount 20 to get 9/20 = 0.45 = 45%

So going from 20 miles to 11 miles is a decrease of 45%

Answer to problem 7 is: 45%

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Problem 13)

An increase of 300% means we added 3 times the original amount onto the original amount.

We take 300% of 25 to get 3*25 = 75

Which is then added onto 25 to get 25+75 = 100

Answer to problem 13 is: 100

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The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i

kirill115 [55]

Answer:

Required largest volume is 0.407114 unit.

Step-by-step explanation:

Given surface area of a right circular cone of radious r and height h is,

$S=\pi r\sqrt{r^2+h^2}$

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$f(r,h)=\frac{1}{3}\pi r^2 h$

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$g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)$

We know for maximum volume $r\neq 0$ . So let $\lambda$ be the Lagranges multipliers be such that,

$f_r=\lambda g_r$

$\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)$

And,

$f_h=\lambda g_h$

$\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}$

$\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)$

Substitute (3) in (2) we get,

$\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)$

$\implies h^2=2r^2$

Substitute this value in (1) we get,

$\pi r\sqrt{h^2+r^2}=8$

$\implies \pi r \sqrt{2r^2+r^2}=8$

$\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252$

Then,

$h=\sqrt{2}(1.21252)\equiv 1.71476$

Hence largest volume,

$V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114$

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3 years ago

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