Hey! So, here's a tip. When writing exponents, an easier way is to write a^b, rather than a to the b power. Besides that, here is your answer!

So-------

9^3=729

3^2=9

6^3=216

15^2=225

Now that we have that figured out, we can add them together, wish is simple. 729 + 9 + 216 + 225= 1,179.

Therefore, your final answer will be 1,174.

If you have any questions on this, I'm happy to help you. :)

3 0

3 years ago

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use the general slicing method to find the volume of The solid whose base is the triangle with vertices (0 comma 0 ), (15 comma

lyudmila [28]

Answer:

volume V of the solid

$\boxed{V=\displaystyle\frac{125\pi}{12}}$

Step-by-step explanation:

The situation is depicted in the picture attached

(see picture)

First, we divide the segment [0, 5] on the X-axis into n equal parts of length 5/n each

[0, 5/n], [5/n, 2(5/n)], [2(5/n), 3(5/n)],..., [(n-1)(5/n), 5]

Now, we slice our solid into n slices.

Each slice is a quarter of cylinder 5/n thick and has a radius of

-k(5/n) + 5 for each k = 1,2,..., n (see picture)

So the volume of each slice is

$\displaystyle\frac{\pi(-k(5/n) + 5 )^2*(5/n)}{4}$

for k=1,2,..., n

We then add up the volumes of all these slices

$\displaystyle\frac{\pi(-(5/n) + 5 )^2*(5/n)}{4}+\displaystyle\frac{\pi(-2(5/n) + 5 )^2*(5/n)}{4}+...+\displaystyle\frac{\pi(-n(5/n) + 5 )^2*(5/n)}{4}$

Notice that the last term of the sum vanishes. After making up the expression a little, we get

$\displaystyle\frac{5\pi}{4n}\left[(-(5/n)+5)^2+(-2(5/n)+5)^2+...+(-(n-1)(5/n)+5)^2\right]=\\\\\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2$

But

$\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2=\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}((5/n)^2k^2-(50/n)k+25)=\\\\\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)$

we also know that

$\displaystyle\sum_{k=1}^{n-1}k^2=\displaystyle\frac{n(n-1)(2n-1)}{6}$

and

$\displaystyle\sum_{k=1}^{n-1}k=\displaystyle\frac{n(n-1)}{2}$

so we have, after replacing and simplifying, the sum of the slices equals

$\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)=\\\\=\displaystyle\frac{5\pi}{4n}\left(\displaystyle\frac{25}{n^2}.\displaystyle\frac{n(n-1)(2n-1)}{6}-\displaystyle\frac{50}{n}.\displaystyle\frac{n(n-1)}{2}+25(n-1)\right)=\\\\=\displaystyle\frac{125\pi}{24}.\displaystyle\frac{n(n-1)(2n-1)}{n^3}$

Now we take the limit when n tends to infinite (the slices get thinner and thinner)

$\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}\displaystyle\frac{n(n-1)(2n-1)}{n^3}=\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}(2-3/n+1/n^2)=\\\\=\displaystyle\frac{125\pi}{24}.2=\displaystyle\frac{125\pi}{12}$

and the volume V of our solid is

$\boxed{V=\displaystyle\frac{125\pi}{12}}$

3 0

3 years ago

a school bought 831 boxes of computer paper for the computer lab. Each box has 59 sheets of paper inside it. how many sheets of

Bess [88]

<span> 831 *59=49,029

hope this helps!</span>

4 0

3 years ago

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