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3 years ago

I WILL GIVE BRAINLIEST TO WHOEVER ANSWERS ALL 5 QUESTIONS AND I WILL ALSO GIVE 11 POINTS. PLEASE HELP ME I NEED IT SO MUCH.

Mathematics

1 answer:

balandron [24]3 years ago

7 0

1. (-5,-11)
2. C, reflected across x-axis then translated.
3. A’(-1,7), B’(-4,6), C’(-2,2)
4. C’(3,-4)
5. B, rotates 90 counter clockwise and then translate

(if any of these are wrong i’m sorry! :<)

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What is the midpoints of a segment whose endpoints are (-3, 8) and (7, -2) ?

Sidana [21]

$M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
$M = (\frac{7 - (-3)}{2}, \frac{-2 + 8}{2})$
$M = (\frac{7 - 3}{2}, \frac{6}{2})$
$M = (\frac{4}{2}, 3)$
$M = (2, 3)$

3 0

3 years ago

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)

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