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2 years ago

A. Car B was 20 miles east of car A

Mathematics

1 answer:

cupoosta [38]2 years ago

4 0

Answer:

Step-by-step explanation:

Its D because If one was going 60 and then the second one went 150 on the same road. They both had to start off low and the lowest number closer to those to are 30.

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What are two integers whose product is 0

guapka [62]

Answer:

At least one of the integer must be equal to zero to obtain the product zero

Step-by-step explanation:

we know that

When multiplying integers , if one or both of the integers is 0, then the product is 0.

The multiplication property states that the product of any number and zero is zero.

a*(0)=0

therefore

At least one of the integer must be equal to zero to obtain the product zero

therefore

7 0

2 years ago

Please help

katrin2010 [14]

$\dfrac{n^2+3n+2}{n^2+6n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n^2+2n+n+2}{n^2+4n+2n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n(n+2)+1(n+2)}{n(n+4)+2(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{(n+2)(n+1)}{(n+2)(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{n+1}{n+4}-\dfrac{2n}{n+4}=\dfrac{n+1-2n}{n+4}=\dfrac{1-n}{n+4}$

$Answer:\ \boxed{B.\ \dfrac{1-n}{n+4}}$

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3 years ago

Which inequality represents this graph?

murzikaleks [220]

The answer is y
$y > 4$

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2 years ago

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777dan777 [17]

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2 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)

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

and the volume V of our solid is

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

3 0

2 years ago