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3 years ago

13

The length of a rectangular garden is 15 less than twice the width. The area is 3375 ft^2. What are the dimensions of the garden

?

1 answer:

tatyana61 [14]3 years ago

8 0

The answer could 156 or 268 Thats what i got

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PLEASE HELP! Solve for q(x) and r(x). (4x^3-3x^2+5x-7)/(x-2) = q(x)+ r(x)/(x-2)

Lerok [7]

Asked and answered elsewhere.
brainly.com/question/9169487

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4 years ago

Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.

Amiraneli [1.4K]

The side x is equal to 12

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3 years ago

Please I need some help!

Schach [20]

(f+g)(x) = 8x+2

(f-g)(x) = 4x-6

Step-by-step explanation:

Given

$f(x) =6x-2\\g(x)=2x+4$

we have to find

<u>a. (f+g)(x)</u>

$(f+g)(x)=f(x)+g(x)\\=(6x-2)+(2x+4)\\=6x-2+2x+4\\=6x+2x-2+4\\=8x+2$

<u>b. (f-g)(x)</u>

$(f-g)(x)=f(x)-g(x)\\=(6x-2)-(2x+4)\\=6x-2-2x-4\\=6x-2x-2-4\\=4x-6$

Hence,

(f+g)(x) = 8x+2

(f-g)(x) = 4x-6

Keywords: Functions, Operations on Functions

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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)

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

and the volume V of our solid is

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

3 0

3 years ago

Tim added two fractions with denominators of 4 but with different numerators. The sum was 3 over 4. Which two fractions did Tim

puteri [66]

Answer:

It could be 2/4+1/4

2+1=3 and they both have a denominator of 4

4 0

3 years ago