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1 year ago

a rectangular garden is fenced on alls ides with 256 feet fencing. the garden is 8 feet longer than it is wide

Mathematics

1 answer:

pickupchik [31]1 year ago

6 0

The length and width of the rectangular garden is estimated as-

Length = 68 ft
width= 60 ft

<h3>How to calculate perimeter of the rectangle?</h3>

A rectangle's perimeter is the total length and distance of its border across all sides. The perimeter of a rectangular box is a linear measurement that is expressed in meters, feet, cm, or yards. Let us first examine the two primary characteristics of a rectangle.

A rectangle's four angles are all 90°.
A rectangle's opposite sides are of equal length.

The perimeter of the rectangle is given as-

Perimeter = 2(length + width)

perimeter= 256

Let the width of the rectangle be x.

Length is 8 feet longer than width, Thus,

length = x+8

Substitute the values in the formula of perimeter;

256 = 2( x+ 8+ x)

256 = 2( 2x + 8)

128= 2x + 8

120= 2x

x=60

Thus, width = 60 ft.

Length = 60 + 8 = 68 ft.

Therefore, the length and width of the rectangular garden is estimated.

To know more about the area of the rectangle, here

brainly.com/question/2607596

#SPJ4

The complete question is -

A rectangular garden is fenced on all sides with 256 feet of fencing. The garden is 8feet longer than it is wide. Find the length and width of the garden.

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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

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Now, we slice our solid into n slices.

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So the volume of each slice is

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

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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

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

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3 years ago

Will ran 120 minutes in the past two days. He ran 20 more minutes the second day than the first day. how many minutes did he run

valentinak56 [21]

Here is the equation: x+(x+20) = 120

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