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Alina [70]
2 years ago
9

the perimeter of a basketball court is 102 meters, and length is 6 meters longer than twice the width, what r length and width.

what is width _m.
Mathematics
1 answer:
stepladder [879]2 years ago
7 0
Perimeter = 2(l + w)

Thus,

102 = 2(l + w)

l = 6 + 2w

102 = 2( 6 + 2w + w)
102 = 12 + 4w +2 w
102 = 12 + 6w
90 = 6w

Thus, w = 15.

Length = 6 + 2w
= 6 + 30
= 36.

Thus, the length is 36 meters and width is 15 meters.
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just olya [345]

Answer & Step-by-step explanation:

In order to solve this problem, it's important that we look at the tiles and the the signs that are in front of them. The top row of tiles represents our first expression and the bottom row of tiles represents our second equation.

The two large tiles are positive so they are going to be positive in our equation.

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The four blue rectangle tiles are also positive, so they are going to be positive in our equation. The two red rectangle tiles are negative, so they are going to be negative in out equation.

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The two red square tiles are negative, so they are going to be negative in our equation. The four blue square tiles are positive, so they are going to be positive in our equation.

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2 years ago
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love history [14]

Answer:

C

Step-by-step explanation:

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we don't include the median in the count, only use it to divide the data into the two parts,

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2 years ago
How do you solve his with working
AlexFokin [52]
Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

\bf \begin{array}{cllll}&#10;\textit{circumference of a circle}\\\\ &#10;2\pi r&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{arc's length}\\\\&#10;s=\cfrac{\theta r\pi }{180}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+&#10;\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}&#10;\\\\\\&#10;15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.


\bf \begin{array}{cllll}&#10;\textit{area of a circle}\\\\ &#10;\pi r^2&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{area of a sector of a circle}\\\\&#10;s=\cfrac{\theta r^2\pi }{360}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}&#10;\\\\\\&#10;90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

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Dennis_Churaev [7]

Answer:

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Step-by-step explanation:

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