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ziro4ka [17]
2 years ago
11

Joe made a scale drawing of the community pool in his town. the pool has a perimeter of 77 meters, what is the length and width

in meters?

Mathematics
1 answer:
viva [34]2 years ago
8 0

Answer:

The length of the pool is 28 meters and the width of the pool is 10.5 meters.

Step-by-step explanation:

In the scale drawing of Joe, the community pool has (length : width) = 8 : 3

Let the actual length of the pool is 8x meters and the actual width is 3x meters.

Now, given that the actual pool has a perimeter of 77 meters.

So, 2(8x + 3x) = 77

⇒ 22x = 77

⇒ x = 3.5

So, the length of the pool is 8x = 8(3.5) = 28 meters and the width of the pool is 3x = 3(3.5) = 10.5 meters. (Answer)

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Dennis_Churaev [7]

Answer:

1.5 kiloleter

Step-by-step explanation:

1 kiloleter = 1000 liters

so

1500 liters = 1.5 kiloleters

pls mark brainliest if possible lol

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*WILLL GIVE BRAINIEST TO COORECT RESPONCE*<br> how do i do this?<br> what is the answer?
Sergio [31]

Answer:

line mo

Step-by-step explanation:

Because the triangle are congrument

RST and MNO are the triangle

RT is a line in the RST triqangle

it is equal to the MO line in MNO

3 0
2 years ago
Find the length of LM. Show your work.
kicyunya [14]

Answer:

<h2>20</h2>

<em>Solution</em><em>,</em>

<em><</em><em>N=</em><em>1</em><em>8</em><em>0</em><em>-</em><em>5</em><em>3</em><em>-</em><em>4</em><em>4</em>

<em> </em><em> </em><em> </em><em> </em><em>=</em><em>8</em><em>3</em>

<h3><em>Apply </em><em>sines </em><em>rule,</em></h3>

<em>\frac{lm}{sin \: 83}  =  \frac{14}{sin44}  \\ lm =  \frac{14 \: sin \: 83}{sin \: 44}  \\ lm = 20</em>

<em>hope </em><em>this </em><em>helps.</em><em>.</em>

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Y_Kistochka [10]

Answer:

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Step-by-step explanation:

6 0
2 years ago
Any volunteer please help
den301095 [7]

Answer:

Part A)

The equation in the point-slope form is:

y-11=\frac{4}{3}\left(x-\left(-2\right)\right)

Part B)

The graph of the equation is attached below.

Step-by-step explanation:

Part A)

Given

  • The point = (-2, 11)
  • m = 4/3

The point-slope form of the line equation is

y-y_1=m\left(x-x_1\right)

Here, m is the slope and (x₁, y₁) is the point

substituting the values m = 4/3 and the point (-2, 11)  in the point-slope form of the line equation

y-y_1=m\left(x-x_1\right)

y-11=\frac{4}{3}\left(x-\left(-2\right)\right)

Thus, the equation in the point-slope form is:

y-11=\frac{4}{3}\left(x-\left(-2\right)\right)

Part B)

As we have determined the point-slope form which passes through the point (-2, 11) and has a slope m = 4/3

The graph of the equation is attached below.

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