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lana [24]
3 years ago
11

Find the perimeter of this figure. (Image down below)

Mathematics
2 answers:
Schach [20]3 years ago
7 0
Since perimeter is adding all the sides together it would be (8×2 = 16) and (14×2 = 28 ) so 28+16=44 ^^ I agree with the answer above :)
Jet001 [13]3 years ago
6 0

Answer:

The answer is

<h2>44 ft</h2>

Step-by-step explanation:

The figure above is a rectangle

To find the perimeter of a rectangle we use the formula

Perimeter of a rectangle = 2l + 2w

where

l is the length of the rectangle

w is the width of the rectangle

From the question

length = 14 ft

width = 8ft

Substituting the values into the above formula we have

Perimeter = 2(14) + 2(8)

= 28 + 16

We have the final answer as

<h3>Perimeter = 44 ft</h3>

Hope this helps you

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Helppppp pls im stuck on thi one problem :(
Ray Of Light [21]

First, find the simplified ratio of the given ratio:

Number of pirates: 22

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Divide 22 with 2 to find the ratio: 22/2 = 11

For every ship there is, there will be 11 pirates.

Now, find the ratio for the other questions:

IF there are 5 ships: 5 x 11 = 55

If there are 5 ships, there will be 55 pirates.

If there are 11 ships: 11 x 11 = 121

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5 0
3 years ago
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Solve the following systems of inequalities and select the correct graph: 2x-y&gt;4 x+y&lt; -1
Serga [27]

Answer:

Step-by-step explanation:

2x - y > 4 ⇒ y < 2x - 4

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3 0
3 years ago
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sertanlavr [38]

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5 0
2 years ago
Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul
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Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

\bf B=\displaystyle\frac{L}{4\pi d^2}

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<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

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Hence the apparent brightness of  Alpha Centauri A is

\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728

According to the inverse square law for light intensity

\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}

where

\bf I_1= light intensity at distance \bf d_1  

\bf I_2= light intensity at distance \bf d_2  

Let \bf d_2  be the distance we would have to place the 50-watt bulb, then replacing in the formula

\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

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