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Andrej [43]
4 years ago
8

A rectangle field has an area of 300 square meters and a perimeter of 80 meters. what are the length and width of the field?

Mathematics
1 answer:
Yakvenalex [24]4 years ago
6 0
By definition the area of a rectangle is:
 A = w * l
 The perimeter is:
 P = 2w + 2l
 Where,
 w: width
 l: long
 Substituting values:
 300 = w * l
 80 = 2w + 2l
 Solving the system of equations we have:
 l = 30 m
 w = 10 m
 Answer:
 
The length and width of the field are:
 
l = 30 m
 
w = 10 m
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Which choice is equivalent to the expression below? 5√10+√40+√90 <br><br> the answer was 10√10
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Answer:

5\sqrt{10}+\sqrt{40} +\sqrt{90}=10\sqrt{10}

Step-by-step explanation:

We have the following expression:

5\sqrt{10}+\sqrt{40} +\sqrt{90}

Remember the property of root multiplication:

\sqrt[n]{x*y}=\sqrt[n]{x}\sqrt[n]{y}

Therefore:

\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}

Also:

\sqrt{90}=\sqrt{9*10}=\sqrt{9}*\sqrt{10}=3\sqrt{10}

Then we have that:

5\sqrt{10}+\sqrt{40} +\sqrt{90}=5\sqrt{10}+2\sqrt{10}+3\sqrt{10}

Remember the property of sum of roots:

a\sqrt{b}+2a\sqrt{b}=3a\sqrt{b}

Finally:

5\sqrt{10}+\sqrt{40} +\sqrt{90}=(5+2+3)\sqrt{10}

5\sqrt{10}+\sqrt{40} +\sqrt{90}=10\sqrt{10}

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