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Lesechka [4]
3 years ago
12

Find the perimeter of a rectangle of which length is 5 more than the width

Mathematics
2 answers:
Leokris [45]3 years ago
8 0

Answer:

<em>the</em><em> </em><em>perim</em><em>eter</em><em>=</em><em> </em><em>1</em><em>0</em><em>+</em><em>4</em><em>W</em><em>.</em>

<em>where</em><em> </em><em>W</em><em>=</em><em> </em><em>the</em><em> </em><em>wid</em><em>th</em><em>.</em>

Step-by-step explanation:

mathematically, perimeter (p)= 2L +2W

where L= length.and W=width

<em>from</em><em> </em><em>the</em><em> </em><em>exp</em><em>ression</em><em> </em><em>;</em><em>length is 5 more than the width</em>

<em>it</em><em> </em><em>is</em><em> </em><em>wri</em><em>tten</em><em> </em><em>as</em><em> </em>L = 5+W

<em>sub</em><em>stitute</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>formu</em><em>la</em><em> </em>

P= 2(5+W)+2W

P=10+2W+2W

P=10+4W as the perimeter of the rectangle.

MA_775_DIABLO [31]3 years ago
5 0

Answer:

P = 4W + 10

Step-by-step explanation:

Perimeter is 2 × (length + width)

let length be L and width be W

if length is 5 more than width

L = W + 5

P = 2(L + W)

P = 2(W + 5 + W)

P = 2(2W + 5)

P = 4W + 10

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