3 years ago

How to solve number 1

Mathematics

1 answer:

Mkey [24]3 years ago

4 0

You obviously solve whatever the problem is

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What is the length and width of a room that is 60 square meters and has a perimeter of 32 meters

AleksAgata [21]

L = lenght
W = width

$L * W = 60 \\
2L + 2W=32\\\\
L = \frac{32-2W}{2}\\
\frac{32-2W}{2}*W = 60\\
(16-W)*W = 60\\
-W^{2} + 16W -60 = 0\\\\

d = 16^{2} - 4*1*60 = 256 - 240 = 16 \\\\
W1 = \frac{-16+\sqrt{16}}{-2} = \frac{-12}{-2} = 6\\
W2 = \frac{-16-\sqrt{16}}{-2} = \frac{-20}{-2} = 10\\

L1 = 10\\
L2 = 6\\
$

so

{L,W} = {6,10} or {10,6}

but L > W so

L = 10
W = 6

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