The length of a rectangle is 10 mm longer than its width. Its perimeter is more than 80 mm. Let w equal the

2 answers:

8 0

We know that the length (L) of the rectangle in question is 7mm longer than its width (W). Let's represent that as the following:
L=7+W

A rectangle's perimeter (the total sum of its sides) will be made my 2 sides representing the length (2L) and 2 sides representing the width (2W). We also know that this rectangle's perimeter is greater than 62. Since eventually we are solving for W, let's state all expressions in terms of W:
2L=2(7+W)
2(7+W)+2W>62
14+2W+2W>62
14+4W>62
4W>62-14
4W>48
W>48/4
W>12
If the rectangle's perimeter is greater than 62, then the width will be greater than 12. Let's confirm this:
Perimeter=2L+2W
P=2(7+12)+2(12)
P=14+24+24
P=62
So we can see that if the perimeter is to surpass 62, W needs to be greater than 12 and L ( which is also 7+W) needs to be greater than 19.

jok3333 [9.3K]3 years ago

3 0

P=L+L+W+W=2(L+W)
P>80
2(L+W)>80
divide 2
L+W>80

L is 10 more than W
L=10+W

A. L=10+W

B. L+W>80

C.
L+W>80
sub L=10+W
10+W+W>80
minus 10
2W>70
divid 2
W>35
W is mor than 35

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