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Sagot :
So,
The problem tells us that the length of the rectangle is 5 feet more than twice the width.
Mathematically:
length = 5 + 2width
l = 5 + 2w
We know that the perimeter is 2l + 2w.
2l + 2w = 130
2(5 + 2w) + 2w = 130
Distribute
10 + 4w + 2w = 130
Collect Like Terms
10 + 6w = 130
Subtract 10 from both sides
6w = 120
Divide both sides by 6
w = 20
The width is 20.
Since
l = 5 + 2w
Substitute
l = 5 + 2(20)
Multiply
l = 5 + 40
Collect Like Terms
l = 45
The length is 45.
Check
2l + 2w = 130
2(45) + 2(20) = 130
90 + 40 = 130
130 = 130
Width: 20 ft.
Length: 45 ft.
The problem tells us that the length of the rectangle is 5 feet more than twice the width.
Mathematically:
length = 5 + 2width
l = 5 + 2w
We know that the perimeter is 2l + 2w.
2l + 2w = 130
2(5 + 2w) + 2w = 130
Distribute
10 + 4w + 2w = 130
Collect Like Terms
10 + 6w = 130
Subtract 10 from both sides
6w = 120
Divide both sides by 6
w = 20
The width is 20.
Since
l = 5 + 2w
Substitute
l = 5 + 2(20)
Multiply
l = 5 + 40
Collect Like Terms
l = 45
The length is 45.
Check
2l + 2w = 130
2(45) + 2(20) = 130
90 + 40 = 130
130 = 130
Width: 20 ft.
Length: 45 ft.
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