A rectangle has a perimeter of 80. if the length is four more than twice the width, what is the length of the rectangle
1 answer:
Answer:
28
Step-by-step explanation:
l = length
w = width
p = perimeter
w = w
l + l + w + w = p (adding all the rectangle's sides together gets us our perimeter)
l = 2w + 4 (four more than twice the width)
Substitute 2w + 4 for 'l' in the original equation
(2w + 4) + (2w + 4) + w + w = 80
Solve equation normally:
6w + 8 = 80
6w = 72
w = 72/6 = 12
(w = 12)
Length = 2w + 4
Substitute 12 for 'w':
2(12) + 4 = l
24 + 4 = l
28 = l
The length is 28
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==jding713==
P=4*a, where P - perimeter, a - length of one side of the square.
if a=x+3 and P=48, then 4(x+3)=48;
x=9.
Answer: 9 units.
Tenths: 1.1
Hundredths: 1.07
Thousandths: 1.068
