What is the length and width of a rectangle with a perimeter of 28 feet and an area of 48 sq feet
2 answers:
Answer:
The rectangle is 8 ft by 6 ft.
Step-by-step explanation:
Let x = length and y = width
Then, 2x+2y = 28 and xy = 48
Since 2x+2y = 28, y = 14 - x
So, x(14 - x) = 48
x2 - 14x + 48 = 0
(x-8)(x-6) = 0
x = 8 or x = 6
Hope this helps!
Answer:
The dimensions are 6 ft by 8 ft
Step-by-step explanation:
Area = l*w
Perimeter = 2(l+w)
48 = lw
28 = 2(l+w)
Divide by 2
28/2 = 2/2(l+w)
14 = l+w
Subtract w from each side
14-w =l+w-w
14-w =l
Substitute this into the first equation for area
48 = lw
48 = (14-w) *w
48 = 14w - w^2
Subtract 14w from each side
48 -14w = 14w-14w -w^2
-14w +48 = -w^2
Add w^2 to each side
w^2 -14w +48 = -w^2+w^2
w^2 -14w +48 =0
Factor
(w-6) (w-8) = 0
Using the zero product property
w-6=0 w-8=0
w=6 w=8
So the width could be 6 or the width could be 8
If the width is 6
14-w =l
14-6 =l
8 =l
The length is 8
If the width is 8
14-w =l
14-8 =l
8 =l
The length is 6
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Step-by-step explanation:
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***
CHECK
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***
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Answer:
negative
Step-by-step explanation:
negative because you are losing something
4 is your answer
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so 4 is the amount
hope this helps
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