Respuesta :
By definition the area of a rectangle is given by:
A = (w) * (l)
Where,
w: width
l: long
We rewrite the expression based on:
"The length of a rectangle is 3m longer than its width"
A = (w) * (w + 3)
We substitute the value of the area:
154 ^ 2 = (w) * (w + 3)
Rewrite:
23716 = w ^ 2 + 3w
w ^ 2 + 3w-23716 = 0
We solve the polynomial:
w = 152.5
Note: the other root is negative.
Answer:
the width of the rectangle is:
w = 152.5 m
A = (w) * (l)
Where,
w: width
l: long
We rewrite the expression based on:
"The length of a rectangle is 3m longer than its width"
A = (w) * (w + 3)
We substitute the value of the area:
154 ^ 2 = (w) * (w + 3)
Rewrite:
23716 = w ^ 2 + 3w
w ^ 2 + 3w-23716 = 0
We solve the polynomial:
w = 152.5
Note: the other root is negative.
Answer:
the width of the rectangle is:
w = 152.5 m
If the length of a rectangle is 3m longer than its width, then:
L=W+3
Is the area really 154^2? Or is it 154m^2? If yes, then:
A=LW
154=(W+3)(W)
154=(W^2+3W)
0=W^2+3W-154
0=(W-11)(W+14)
This means either (W-11) or (W+14) is equal to zero so:
W=11 and W=-14
To find out let's substitute the numbers:
154=(11+3)(11)
154=154
Therefore, the width of the rectangle is 11m
L=W+3
Is the area really 154^2? Or is it 154m^2? If yes, then:
A=LW
154=(W+3)(W)
154=(W^2+3W)
0=W^2+3W-154
0=(W-11)(W+14)
This means either (W-11) or (W+14) is equal to zero so:
W=11 and W=-14
To find out let's substitute the numbers:
154=(11+3)(11)
154=154
Therefore, the width of the rectangle is 11m