Respuesta :
Answer:
a) Length = [tex]8x[/tex]
b) Width = 27 units
Step-by-step explanation:
We are given the following in the question:
Area of rectangle, A =
[tex]A = 16x^2 + 56x[/tex]
Width of rectangle, w =
[tex]w = 2x + 7[/tex]
a) Length of the rectangle.
Area of rectangle is given by:
[tex]A = l\times w[/tex]
where l is the length of rectangle.
Putting values, we get,
[tex]16x^2 + 56x = a(2x+7)\\\\l = \dfrac{16x^2 + 56x}{2x + 7}\\\\l = \dfrac{8x(2x+7)}{2x+7}=8x[/tex]
b) Width of rectangle
Length of rectangle = 80
[tex]8x = 80\\\Rightarrow x = 10[/tex]
Thus, width of rectangle is
[tex]2x + 7\\=2(10) + 7= 27[/tex]
Thus, the width of rectangle is 27 units.
Part(a): The required expression is [tex]8x[/tex]
Part(b): The required length is 10 units and breadth is 27 units.
Area of the rectangle:
The formula of the area of the rectangle is [tex]A=l\times b[/tex]
Given that.
The area of the rectangle =[tex]16x^2+56x[/tex]
Width of the rectangle=[tex]2x+7[/tex]
Part(a):
Now, the area of the rectangle is equal to [tex](l\times b)[/tex] then,
[tex]16x^2+56x=l(2x+7)\\8x(2x+7)=l(2x+7)\\l=8x[/tex]
Part(b):
As the length of the rectangle is 80 then,
[tex]8x=80\\x=\frac{80}{8} \\x=10[/tex]
And the width of the rectangle is ,
[tex]2x+7=2(10)+7\\=27[/tex]
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