Question

CAN ANYONE PLEASE HELP WITH MATH?

The rectangle has an area of 4(x+3) square units.
A- If the dimensions of the rectangle are doubled, what is the area of the new rectangle in terms of x? Show your work.


B- Will the ratio of the area of the original rectangle to the area of the larger rectangle be the same for any positive value of x? Explain.

Answer 1

A-
`a_(1) = length , b_(2)=width, A_(1)= Area;`

`a_(2)=na_(1) , b_(2)=nb_(1) ==> A_(2)= 2^(n) A_(1)`
if we double the dimensions of the rectangle, the area will be fourfold:

`A_(2) = 2^2[4(x+3)]=16x+48`

B- yes, it will always be the same because:

`(A_(2))/(A_(1)) = (4(4(x+3)))/(4(x+3)) =4`

Answer 2

A)
If it has an area of 4(x+3) we can think that one side has a length of 4, and the other has a length of (x+3).

So, if the dimensions were doubled, 4 x 2 = 8. And 2(x+3) = 2x+6.

The new area would be:

8(2x+6) = 16x+48.

B)
The ratio will be the same. For example lets plug in some points:

x=0
4(0+3) = 4(3) = 12

And
16(0)+48 = 0+48 = 48

So the ratio is 48/12 = 4

Lets plug in another point.

x=2
4(2+3)= 4(5) = 20

And
x=2
16(2)+48 = 32 + 48 = 80

80/20 = 4

So the ratio is the same :)