Question

CAN ANYONE PLEASE HELP WITH MATH?

Answer 1

A- $a_{1} = length , b_{2}=width, A_{1}= Area;$
<u />$a_{2}=na_{1} , b_{2}=nb_{1} ==> A_{2}= 2^{n} A_{1}$
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:
$\frac{A_{2}}{A_{1}} = \frac{4(4(x+3))}{4(x+3)} =4$

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 :)