Sagot :
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 :)
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 :)
A- [tex]a_{1} = length , b_{2}=width, A_{1}= Area; [/tex]
[tex]a_{2}=na_{1} , b_{2}=nb_{1} ==> A_{2}= 2^{n} A_{1}[/tex]
if we double the dimensions of the rectangle, the area will be fourfold:
[tex]A_{2} = 2^2[4(x+3)]=16x+48[/tex]
B- yes, it will always be the same because:
[tex] \frac{A_{2}}{A_{1}} = \frac{4(4(x+3))}{4(x+3)} =4[/tex]
[tex]a_{2}=na_{1} , b_{2}=nb_{1} ==> A_{2}= 2^{n} A_{1}[/tex]
if we double the dimensions of the rectangle, the area will be fourfold:
[tex]A_{2} = 2^2[4(x+3)]=16x+48[/tex]
B- yes, it will always be the same because:
[tex] \frac{A_{2}}{A_{1}} = \frac{4(4(x+3))}{4(x+3)} =4[/tex]