Answer:

Step-by-step explanation:
Let w and l be the width and length of the garage.
Let
and
be the area of garage at present and new garage.
Given:
The area of the garage at present

And he planed to tripling the dimensions of the garage.
We need to find the area of the new garage.
Solution:
We know the area of the rectangular garage.


---------------(1)
The dimension of the new garage is triple, so the area of the new garage is.


Substitute
from equation 1.


Therefore, the area of the new garage 
The earth is approximately 9.2*10^7 farther than mercury
Answer:
23/16
Just set the missing fraction has X, and multiply the numerator and denominator of 3/2 by 8 so that it equals 24/16. This way you can isolate X and get x=23/16. It can't be simplified further, so that should be your final answer.
Answer:
9x2+4x−14
Step-by-step explanation:
Let x be the width of the sidewalk and the area becomes:
A=LW and L=6+2x and W=4+2x now we have
A=(6+2x)(4+2x) and we are told that A=48ft^2
48=24+20x+4x^2
4x^2+20x-24=0
4(x^2+5x-6)=0
x^2+5x-6=0
x^2-x+6x-6=0
x(x-1)+6(x-1)=0
(x+6)(x-1)=0
So x=-6, 1, however since x is a measurement it must be positive thus
x=width=1 ft is the only possible solution.