Question

MATH HELP PLEASE
QUESTION IS ATTACHED
MEDAL!

Answer 1

Are my answers correct? Most importantly, how do you do Part C?

Answer 2

yes, but you could simplify A) to 4x+12

Part C asks for the ratio of the area of the original to the bigger :
\[ \frac{4(x+3)}{16(x+3)} \]
what do you think?

Answer 3

I don't really understand that question because of what it says about "be the same for any positive value of x?"

Answer 4

Can you help me understand that about what it means really?

Answer 5

If you look at the ratio of areas you have the formula
\[ \frac{4(x+3)}{16(x+3)} \]

notice that even if you change x, that ratio always simplifies to ¼

well, *almost* any x. There is a problem if x=-3, which causes a divide by 0, and we get an undefined number. So we can't say the ratio of the areas will *always* be the same, no matter what x is. But if x is positive, we will never divide by 0, so it is ok to claim the ratio will be the same, no matter what x is (as long as x>0)

Answer 6

So are you saying that the answer to Part C is "yes"?

Answer 7

yes, the answer is yes. the ratio of the areas will be ¼
this will be true for all values of x>0 (because we will always be able to simplify (x-3)/(x-3) to 1 if x > 0 (or any x ≠ -3, for that matter)

Answer 8

not clear ?

Answer 9

Well kind of. I am just a middle schooler so I don't really get that much about what you're talking about.

Answer 10

First, do you understand how to write down the ratio of the area of the original rectangle to the bigger rectangle ?

Answer 11

yes got tat

Answer 12

I assume you know that you can simplify
\[ \frac{4(x+3)}{16(x+3)} = \frac{1(x+3)}{4(x+3)} \]

Answer 13

yes (pretty much)

Answer 14

by dividing top and bottom by 4

Answer 15

yes got that also

Answer 16

last step. anything divided by itself is 1
so \[ \frac{1(x+3)}{4(x+3)} = \frac{1}{4} \]

Answer 17

and it does not matter what x is... you always get ¼
so the answer to the question
" Will the ratio.... always be the same?" is Yes, it will always be ¼

Answer 18

ok. got that thanks @phi

Answer 19

The reason they toss in the "for any positive value of x" is that if they did NOT mention that, you would have to say NO, the ratio is not always ¼, because when x= -3
you would have
\[ \frac{1(x+3)}{4(x+3)} = \frac{1(-3+3)}{4(-3+3)} = \frac{0}{0} \]
and that is undefined. Who ever asked the question did not want you to answer no (based on a technicality) so they mentioned x>0

Answer 20

one last observation. The idea is that if you double the sides of a rectangle you will cause its area to go up not by 2, but by a factor of 4. The shape does not matter. So if you are told a rectangle has area 1, and you double the length of both its sides (whatever they are), you will get an area of 4.