ilorty

Thanks heureka!

 I suggest you first draw it:

Ive also been getting a lot of the same problems, and it's pretty frustrating. But, oh well, the rest of the site seems to be fine.....

Hello! Are you able to see the picture,or was it blocked by moderator? Thanks!

First, multiply both sides by 3:

15c-12=21-(4-c)

15c-12=21-4+c

          =17+c

From here, subtract c from both sides:

14c-12=17

Add 12 to both sides:

14c=29

Can you solve it from here?

:)

acctually, you don't need to find sin(270) or log_2 2048: since cos(90) is 0, everything multiplied together is also 0

:)

Okay!

Since both 2 and 3 are prime numbers, we should find the prime factorization of \(12^2\cdot18^3\). So:

12 = 2*2*3

12=2*2*3

18=2*3*3

18=2*3*3

18=2*3*3

As you see, we have 8 3's and 7 2'. So, 8+7 is \(\boxed{15}\)

:)