Question

http://prntscr.com/6vicud
http://prntscr.com/6viczg

@lυἶცἶ0210

Answer 1

Lel xDDD

Answer 2

That same type of problem :P

Answer 3

Have you tried plugging in simple values? e.e

Answer 4

If they're equivalent they'll give you the same answer like the last one.

Answer 5

Hmm ok.

Answer 6

No I haven't plugged them (since I'm a type of person to solve and see)

Answer 7

Make it easy, try -4. the 0 will make it a 1

and if you want to actually solve.. ask Perl xD

Answer 8

But yea, plugging in -4 \(\Large m_a=2000(1.05)^{(-4+4)} \)

\(\Large m_a=2000(1.05)^{0} \)


\(\Large m_a=2000 \)

Answer 9

For checking equivalency I suggest option C, try it and see what happenss :3

Answer 10

Mkay I'll try that

Answer 11

Oh, btw, if you want to actually solve it, try using the exponent rules :)

Answer 12

Use: \(\Large x^{a+b} =a^a*a^b \)

Answer 13

Those a's should be x. sorry :P

Answer 14

It's fine lol I see it xD

Answer 15

Wait so it's somewhat similar to PEMDAS rules right? I have a headache right now so firgive me if I ask a trivial question.

Answer 16

*forgive

Answer 17

For order in what to do.

Answer 18

yeah

Answer 19

Thanks I got it. :)