Question

See attachment.

Answer 1

two theorems here\[\sqrt[b]{x^a}=x^{a/b}\]and\[x^ax^b=x^{a+b}\]

Answer 2

4y^7 is the answer i got from using the rules ceasefire3 wrote above

Answer 3

Okay, what that is, is

y^( 10/3 ) * 16y^( 11/3)

When we are multiplying indicies, we add them.

10/3 + 11/3 = 21/3, which is 7. So quite simply the answer is;

16y^7.

Answer 4

actually the 16 is cube rooted

Answer 5

thats why I got 4y^7 instead. also, i meant turingtest, not ceasefire, in my last comment

Answer 6

It isn't. The y is exclusive from the 16. If it were (16y)^11/3, then yes. But it isn't.

Answer 7

cube root 16 is not 4
4^3=64

Answer 8

oh yeah, did the sqrt, its 2y^7

Answer 9

Ignore me. The 16 should be cube rooted. I just looked at the equation again -.-

Answer 10

cube root 16 is not 2 either

Answer 11

\[\sqrt[3]{16}=2\sqrt[3]2\]

Answer 12

lol ok I'm Realy Confused!
Whats that answer.

Answer 13

\[2\sqrt[3]2x^7\]

Answer 14

I'm only telling because you got confused, normally I don't give answers like that.

Answer 15

x=y

Answer 16

Like this?

Answer 17

yes

Answer 18

lol x=y

Answer 19

there are no x's in your expression
only y'
that is why i said x=y
i wanted you to replace the x with y

Answer 20

y's*

Answer 21

I'm sorry I need to start reading more carefully :P
yes put a y there of course not x

Answer 22

ok THX