Question

Assuming x ≠ 0 and y ≠ 0, what is the quotient of 8x^(9y³+2x^(6y²+8x³))/4y³

Answer 1

\[(8x ^{9y ^{3}+2x ^{6y ^{2}+8x ^{3}}})/4y ^{3}\]
is this it?

Answer 2

yes

Answer 3

plz help!

Answer 4

ireally need some help

Answer 5

\[8^{9y^3+2x^{6y^2+8x^3}}/ 4y^3\]
\[2^{3({9y^3+2x^{6y^2+8x^3})}}/ 2^2y^3\]
\[2^{{27y^3+6x^{6y^2+8x^3}}}/ 2^2y^3\]

Answer 6

i think all you can can do is eliminate the 4s

Answer 7

Yea, you can't simplify it farther. You can expand the exponent like I did above, but you can't simplify further.

Answer 8

@gatorgalvet what kind of answer are you ;looking for?

Answer 9

\[2^{{27y^3+6x^{6y^2+8x^3}}}/ 2^2y^3\]
Question says for x not= 1 and y not= 1. But x can = 1 and y = 1 . And so we can substitute 1 for both x and y and solve for the quotient:

\[2^{{27+6}}/ 2^2y^3\]
\[2^{{33}}/ 2^2y^3\]
From this, you could say the following, but this doesn't follow the rules of mathematics:

\[2^{{31}}/ y^3\]

but straightforward, with what we know,
\[2^{{33}}/ 2^2y^3\]
the above should be the simplest quotient we can get that follows the rules of mathematics.

Answer 10

@genius12 i don't think that's how this works, you can't substitute 1s without any indication that substitution is required or desired

Answer 11

Since we substituted 1 for y, it becomes this:
\[2^{{33}}/ 2^2\]
\[2^{31}\]
That would be your answer.