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Mathematics 15 Online
OpenStudy (anonymous):

simplify 6n^6/3n^4 help please !!!!!

jimthompson5910 (jim_thompson5910):

the coefficients are 6 and 3 you divide them to get 6/3 = 2

jimthompson5910 (jim_thompson5910):

the variable expressions are: n^6 and n^4 you divide them by subtracting the exponents: 6 - 4 = 2 the final exponent is 2

jimthompson5910 (jim_thompson5910):

which means that 6n^6/3n^4 = 2n^2

OpenStudy (anonymous):

thank you so so much you are very nice C:

jimthompson5910 (jim_thompson5910):

you're welcome

OpenStudy (anonymous):

hey can you help me with another one ? @jim_thompson5910

jimthompson5910 (jim_thompson5910):

sure one more

OpenStudy (anonymous):

\[[6x ^{2}y ^{4}/3x ^{4}y ^{3} ]^{3}\]

OpenStudy (anonymous):

help please c; @jim_thompson5910

jimthompson5910 (jim_thompson5910):

first i would simplify the stuff on the inside of the brackets

jimthompson5910 (jim_thompson5910):

so using what I showed you above, what do you get?

OpenStudy (anonymous):

2x^-2

jimthompson5910 (jim_thompson5910):

close, but you're forgetting about the y term

jimthompson5910 (jim_thompson5910):

y^4/y^3 = ???

OpenStudy (anonymous):

idk the answer to that

jimthompson5910 (jim_thompson5910):

subtract the exponents

OpenStudy (anonymous):

1.3?

jimthompson5910 (jim_thompson5910):

the exponents are 4 and 3, you subtract them to get 4-3 = 1

jimthompson5910 (jim_thompson5910):

so that means y^4/y^3 = y^1 = y make sense?

OpenStudy (anonymous):

oh ok i got it

jimthompson5910 (jim_thompson5910):

so this means that \[\large \left[\frac{6x ^{2}y ^{4}}{3x ^{4}y ^{3}} \right]^{3}\] turns into \[\large \left[2x^{-2}y \right]^{3}\]

jimthompson5910 (jim_thompson5910):

what's next?

OpenStudy (anonymous):

multiply them all by 3?

jimthompson5910 (jim_thompson5910):

close, very close

jimthompson5910 (jim_thompson5910):

you raise each term to the 3rd power

jimthompson5910 (jim_thompson5910):

you do that to the following terms: \(\large 2\), \(\large x^{-2}\), and \(\large y\)

OpenStudy (anonymous):

so what would be the answer ?

OpenStudy (skullpatrol):

$$2^3*(x^{-2})^3*y^3$$

jimthompson5910 (jim_thompson5910):

what do you get when you raise 2 to the 3rd power?

OpenStudy (anonymous):

8

OpenStudy (skullpatrol):

Yes.

jimthompson5910 (jim_thompson5910):

what happens when you raise \(\large x^{-2}\) to the 3rd power

OpenStudy (anonymous):

would it be 2n^2

OpenStudy (skullpatrol):

$$b^n=b*b*b*...*b$$n factors the base is b and the exponent is n

OpenStudy (skullpatrol):

$$(x^{-2})^3=x^{-2}*x^{-2}*x^{-2}$$

OpenStudy (anonymous):

would it be 2n^2

OpenStudy (skullpatrol):

No.

OpenStudy (anonymous):

2n^10 ?

OpenStudy (skullpatrol):

No, try and use the definition I posed above.

OpenStudy (anonymous):

n^2/2

OpenStudy (skullpatrol):

No,recall when you multiply two powers having the same base you add the exponents.

OpenStudy (anonymous):

n^2

jimthompson5910 (jim_thompson5910):

or you could multiply exponents

jimthompson5910 (jim_thompson5910):

(n^-2)^3 = n^(-2*3) = ??

OpenStudy (anonymous):

so it would be n^2

OpenStudy (skullpatrol):

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