Question

someone please help me please

Answer 1

divide assume the denominator does not equal zero
(18x^5y^3 + 6x^2y^4 + 24x^6y) ÷ 3xy

Answer 2

Do you know how to divide polynomials?

Answer 3

sorta

Answer 4

X_X I have to get off but we can do the first one,

\(\ \dfrac{18x^5y^3 + 6x^2y^4 + 24x^6y}{3xy} = \dfrac{18x^5y^3}{3xy} + \dfrac{6x^2y^4}{3xy} + \dfrac{24x^6y}{3xy}\)

\(\ \dfrac{18x^5y^2}{3xy} = \dfrac{18}{3} * \dfrac{x^5}{x} * \dfrac{y^2}{y} = 6*x^4 * y = 6x^4y \)

Answer 5

ok so is this how to do it

Answer 6

Yes this is how you're suppose to do the last 2. I showed how the first one is going to look like, in the end you should get something like: 6x^4y + ???? + ???

Answer 7

ok thank you sooo much:)

Answer 8

Yw, remember that when you divide you subtract exponent

Ex: \(\ \dfrac{a^3}{a^2} = a^{3-2} = a^1 = a \)

Answer 9

Maybe @jigglypuff314 can help you :P baii

Answer 10

i got it lol :)

Answer 11

kk hf