2y^{5 } Divided by 5y^{5}z^{4}

z^{4 }7

2y^{5 } Divided by 5y^{5}z^{4}

z^{4 }7

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so the question becomes

2y^{5 } x 7

z^{4 } 5y^{5}z^{4}

then we multiply across to get

14y^{5 }

5y^{5}z^{8}

we can cross out the y^{5 }in the numerator and denominator as a simplifying factor.

so our answer would be

14

5z^{8}

When dividing by a fraction, you multiply by the reciprocal. This means that you "flip" the second fraction (the numerator becomes the denominator and the nominator becomes the numerator)

2y^{5} divided by 5y^{5}z^{4} Becomes
2y^{5} Times 7

z^{4} 7 z^{4} 5y^{5}z^{4}

Since the numerator of the first fraction and the denominator of the second fraction both have a y^{5} term, they will cancel. No other factors occur in both a numerator and a denominator, so no other terms cancel.

When we multiply straight across, we get the answer:

14

5z^{8}

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