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 Post subject: Exponentials Drill P.102Posted: Wed Mar 02, 2011 3:59 pm

Joined: Fri Feb 04, 2011 7:55 am
Posts: 8
Hello,

How you explain (48z^4y^-3)/(12zy^2) = (4z^3)/y^2 ?

Thank you

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 Post subject: Re: Exponentials Drill P.102Posted: Wed Mar 02, 2011 6:25 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
Let's take this one in pieces.

The 48 and 12 reduce to become a 4 on top.
The z^4 on top and z on bottom reduce to become z^3 on top.
The y^-3 on top and y^2 on bottom reduce to become y^5 on bottom.

This leaves us with (4z^3)/y^5

Is there a particular step in this process that's giving you particular trouble?

Veritas Help

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 Post subject: Re: Exponentials Drill P.102Posted: Thu Mar 03, 2011 7:12 am

Joined: Fri Feb 04, 2011 7:55 am
Posts: 8
Not a problem for 48 and 12 but:

1. The rules for know how to reduce: z^4 in z^3 with z?
2. Particulary how y^-3 become y^5 with y^2?
3. How to know which on stay on the top and the bottom?

Thank you

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 Post subject: Re: Exponentials Drill P.102Posted: Fri Mar 04, 2011 9:04 am

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
Go back and look at Math Essentials, page 99, and you will see the big table of Exponent rules --

1. z^4 over z (which is really z^1) is z^(4-1) which is z^3.
2. y^(-3) over y^3 is y^(-3-2) which is y^(-5). y^(-5) on top is the same as y^5 on bottom.
3. Generally, pieces stay where they are, unless we have a negative exponent, in which case, we switch from top to bottom. If we are reducing / canceling, the one with the bigger exponent is the one that stays and the one with the smaller exponent is the one that gets canceled.

Veritas Help

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