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Algebra I help

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Tue Dec 09, 2008 10:09 pm
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Winter's Twelfth Night says...



I am a bit confused about the exponent rules... I just need to simplify these problems.

ab²∗a³b=

(ab)⁴=

x⁵∕y⁴∗(y∕x)³=

Thank you so much for your help!!!
~Winter
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Mamillius: A sad tale’s best for winter. I have one
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Tue Dec 09, 2008 10:14 pm
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Rosey Unicorn says...



When multiplying exponents, just add them. When dividing them, subtract. When your multiplying the group [the (ab)4 problem] it's just a case of figuring out how many a's need to be multiplied. So, in that problem, it's four.

Get it?
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Wed Dec 10, 2008 12:04 am
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Winter's Twelfth Night says...



Thank you so much for your help! I'm still a little confused about multiplying the four a's. Would the answer just be ab⁴? Thanks again for replying!
~WInter
Mamillius: Merry or sad shall’t be?
Hermione: As merry as you will.
Mamillius: A sad tale’s best for winter. I have one
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Wed Dec 10, 2008 5:19 pm
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Rosey Unicorn says...



Close. It'd be a4b4 (Pretend those are subscripts, my computer won't let me do it) Since your multiplying both the a and the b.
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Fri Dec 12, 2008 1:21 am
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Winter's Twelfth Night says...



Oh ok! I get it now. Thank you sooooo much for the help!
Mamillius: Merry or sad shall’t be?
Hermione: As merry as you will.
Mamillius: A sad tale’s best for winter. I have one
Of sprites and goblins.

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Fri Dec 12, 2008 1:22 am
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Rosey Unicorn says...



Your welcome!

This stuff took me forever to learn too.
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Fri Dec 12, 2008 10:33 pm
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Winter's Twelfth Night says...



Yes, I definitely don't have a math mind.
Mamillius: Merry or sad shall’t be?
Hermione: As merry as you will.
Mamillius: A sad tale’s best for winter. I have one
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Fri Dec 12, 2008 11:00 pm
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Incandescence says...



Winter:


I'll go ahead and carry the first problem out in full, if you don't mind.


ab²∗a³b=a¹∙b²∙a³∙b¹

By the associative property (i.e., abc=(ab)c=a(bc)), we can rewrite this as:

ab²∗a³b=a¹∙(b²∙a³)∙b¹

Since associative multiplication is commutative (i.e., ab=ba), we can write this as:

ab²∗a³b=a¹∙(b²∙a³)∙b¹=a¹∙(a³∙b²)∙b¹

and by associativity we can say:

ab²∗a³b=a¹∙(b²∙a³)∙b¹=a¹∙(a³∙b²)∙b¹=(a¹∙a³)∙(b²∙b¹)

Since the rule of exponent multiplication is addition, we get:
ab²∗a³b=a¹∙(b²∙a³)∙b¹=a¹∙(a³∙b²)∙b¹=(a¹∙a³)∙(b²∙b¹)=(a¹⁺³)(b²⁺¹)=a⁴b³

The second problem is identical to the first problem: just make sure you understand that everything inside the parentheses is put to the exponent. That is, if I had (ab)³, I would say, (ab)³=(ab)(ab)(ab), by definition of exponentiation. From there, I would follow the procedure I gave you above, changing the parentheses around and use commutativity to get my result.


Best,
Brad
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Sat Dec 13, 2008 10:52 pm
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Winter's Twelfth Night says...



Thank you for writing it out! That makes a lot of sense. So you just write it out as a longer problem so that you can add up the like bases. That's what you were saying, right? At least that's the way that makes the most sense to me. Anyway, thanks again!
-Winter
Mamillius: Merry or sad shall’t be?
Hermione: As merry as you will.
Mamillius: A sad tale’s best for winter. I have one
Of sprites and goblins.

The Winter's Tale




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Sat Dec 27, 2008 4:57 am
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thunder_dude7 says...



It seems that you've been helped, but I'm bored.

ab²∗a³b=

Start by looking at the variables individually. Let's start with "a". a^3 multiplied by a, written out, is...

a x a x a x a

That equals a^4.

The same with the Bs means that they're equal to b^3.

(ab)⁴=

Well, this can be written as "(ab) x (ab)". This is a double distributive property thing. First, distribute the a. That makes the second part a^2 x ab. Next, distribute the b. That makes it a^2b x ab^2.

Well, I have to go. Can't do the third one. Sorry.
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