anonymous
  • anonymous
(-4x^2y)^3 I really need help, I still don't understand.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[ (-4x^{2y})^{3} is this what you mean?\]
anonymous
  • anonymous
\[(ab^{c})^{d} = a^{d} b ^{cd}\]
anonymous
  • anonymous
so it will be \[-4^{3}x ^{6y}\]

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anonymous
  • anonymous
= -64x^6y
anonymous
  • anonymous
remember that y has an implicit exponent of 1, and 1*3 = 3.
anonymous
  • anonymous
no i meant (-4x to the second power than y)^3 the y is not part of the exponent
anonymous
  • anonymous
oh, so it's \[(-4x ^{2}y)^{?}\]
anonymous
  • anonymous
the concept is the same, -4^3 x^6 y^3
anonymous
  • anonymous
or -64x^6 y^3
anonymous
  • anonymous
I think the question is :\[(-4x^2y)^3\] right?
anonymous
  • anonymous
yes thats right, i just dont understand how to solve it
anonymous
  • anonymous
As I said, when raising a product to a power, you multiply the exponents of the factors. So the 3 would be multiplied to the power of the -4 factor, and the power of the x factor, and the power of the y factor.
anonymous
  • anonymous
alright, first of all you've got to mulitply the power 3 with the given between the brackets and you'll get ^_^: \[=(-4)^3 x^6 y^3 = 48x^6y^3\] is it clearer now ? :)
anonymous
  • anonymous
\[(-4)^3 \ne 48 \] \[(-4)^3 = -64 \]
anonymous
  • anonymous
lol! sry >_< silly mistake , thank you polpak :)
anonymous
  • anonymous
do you understand it now courtney ^_^?
anonymous
  • anonymous
just substitute the power to each :)
anonymous
  • anonymous
just pass the power out to each*
anonymous
  • anonymous
yes, kind of but there is another problem now that is in the same format but is a fraction and has that equal sign with a / through it. grr i hate math lol
anonymous
  • anonymous
don't say that, look at the problem , write it down on paper and try to use tha same concept like the one here :) The problem never gets harder, it just gets more detailed ^_^
anonymous
  • anonymous
nothing changes, just apply the same rule :)
anonymous
  • anonymous
The rule applies equally to division. The only part that gets messy is when you have a sum that you're raising to a power.
anonymous
  • anonymous
im trying to stay positive, thank you so much for your help you two are life savers! im writing all the work you just sent as my work shown but what part is the part that i put in the space for the answer to the problem?
anonymous
  • anonymous
Try to see if you can figure it out from what we've said. We'll check your answer.
anonymous
  • anonymous
np :)
anonymous
  • anonymous
ok im working it out now than ill post it on here, thank you again.
anonymous
  • anonymous
\[(((3p ^{3}v ^{4} \ s ^{4}))^{2}\]
anonymous
  • anonymous
ok this is a fraction the s^4 is on the bottom and it also says s with the equal sign / 0
anonymous
  • anonymous
i got 9p^6y^8 over s^8
anonymous
  • anonymous
So.. \[(3p^3v^4/s^4)^2, s \ne 0\]
anonymous
  • anonymous
Then yes, that is the correct answer. Good job!
anonymous
  • anonymous
yayyy thank you, even my professor cant teach me math and over the computer you got right into my head lol
anonymous
  • anonymous
It may be somewhat helpful to expand the exponents for a while to see what is happening. \[(3p^3v^4/s^4)^2 = [(3pppvvvv)/(ssss)]^2 \] \[= [(3pppvvvv)(3pppvvvv)]/[(ssss)(ssss)]\] \[= [(3*3)ppppppvvvvvvvv]/[ssssssss]\] \[= 9p^6v^8/s^8\]
anonymous
  • anonymous
that acutally helps alot, can i ask you on more question?
anonymous
  • anonymous
Of course!
anonymous
  • anonymous
ok so i know how to do this in a normal way but they want it in exponential form and im confused. the first is (5^3)^3 and the next is in fraction form which is even worse lol (r/s)^6 (s\[\neq\]0)
anonymous
  • anonymous
So try to expand it out. \[(5^3)^3 = (5^3)(5^3)(5^3)\] Now how would you expand that further?
anonymous
  • anonymous
five times five times five than add the exponents?
anonymous
  • anonymous
\[5^3 = 5*5*5\] \[\rightarrow 5^3*5^3*5^3 = (5*5*5)*(5*5*5)*(5*5*5)\] \[=5^?\]
anonymous
  • anonymous
5^9?
anonymous
  • anonymous
Yep.
anonymous
  • anonymous
awesome now to the fraction, can you help me with that one?
anonymous
  • anonymous
Now we can see that the faster way to arrive at the same result would be to simply : \[(5^3)^3 = 5^{3*3} = 5^9\]
anonymous
  • anonymous
okay so you multiply the exponents too?
anonymous
  • anonymous
Well, remember when we expanded it out, we said that we had three groups with three 5's each. So how many 5's do we have multiplied together? 3*3 right?
anonymous
  • anonymous
right, im trying to keep up im just so lost
anonymous
  • anonymous
So if you're not comfortable with multiplying the exponents, just try expanding the next one out for yourself. \[a^6 = a*a*a*a*a*a\] \[ \rightarrow (r/s)^6 =\ ?\]
anonymous
  • anonymous
We'll just go one step at a time.
anonymous
  • anonymous
would i seperate them and do r*r*r*r*r*r and s*s*s*s*s*s?
anonymous
  • anonymous
No, just multiply the whole thing.
anonymous
  • anonymous
so r/s*r/s*r/s*r/s*r/s*r/s
anonymous
  • anonymous
Yes, but keep the parens so it's more readable. \[(r/s)*(r/s)*(r/s)*(r/s)*(r/s)*(r/s)\] And when we multiply two fractions, we do what?
anonymous
  • anonymous
top times top bottom times bottom?
anonymous
  • anonymous
yep
anonymous
  • anonymous
so its (r^6)/(s^6)
anonymous
  • anonymous
Yes. And here again we can arrive at this answer in a more direct way once we understand the process \[ (r/s)^6 = (r^1/s^1)^6 = r^{1*6}/s^{1*6} = r^6/s^6\]
anonymous
  • anonymous
that would be the answer correct since we do not have the values for r and s? what about the =/ thing?
anonymous
  • anonymous
They have to make the restriction \[s \ne 0\] because if s is 0 then you're dividing by 0 which is not defined. That answer is correct regardless of what r and s is. If we know more about them we might be able to simplify further (if s was 2 and r was 4 for example). But even in that case our answer is correct. It just might not be in the form they're looking for.
anonymous
  • anonymous
yayyy you are so amazing and kind hearted. that you so so much!
anonymous
  • anonymous
You're very welcome.

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