Question

(-4x^2y)^3

I really need help, I still don't understand.

Answer 1

\[ (-4x^{2y})^{3} is this what you mean?\]

Answer 2

\[(ab^{c})^{d} = a^{d} b ^{cd}\]

Answer 3

so it will be \[-4^{3}x ^{6y}\]

Answer 4

= -64x^6y

Answer 5

remember that y has an implicit exponent of 1, and 1*3 = 3.

Answer 6

no i meant (-4x to the second power than y)^3

the y is not part of the exponent

Answer 7

oh, so it's \[(-4x ^{2}y)^{?}\]

Answer 8

the concept is the same, -4^3 x^6 y^3

Answer 9

or -64x^6 y^3

Answer 10

I think the question is :\[(-4x^2y)^3\]


right?

Answer 11

yes thats right, i just dont understand how to solve it

Answer 12

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.

Answer 13

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 ? :)

Answer 14

\[(-4)^3 \ne 48 \]
\[(-4)^3 = -64 \]

Answer 15

lol! sry >_< silly mistake , thank you polpak :)

Answer 16

do you understand it now courtney ^_^?

Answer 17

just substitute the power to each :)

Answer 18

just pass the power out to each*

Answer 19

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

Answer 20

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 ^_^

Answer 21

nothing changes, just apply the same rule :)

Answer 22

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.

Answer 23

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?

Answer 24

Try to see if you can figure it out from what we've said. We'll check your answer.

Answer 25

np :)

Answer 26

ok im working it out now than ill post it on here, thank you again.

Answer 27

\[(((3p ^{3}v ^{4} \ s ^{4}))^{2}\]

Answer 28

ok this is a fraction the s^4 is on the bottom and it also says s with the equal sign / 0

Answer 29

i got 9p^6y^8 over s^8

Answer 30

So..
\[(3p^3v^4/s^4)^2, s \ne 0\]

Answer 31

Then yes, that is the correct answer. Good job!

Answer 32

yayyy thank you, even my professor cant teach me math and over the computer you got right into my head lol

Answer 33

It may be somewhat helpful to expand the exponents for a while to see what is happening.

\[(3p^3v^4/s^4)^2 = [(3pppvvvv)/(ssss)]^2 \]
\[= [(3pppvvvv)(3pppvvvv)]/[(ssss)(ssss)]\]
\[= [(3*3)ppppppvvvvvvvv]/[ssssssss]\]
\[= 9p^6v^8/s^8\]

Answer 34

that acutally helps alot, can i ask you on more question?

Answer 35

Of course!

Answer 36

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)

Answer 37

So try to expand it out.
\[(5^3)^3 = (5^3)(5^3)(5^3)\]
Now how would you expand that further?

Answer 38

five times five times five than add the exponents?

Answer 39

\[5^3 = 5*5*5\]
\[\rightarrow 5^3*5^3*5^3 = (5*5*5)*(5*5*5)*(5*5*5)\]
\[=5^?\]

Answer 40

5^9?

Answer 41

Yep.

Answer 42

awesome now to the fraction, can you help me with that one?

Answer 43

Now we can see that the faster way to arrive at the same result would be to simply :

\[(5^3)^3 = 5^{3*3} = 5^9\]

Answer 44

okay so you multiply the exponents too?

Answer 45

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?

Answer 46

right, im trying to keep up im just so lost

Answer 47

So if you're not comfortable with multiplying the exponents, just try expanding the next one out for yourself.

\[a^6 = a*a*a*a*a*a\]
\[ \rightarrow (r/s)^6 =\ ?\]

Answer 48

We'll just go one step at a time.

Answer 49

would i seperate them and do r*r*r*r*r*r and s*s*s*s*s*s?

Answer 50

No, just multiply the whole thing.

Answer 51

so r/s*r/s*r/s*r/s*r/s*r/s

Answer 52

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?

Answer 53

top times top bottom times bottom?

Answer 54

yep

Answer 55

so its (r^6)/(s^6)

Answer 56

Yes. And here again we can arrive at this answer in a more direct way once we understand the process

\[ (r/s)^6 = (r^1/s^1)^6 = r^{1*6}/s^{1*6} = r^6/s^6\]

Answer 57

that would be the answer correct since we do not have the values for r and s?
what about the =/ thing?

Answer 58

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.

Answer 59

yayyy you are so amazing and kind hearted. that you so so much!

Answer 60

You're very welcome.