anonymous
  • anonymous
Okay, lets try this again... [(3^-1)(x)(y^-4)(z^5)/(7x^-4)(y^2)(z)]^-2
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
Haha this looks awfully familiar. Did you have a problem with the other answer?
anonymous
  • anonymous
I got this: \[[(x^5z^7)/(21y^6)]^-2\] so far
anonymous
  • anonymous
Yes I did, but now I got stuck

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anonymous
  • anonymous
Mine was slightly different, the z value ended up being different from what you gave me, and I don't understand why.
anonymous
  • anonymous
Z doesn't move. You just cancel it normally. 5-1 is 4. I didn't even pay attention to z until the very end.
anonymous
  • anonymous
oh yea, thats right
anonymous
  • anonymous
3^-1 right? so, where is 21 come from? I got: (x5z7)/(3y6)
anonymous
  • anonymous
Inik the 3 moves to the bottom with the z, because it has the negative power. Then you multiply them to get 21.
anonymous
  • anonymous
ooo... I see 7 as well. agree with 21
anonymous
  • anonymous
Alright, l2theM, do you understand/have any more questions?
anonymous
  • anonymous
Well, I still haven't simplified all of the way. Now that I got the above answer, how do I handle the ^-2? Do I move the top portion to the bottom, and the bottom portion to the top?
anonymous
  • anonymous
*I did fix the z
anonymous
  • anonymous
sorry it's messy
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anonymous
  • anonymous
Rain, how did you get the 3 on the bottom and 39 on the top?
anonymous
  • anonymous
eh, 49 rather
anonymous
  • anonymous
Yes. Everything gets flipped because they all become negative. Then you multiply all exponents by 2 and square the 21. You should get: \[x ^{10} z ^{8} \div 441 y ^{12}\]
anonymous
  • anonymous
Rain, one that makes it incredibly complicated and confusing... And you also forgot to square the 3.
anonymous
  • anonymous
I got 441y^12z^2/x^10z^10
anonymous
  • anonymous
Yeah sorry I typed it backwards. Yours is right.
anonymous
  • anonymous
I don't understand what you are doing with the z though. The z^5 should go to the bottom, and the z should go to the top. Why is it z^8 then?
anonymous
  • anonymous
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anonymous
  • anonymous
and also, since I have a z^2 on the top, and a z^10 on the bottom, do I simplify them to z^1/z^5?
anonymous
  • anonymous
Because you still cancel them. 5-1 is still 4 lol Once it is cancelled you square it, and since it is an exponent you multiply them, so 4x2=8.
anonymous
  • anonymous
Rain, I don't think that is right.
anonymous
  • anonymous
oh it's the whole thing to the power of -2. I only did the bottom. My bad, sorry!
anonymous
  • anonymous
You can do it either way. If you multiply first, you get the z^2/z^10 and you cancel by doing 10-2 which is still 8 lol.
anonymous
  • anonymous
Hahaha no worries. Just remember it applies to the whole thing. I like doing things the easy way, so thats why I did it in that order.
anonymous
  • anonymous
but why are you subtracting them if they are on opposite sides of the equation? are you changing the z^2 to z^-2 and subtracting it from the larger z^10?
anonymous
  • anonymous
Here, let me explain another way. How would you simplify \[x ^{2} \div x\] ?
anonymous
  • anonymous
Okay, what I did was I simplified everything in the equation and made it so all of the exponents were positive. Then I switch their locations so the ^-2 would become ^2, then I squared everything and got this: 441y^12z^2/x^10z^10
anonymous
  • anonymous
That would be just x
anonymous
  • anonymous
and I am sorry for the trouble, I just want to make sure I am doing this right. I have a final coming up, and I need to make a 95 or better, so really want to fully understand this.
anonymous
  • anonymous
No worries at all. That's that this site is for! Well, you have that down, so, explain to me how you got that. And how would you do x^87/x?
anonymous
  • anonymous
it would be x^87 with no denominator, right? I think I did the first sub-question wrong
anonymous
  • anonymous
No you did the first one right. But what did you DO? How did you know that it was just x?
anonymous
  • anonymous
And no, it would be x^86. so what did I do there? How did I get that answer?
anonymous
  • anonymous
You must have just subtracted them. But I am thinking of an equivalent equation with the same bases like this: 4/2. 4 is the same as 2^2. So the equivalent equation would be 2^2/2. The answer would be 2 no matter what.
anonymous
  • anonymous
Oh, you DID subtract the base from the numerator. x*x/x would mean 1 x goes away, which would be subtracting, right?
anonymous
  • anonymous
Quantum, I got your answer but flipped. \[411y ^{12} \div x ^{10} z ^{8}\]What did I do wrong?
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anonymous
  • anonymous
I think you got it right this time.
anonymous
  • anonymous
Normally I get an answer sheet with my practice tests, but my final doesn't give me one so I have no idea what the true answer is.
anonymous
  • anonymous
Hm it should be right. http://www.wolframalpha.com/input/?i=%5B%5B%283^-1%29%28x%29%28y^-4%29%28z^5%29%5D%2F%5B%287x^-4%29%28y^2%29%28z%29%5D%5D^-2 Copy&paste that link
anonymous
  • anonymous
I don't know how to do it by that method Rain, that looks completely foreign to me.
anonymous
  • anonymous
Yes, that's right. You subtract them! Does that make sense now?
anonymous
  • anonymous
Yes, it makes sense when you have x^2/x, but when you have x/x^2, that would make the problem equal x too, right?
anonymous
  • anonymous
x/x=x?
anonymous
  • anonymous
You subtract the bigger number by the smaller number, and whichever one was bigger, that's where it goes. Like if the x^87 was on bottom and x was on top, it would be 1/x^86. it's the same thing.
anonymous
  • anonymous
No, x/x is one. Because both cancel each other. It's like 4/4 or 7/7.
anonymous
  • anonymous
Ah, okay, you are using x^2 and applying a rule that the inverse would be negative in order to isolate the variables?
anonymous
  • anonymous
oh
anonymous
  • anonymous
Yeah so take like, (x^2)/(x^5). what do you get?
anonymous
  • anonymous
But if you had x/x^2, you can't subtract the top x from the bottom, because you would have a 0 in the numerator, right?
anonymous
  • anonymous
0/x^3?
anonymous
  • anonymous
or x^-3/0? No, that can't be right
anonymous
  • anonymous
Yes, which would make it one. Any number^0 is one. Like y^0 is one, z^0 is one... It's always one.
anonymous
  • anonymous
You had it right the first time, but it cancels to one. So it would be 1/x^3.
anonymous
  • anonymous
Okay, so if you take x^2/x^5 and turned the x^2 into x^-2 and subtract it by the denominator, you assume that there is a x^0 left over, and that equals 1?
anonymous
  • anonymous
Yes, but that could get confusing quickly. Try this. Write out all of the x's on top and bottom. Then cross out as many as you can. Then explain to me what you did and what it means.
anonymous
  • anonymous
But essentially yes, you are correct.
anonymous
  • anonymous
x*x/x*x*x*x*x would be 0/x^3 or 1/x^3?
anonymous
  • anonymous
It's one. You do NOT cancel to be zero. The only way to get zero would be to subtract something. Like, you have 4 quarters in a dollar. You cancel the 4 and you got 1 dollar right? You cant just have 4 quarters and then suddenly have nothing unless someone took the money from you, ie subtracted them. Does that make sense?
anonymous
  • anonymous
Okay yes, that makes PERFECT sense :D I think I get it now :D
anonymous
  • anonymous
I just wanted to be sure I completely understood how you were going about this.
anonymous
  • anonymous
Alright, make sure you got this down. Make up a problem just like the one you originally asked and simplify it. I'll tell you if you did it right.
anonymous
  • anonymous
And again, sorry for the trouble, but I think I completely get how to go about these problems now :D I can do alot of this math, but I missed a few basic concepts back in high school, and this helps me bridge those gaps :D
anonymous
  • anonymous
Okay
anonymous
  • anonymous
And no worries. Haha I'm Glad to help.
anonymous
  • anonymous
Okay, making this up... x^-5y^9z^-2/x^3y^-5z^13, going to work it out now.
anonymous
  • anonymous
so far, x^3y^14/x^5z^15
anonymous
  • anonymous
y^14/x^2z^15 would be the final answer
anonymous
  • anonymous
and if I multiplied everything by ^-2, that would make it x^4z^17/y^16?
anonymous
  • anonymous
Almost perfect. But can you tell me what is wrong with the exponent on z and y?
anonymous
  • anonymous
for which problem?
anonymous
  • anonymous
the first one?
anonymous
  • anonymous
do you multiply those exponents by 2 instead of adding 2 to them?
anonymous
  • anonymous
oh yea, x^4z^34/y^32
anonymous
  • anonymous
thats it, got it.
anonymous
  • anonymous
Yes, you do multiply them, you only add when you are Multiplying. Ie,( x^2)^3 = x^6 (x^2)(x^2)= x^4
anonymous
  • anonymous
\[((3{}^{\wedge}-1)(x)(y{}^{\wedge}-4)(z{}^{\wedge}5)/(7x{}^{\wedge}-4)(y{}^{\wedge}2)(z)){}^{\wedge}-2 \] \[\left(\left(3^{-1}\right)(x)\left(y^{-4}\right)\left(z^5\right)/\left(7x^{-4}\right)\left(y^2\right)(z)\right)^{-2} \] \[\left(\left(\frac{x}{3}\right)\frac{z^5}{y^4}/\frac{7}{x^4}\left(y^2 z\right)\right)^{-2} \] \[\left(\frac{x z^5}{3 y^4}/\frac{7 y^2 z}{x^4}\right)^{-2} \] \[\left(\frac{x^5 z^4}{21 y^6}\right)^{-2} \] \[\frac{441 y^{12}}{x^{10} z^8} \]
anonymous
  • anonymous
Lol thanks, that's been established.
anonymous
  • anonymous
This was done with Mathmatica 8. I over looked the reference to WolframAlpha.com in a prior post above. Sorry.

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