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anonymous
 5 years ago
Okay, lets try this again...
[(3^1)(x)(y^4)(z^5)/(7x^4)(y^2)(z)]^2
anonymous
 5 years ago
Okay, lets try this again... [(3^1)(x)(y^4)(z^5)/(7x^4)(y^2)(z)]^2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haha this looks awfully familiar. Did you have a problem with the other answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got this: \[[(x^5z^7)/(21y^6)]^2\] so far

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes I did, but now I got stuck

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mine was slightly different, the z value ended up being different from what you gave me, and I don't understand why.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Z doesn't move. You just cancel it normally. 51 is 4. I didn't even pay attention to z until the very end.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.03^1 right? so, where is 21 come from? I got: (x5z7)/(3y6)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Inik the 3 moves to the bottom with the z, because it has the negative power. Then you multiply them to get 21.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ooo... I see 7 as well. agree with 21

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, l2theM, do you understand/have any more questions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Rain, how did you get the 3 on the bottom and 39 on the top?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes. 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
 5 years ago
Best ResponseYou've already chosen the best response.0Rain, one that makes it incredibly complicated and confusing... And you also forgot to square the 3.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got 441y^12z^2/x^10z^10

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah sorry I typed it backwards. Yours is right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0and 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
 5 years ago
Best ResponseYou've already chosen the best response.0Because you still cancel them. 51 is still 4 lol Once it is cancelled you square it, and since it is an exponent you multiply them, so 4x2=8.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Rain, I don't think that is right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh it's the whole thing to the power of 2. I only did the bottom. My bad, sorry!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can do it either way. If you multiply first, you get the z^2/z^10 and you cancel by doing 102 which is still 8 lol.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hahaha 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
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0Here, let me explain another way. How would you simplify \[x ^{2} \div x\] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, 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
 5 years ago
Best ResponseYou've already chosen the best response.0and 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
 5 years ago
Best ResponseYou've already chosen the best response.0No 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
 5 years ago
Best ResponseYou've already chosen the best response.0it would be x^87 with no denominator, right? I think I did the first subquestion wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No you did the first one right. But what did you DO? How did you know that it was just x?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And no, it would be x^86. so what did I do there? How did I get that answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, you DID subtract the base from the numerator. x*x/x would mean 1 x goes away, which would be subtracting, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Quantum, I got your answer but flipped. \[411y ^{12} \div x ^{10} z ^{8}\]What did I do wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you got it right this time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Normally 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
 5 years ago
Best ResponseYou've already chosen the best response.0Hm 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
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know how to do it by that method Rain, that looks completely foreign to me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, that's right. You subtract them! Does that make sense now?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0No, x/x is one. Because both cancel each other. It's like 4/4 or 7/7.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, okay, you are using x^2 and applying a rule that the inverse would be negative in order to isolate the variables?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah so take like, (x^2)/(x^5). what do you get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But 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
 5 years ago
Best ResponseYou've already chosen the best response.0or x^3/0? No, that can't be right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, which would make it one. Any number^0 is one. Like y^0 is one, z^0 is one... It's always one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You had it right the first time, but it cancels to one. So it would be 1/x^3.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 5 years ago
Best ResponseYou've already chosen the best response.0But essentially yes, you are correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x*x/x*x*x*x*x would be 0/x^3 or 1/x^3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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
 5 years ago
Best ResponseYou've already chosen the best response.0Okay yes, that makes PERFECT sense :D I think I get it now :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just wanted to be sure I completely understood how you were going about this.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, 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
 5 years ago
Best ResponseYou've already chosen the best response.0And 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
 5 years ago
Best ResponseYou've already chosen the best response.0And no worries. Haha I'm Glad to help.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, making this up... x^5y^9z^2/x^3y^5z^13, going to work it out now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so far, x^3y^14/x^5z^15

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y^14/x^2z^15 would be the final answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and if I multiplied everything by ^2, that would make it x^4z^17/y^16?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Almost perfect. But can you tell me what is wrong with the exponent on z and y?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you multiply those exponents by 2 instead of adding 2 to them?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, you do multiply them, you only add when you are Multiplying. Ie,( x^2)^3 = x^6 (x^2)(x^2)= x^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[((3{}^{\wedge}1)(x)(y{}^{\wedge}4)(z{}^{\wedge}5)/(7x{}^{\wedge}4)(y{}^{\wedge}2)(z)){}^{\wedge}2 \] \[\left(\left(3^{1}\right)(x)\left(y^{4}\right)\left(z^5\right)/\left(7x^{4}\right)\left(y^2\right)(z)\right)^{2} \] \[\left(\left(\frac{x}{3}\right)\frac{z^5}{y^4}/\frac{7}{x^4}\left(y^2 z\right)\right)^{2} \] \[\left(\frac{x z^5}{3 y^4}/\frac{7 y^2 z}{x^4}\right)^{2} \] \[\left(\frac{x^5 z^4}{21 y^6}\right)^{2} \] \[\frac{441 y^{12}}{x^{10} z^8} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol thanks, that's been established.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This was done with Mathmatica 8. I over looked the reference to WolframAlpha.com in a prior post above. Sorry.
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