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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

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  1. anonymous
    • 5 years ago
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    Haha this looks awfully familiar. Did you have a problem with the other answer?

  2. anonymous
    • 5 years ago
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    I got this: \[[(x^5z^7)/(21y^6)]^-2\] so far

  3. anonymous
    • 5 years ago
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    Yes I did, but now I got stuck

  4. anonymous
    • 5 years ago
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    Mine was slightly different, the z value ended up being different from what you gave me, and I don't understand why.

  5. anonymous
    • 5 years ago
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    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.

  6. anonymous
    • 5 years ago
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    oh yea, thats right

  7. anonymous
    • 5 years ago
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    3^-1 right? so, where is 21 come from? I got: (x5z7)/(3y6)

  8. anonymous
    • 5 years ago
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    Inik the 3 moves to the bottom with the z, because it has the negative power. Then you multiply them to get 21.

  9. anonymous
    • 5 years ago
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    ooo... I see 7 as well. agree with 21

  10. anonymous
    • 5 years ago
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    Alright, l2theM, do you understand/have any more questions?

  11. anonymous
    • 5 years ago
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    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?

  12. anonymous
    • 5 years ago
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    *I did fix the z

  13. anonymous
    • 5 years ago
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    sorry it's messy

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  14. anonymous
    • 5 years ago
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    Rain, how did you get the 3 on the bottom and 39 on the top?

  15. anonymous
    • 5 years ago
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    eh, 49 rather

  16. anonymous
    • 5 years ago
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    Yes. Everything gets flipped because they all become negative. Then you multiply all exponents by 2 and square the 21. You should get: \[x ^{10} z ^{8} \div 441 y ^{12}\]

  17. anonymous
    • 5 years ago
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    Rain, one that makes it incredibly complicated and confusing... And you also forgot to square the 3.

  18. anonymous
    • 5 years ago
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    I got 441y^12z^2/x^10z^10

  19. anonymous
    • 5 years ago
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    Yeah sorry I typed it backwards. Yours is right.

  20. anonymous
    • 5 years ago
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    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?

  21. anonymous
    • 5 years ago
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  22. anonymous
    • 5 years ago
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    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?

  23. anonymous
    • 5 years ago
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    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.

  24. anonymous
    • 5 years ago
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    Rain, I don't think that is right.

  25. anonymous
    • 5 years ago
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    oh it's the whole thing to the power of -2. I only did the bottom. My bad, sorry!

  26. anonymous
    • 5 years ago
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    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.

  27. anonymous
    • 5 years ago
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    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.

  28. anonymous
    • 5 years ago
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    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?

  29. anonymous
    • 5 years ago
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    Here, let me explain another way. How would you simplify \[x ^{2} \div x\] ?

  30. anonymous
    • 5 years ago
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    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

  31. anonymous
    • 5 years ago
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    That would be just x

  32. anonymous
    • 5 years ago
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    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.

  33. anonymous
    • 5 years ago
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    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?

  34. anonymous
    • 5 years ago
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    it would be x^87 with no denominator, right? I think I did the first sub-question wrong

  35. anonymous
    • 5 years ago
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    No you did the first one right. But what did you DO? How did you know that it was just x?

  36. anonymous
    • 5 years ago
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    And no, it would be x^86. so what did I do there? How did I get that answer?

  37. anonymous
    • 5 years ago
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    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.

  38. anonymous
    • 5 years ago
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    Oh, you DID subtract the base from the numerator. x*x/x would mean 1 x goes away, which would be subtracting, right?

  39. anonymous
    • 5 years ago
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    Quantum, I got your answer but flipped. \[411y ^{12} \div x ^{10} z ^{8}\]What did I do wrong?

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  40. anonymous
    • 5 years ago
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    I think you got it right this time.

  41. anonymous
    • 5 years ago
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    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.

  42. anonymous
    • 5 years ago
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    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

  43. anonymous
    • 5 years ago
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    I don't know how to do it by that method Rain, that looks completely foreign to me.

  44. anonymous
    • 5 years ago
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    Yes, that's right. You subtract them! Does that make sense now?

  45. anonymous
    • 5 years ago
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    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?

  46. anonymous
    • 5 years ago
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    x/x=x?

  47. anonymous
    • 5 years ago
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    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.

  48. anonymous
    • 5 years ago
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    No, x/x is one. Because both cancel each other. It's like 4/4 or 7/7.

  49. anonymous
    • 5 years ago
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    Ah, okay, you are using x^2 and applying a rule that the inverse would be negative in order to isolate the variables?

  50. anonymous
    • 5 years ago
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    oh

  51. anonymous
    • 5 years ago
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    Yeah so take like, (x^2)/(x^5). what do you get?

  52. anonymous
    • 5 years ago
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    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?

  53. anonymous
    • 5 years ago
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    0/x^3?

  54. anonymous
    • 5 years ago
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    or x^-3/0? No, that can't be right

  55. anonymous
    • 5 years ago
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    Yes, which would make it one. Any number^0 is one. Like y^0 is one, z^0 is one... It's always one.

  56. anonymous
    • 5 years ago
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    You had it right the first time, but it cancels to one. So it would be 1/x^3.

  57. anonymous
    • 5 years ago
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    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?

  58. anonymous
    • 5 years ago
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    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.

  59. anonymous
    • 5 years ago
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    But essentially yes, you are correct.

  60. anonymous
    • 5 years ago
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    x*x/x*x*x*x*x would be 0/x^3 or 1/x^3?

  61. anonymous
    • 5 years ago
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    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?

  62. anonymous
    • 5 years ago
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    Okay yes, that makes PERFECT sense :D I think I get it now :D

  63. anonymous
    • 5 years ago
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    I just wanted to be sure I completely understood how you were going about this.

  64. anonymous
    • 5 years ago
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    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.

  65. anonymous
    • 5 years ago
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    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

  66. anonymous
    • 5 years ago
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    Okay

  67. anonymous
    • 5 years ago
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    And no worries. Haha I'm Glad to help.

  68. anonymous
    • 5 years ago
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    Okay, making this up... x^-5y^9z^-2/x^3y^-5z^13, going to work it out now.

  69. anonymous
    • 5 years ago
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    so far, x^3y^14/x^5z^15

  70. anonymous
    • 5 years ago
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    y^14/x^2z^15 would be the final answer

  71. anonymous
    • 5 years ago
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    and if I multiplied everything by ^-2, that would make it x^4z^17/y^16?

  72. anonymous
    • 5 years ago
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    Almost perfect. But can you tell me what is wrong with the exponent on z and y?

  73. anonymous
    • 5 years ago
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    for which problem?

  74. anonymous
    • 5 years ago
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    the first one?

  75. anonymous
    • 5 years ago
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    do you multiply those exponents by 2 instead of adding 2 to them?

  76. anonymous
    • 5 years ago
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    oh yea, x^4z^34/y^32

  77. anonymous
    • 5 years ago
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    thats it, got it.

  78. anonymous
    • 5 years ago
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    Yes, you do multiply them, you only add when you are Multiplying. Ie,( x^2)^3 = x^6 (x^2)(x^2)= x^4

  79. anonymous
    • 5 years ago
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    \[((3{}^{\wedge}-1)(x)(y{}^{\wedge}-4)(z{}^{\wedge}5)/(7x{}^{\wedge}-4)(y{}^{\wedge}2)(z)){}^{\wedge}-2 \] \[\left(\left(3^{-1}\right)(x)\left(y^{-4}\right)\left(z^5\right)/\left(7x^{-4}\right)\left(y^2\right)(z)\right)^{-2} \] \[\left(\left(\frac{x}{3}\right)\frac{z^5}{y^4}/\frac{7}{x^4}\left(y^2 z\right)\right)^{-2} \] \[\left(\frac{x z^5}{3 y^4}/\frac{7 y^2 z}{x^4}\right)^{-2} \] \[\left(\frac{x^5 z^4}{21 y^6}\right)^{-2} \] \[\frac{441 y^{12}}{x^{10} z^8} \]

  80. anonymous
    • 5 years ago
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    Lol thanks, that's been established.

  81. anonymous
    • 5 years ago
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    This was done with Mathmatica 8. I over looked the reference to WolframAlpha.com in a prior post above. Sorry.

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