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Inopeki

ab=c Solve for a Do i need to divide both sides by b?

  • 2 years ago
  • 2 years ago

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  1. Zed
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    yes

    • 2 years ago
  2. Inopeki
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    And that is all?

    • 2 years ago
  3. Zed
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    yes, so when you do that what will you get?

    • 2 years ago
  4. Inopeki
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    a=a?

    • 2 years ago
  5. Zed
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    \[\frac{ab}{b}=\frac{c}{b}\] \[a=\frac{c}{b}\]

    • 2 years ago
  6. Inopeki
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    Oh! now i get it, can you set up another example so i can try?

    • 2 years ago
  7. Zed
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    Sure ax+y=c, solve for x

    • 2 years ago
  8. Inopeki
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    Ok, divide ax with a, then divide c with a getting ax/a+y=c/a. You then simplify that getting x+y=c/a. Then you take both sides minus y getting x=c/a-y

    • 2 years ago
  9. Inopeki
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    Is this correct?

    • 2 years ago
  10. Zed
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    Really close but we need to either get rid of the y first before divide or also divide the y by a

    • 2 years ago
  11. Inopeki
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    YES! Thanks for teaching me :)

    • 2 years ago
  12. Inopeki
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    aw :(

    • 2 years ago
  13. abdul_shabeer
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    No, first start by subtracting y from both sides

    • 2 years ago
  14. Inopeki
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    Alright, so ax+y=c ax=c-y x/a=c/a-y x=c/a-y

    • 2 years ago
  15. Zed
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    remeber that when you divide the a you need to do it to all the terms

    • 2 years ago
  16. Inopeki
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    Where did you get z?

    • 2 years ago
  17. Inopeki
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    Oh right.. so x=c-y/a

    • 2 years ago
  18. zbay
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    \[x=\frac{c-y}{a}\]

    • 2 years ago
  19. Zed
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    zbay is correct

    • 2 years ago
  20. zbay
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    Inopeki got the answer i just put it in the equation maker for ease of the viewers

    • 2 years ago
  21. Inopeki
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    Yay! :D

    • 2 years ago
  22. Zed
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    inopeki you need to write x=(c-y)/a just to make it clearer. i see it now good work

    • 2 years ago
  23. Inopeki
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    Thanks for the help :) Is there any way you can make it a little harder to give me a challenge?

    • 2 years ago
  24. zbay
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    yea i will come up with one for you

    • 2 years ago
  25. Inopeki
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    Thanks.

    • 2 years ago
  26. zbay
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    \[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\] Solve for T_2

    • 2 years ago
  27. Inopeki
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    umm.. ill try

    • 2 years ago
  28. FoolForMath
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    Nobody mentioned division by zero ?!!

    • 2 years ago
  29. abdul_shabeer
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    Good observation @ Foolformath

    • 2 years ago
  30. Inopeki
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    Well to begin with i need to multiply both sides with P2V2, getting P1V1/T1*P2V2, right?\[P _{1}V _{1}divT _{1}+P _{2}V _{2}=T2\]

    • 2 years ago
  31. zbay
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    I'll lay it out by steps for you

    • 2 years ago
  32. Inopeki
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    NOO. wait. Its

    • 2 years ago
  33. abdul_shabeer
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    Cross multiplication

    • 2 years ago
  34. Inopeki
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    P1V1/T1*P2V2=T2

    • 2 years ago
  35. zbay
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    lcd = T_1T_2 \[T_2P_1V_1=V_2P_2T_1\] \[T_2=\frac{V_2P_2T_1}{P_1V_1}\]

    • 2 years ago
  36. Inopeki
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    Oh right, i see it now. You wanted me to put the things i was multiplying by(V2P2) in by the other things that were multiplied. That would mean my answer is right, but yours is simplified.

    • 2 years ago
  37. Inopeki
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    No?

    • 2 years ago
  38. zbay
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    No because you need to multiply both sides by the lcd of \[T_1T_2\] first then we cancel one of the varibles out because they are the same on top and bottom. then we can break T_2 apart from the rest of the left side by division, your answer would look close but in the end it would be wrong

    • 2 years ago
  39. Inopeki
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    What is an lcd? What makes T1 jump up?

    • 2 years ago
  40. zbay
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    lcd= lowest common denominator

    • 2 years ago
  41. Inopeki
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    What is that?

    • 2 years ago
  42. Inopeki
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    Is a denominator the 1 in T1?

    • 2 years ago
  43. zbay
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    The denominator is what is on the bottom of the fraction, so when we do the left side we get \[\frac{T_1T_2P_11V_1}{T_1}\] and because we have the same thing on the top and bottom we can cancel out the denominator leving us with \[T_2P_1V_1\] and we do the same operation on the right side

    • 2 years ago
  44. Inopeki
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    Why do we get T1 on both sides of the line?

    • 2 years ago
  45. zbay
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    because you are multiplying by t1t2 but on each side one of them will cancell

    • 2 years ago
  46. Inopeki
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    Ah

    • 2 years ago
  47. zbay
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    Good job that was a hard one

    • 2 years ago
  48. Inopeki
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    Thanks, when am i going to get to learn this in school?

    • 2 years ago
  49. Zed
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    what grade are you?

    • 2 years ago
  50. Inopeki
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    8th

    • 2 years ago
  51. zbay
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    Not sure when you will cover that but it's a chemistry equation compariing similar gasses to determine preasure, temp, or volume. You might touch it in highschool sometime.

    • 2 years ago
  52. Inopeki
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    Oh, cool. What should i do now?

    • 2 years ago
  53. zbay
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    PV=nrt solve for n

    • 2 years ago
  54. Zed
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    And then try this \[V=\frac{4}{3}\pi r^3\] Solve for r

    • 2 years ago
  55. Inopeki
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    Divide both sides by rt, getting PV/rt=n, am i right? By the way, why is PV upper case?

    • 2 years ago
  56. Inopeki
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    Zed, isnt that for volúmes of spheres or something? exept r should be ^3

    • 2 years ago
  57. zbay
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    good question and i don't have an answer for you but in that equation they are all upercase but the little n but i made a mistake and didn't capitilize the RT

    • 2 years ago
  58. Inopeki
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    But was i right?

    • 2 years ago
  59. Inopeki
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    Zed, im going to do yours now.

    • 2 years ago
  60. zbay
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    yes you were right and that was the ideal gas law by the way

    • 2 years ago
  61. Inopeki
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    Oh, that was a ^3! Divide both sides with (pi)r^2 getting V/(pi)r=4/3r, am i right sofar?

    • 2 years ago
  62. Inopeki
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    What does the law state?

    • 2 years ago
  63. Zed
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    okay so first step is to get r^3 by itself. what do you get?

    • 2 years ago
  64. zbay
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    http://en.wikipedia.org/wiki/Ideal_gas_law

    • 2 years ago
  65. Inopeki
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    Wait. I take back my first answer. First you divide both sides by 4/3(pi) to isolate r^3

    • 2 years ago
  66. Zed
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    good. have you done powers before and their inverses?

    • 2 years ago
  67. Inopeki
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    I think so, what do you mean?

    • 2 years ago
  68. Zed
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    okay so we have \[r^3=\frac{3V}{4}\] how do I get rid of the cubed part?

    • 2 years ago
  69. Inopeki
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    Oh, i have done powers but i dont recall doing inverses, is that like squareroot and cubicroot? I guess we need to get the "cubicalroot" out of r^3.

    • 2 years ago
  70. Inopeki
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    We should be able to just divide r by r 2 times.

    • 2 years ago
  71. Inopeki
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    3* and by itself i mean the root.

    • 2 years ago
  72. Zed
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    not quite see let's try this example to help explain it \[x^2=4\]\[x=\sqrt{4}\]\[x=4^{\frac{1}{2}}\]\[x=2\]

    • 2 years ago
  73. Inopeki
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    What? Can you explain that?

    • 2 years ago
  74. Zed
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    okay so when we bring powers (eg 3) over the equation sign they turn to 1/powers (eg 1/3). if we look at squared when we bring it over it turns to 1/2 which is also known as the square root.

    • 2 years ago
  75. Inopeki
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    Why does it become 1/2?

    • 2 years ago
  76. Zed
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    the may be abit hard to follow but this is why. Let f(x)=y \[y=x^7\] \[\ln y=7 \ln x\] \[lnx=(lny)/7\] \[x=e^{(lny)/7}\] \[Inverse = f^{-1}(x)=e^{(lnx)/7}\] \[ f^{-1}(x)=e^{lnx^\frac{1}{7}}\]\[=x^\frac{1}{7}\]

    • 2 years ago
  77. Zed
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    Now usually when learning this we just accept this as fact that \[x^n=x^{\frac{1}{n}}\] where n is a positive integer

    • 2 years ago
  78. Zed
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    That was probably way too much information at once just use the rule in the above post and try to solve \[r^3=\frac{3V}{4}\]

    • 2 years ago
  79. Inopeki
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    So if r^2 becomes 1/2, then r^3 should become 1/3, right?

    • 2 years ago
  80. Zed
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    Correct

    • 2 years ago
  81. Inopeki
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    That means that i have to maake r^3 to 1/3 and convert 1/3 to 4/12(to merge it with the right side). i then convert the right sides fractions too, getting 9V/12. Now i should be able to merge them, right? so i get r=13/12?

    • 2 years ago
  82. Inopeki
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    I screwed that up right...

    • 2 years ago
  83. Inopeki
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    What?

    • 2 years ago
  84. Zed
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    Sorry we want to raise both sides by the power of (1/3)

    • 2 years ago
  85. Inopeki
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    r^1/3=3V^1/3/4?

    • 2 years ago
  86. Zed
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    so the left hand side will be \[(r^3)^{\frac{1}{3}}=r^{\frac{3}{3}}=r^1=r\]

    • 2 years ago
  87. Inopeki
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    I dont understand.

    • 2 years ago
  88. Zed
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    okay the right side is good. we have \[(\frac{3V}{4})^\frac{1}{3}\]

    • 2 years ago
  89. Zed
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    remember on the left hand side we had r^3 and when raised it to the power of 1/3 the powers need to be multiplied.

    • 2 years ago
  90. Inopeki
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    Oh! So its r^3*1/3?

    • 2 years ago
  91. Zed
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    yes exactly it's one of the rules for powers. here are others if you wanted to learn and practice http://www.math.com/school/subject2/lessons/S2U2L2DP.html

    • 2 years ago
  92. Inopeki
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    So the answer is r^3*1/3=3V^3/4

    • 2 years ago
  93. Zed
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    yes and that simplifies to r=(3V/4)^1/3

    • 2 years ago
  94. Inopeki
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    Oh!

    • 2 years ago
  95. Zed
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    Well I'm off to sleep. I'll be on tomorrow if you want more examples or help :)

    • 2 years ago
  96. Inopeki
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    Thanks for everything!

    • 2 years ago
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