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KE8717504

  • 3 years ago

Solve the following formula for the specified variable. C=1/4h(s+k) solve for s???

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  1. etemplin
    • 3 years ago
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    if this was a normal equation, and all those variables were numbers, how would you solve for s? hint: try to get s by itself

  2. KE8717504
    • 3 years ago
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    It's been so long since i actually did this type of problem

  3. etemplin
    • 3 years ago
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    i also have a question about the original question is it C=(1/4)h(s+k) or C=1/(4h(s+k))

  4. KE8717504
    • 3 years ago
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    it's the first one that you put on there

  5. KE8717504
    • 3 years ago
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    c=1/4h(s+k)

  6. etemplin
    • 3 years ago
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    the first and easiest thing to get rid of is the 1/4 if we want to re-write the equation, we could like this: c=(h(s+k))/4 since we are dividing by 4 on the right, we can multiply both sides by 4 to get closer to having s by itself (by moving the 4 over) we get: 4c=h(s+k) now, to get rid of that h notice its being multiplied by the quantity inside the parenthesis to move it to the other side, we need to divide both sides by h (4c)/h=s+k finally, subtract k from both sides to get your answer

  7. KE8717504
    • 3 years ago
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    ok im confused

  8. etemplin
    • 3 years ago
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    |dw:1351648217904:dw|

  9. etemplin
    • 3 years ago
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    |dw:1351648246343:dw|

  10. KE8717504
    • 3 years ago
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    ok so when i solved that i got 4c/h-k

  11. KE8717504
    • 3 years ago
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    is that right ?

  12. etemplin
    • 3 years ago
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    you should get 4c=h(s+k)

  13. KE8717504
    • 3 years ago
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    ok

  14. KE8717504
    • 3 years ago
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    That's what you got as the answer??

  15. etemplin
    • 3 years ago
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    now, divide both sides by h

  16. KE8717504
    • 3 years ago
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    so the h on the one side would cancel and just leave me with s+k but i don't know what i would get on the other side

  17. etemplin
    • 3 years ago
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    (4c)/h

  18. KE8717504
    • 3 years ago
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    so my answr is going to be 4c/h=s +k??

  19. etemplin
    • 3 years ago
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    you have to get s by itself subtract k from both sides

  20. KE8717504
    • 3 years ago
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    OMG!! I'm sooo lost

  21. KE8717504
    • 3 years ago
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    I thought that I just subracted something

  22. etemplin
    • 3 years ago
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    it says solve for s that means you have to get s by itself

  23. etemplin
    • 3 years ago
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    think of it this way if you have s+k and you want to get s by itself, wouldnt it be necessary to subtract k s+k-k=s

  24. KE8717504
    • 3 years ago
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    yes

  25. etemplin
    • 3 years ago
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    since you subtracted k from one side, you MUST subtract it from the other

  26. KE8717504
    • 3 years ago
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    ok im just confused on what im going to get i thought what i typed a minute ago was my answr

  27. etemplin
    • 3 years ago
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    your answer is going to have s by itself on one side and a whole bunch of stuff on the other after you subtract k from both sides, the final answer is s=((4c)/h)-k

  28. KE8717504
    • 3 years ago
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    4c/h-k?

  29. KE8717504
    • 3 years ago
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    I thought that is what i put a minute ago

  30. etemplin
    • 3 years ago
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    it was not. you hadnt subtracted k yet

  31. KE8717504
    • 3 years ago
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    Ok well i got that one wrong because the C was not capital

  32. KE8717504
    • 3 years ago
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    b=1/2f(r+k)

  33. KE8717504
    • 3 years ago
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    for r

  34. KE8717504
    • 3 years ago
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    I've got this stuff due tommorrow and im still not doen after this

  35. KE8717504
    • 3 years ago
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    r u there?

  36. etemplin
    • 3 years ago
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    yes. sorry

  37. KE8717504
    • 3 years ago
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    its ok. r u working it or something

  38. etemplin
    • 3 years ago
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    i got distracted

  39. etemplin
    • 3 years ago
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    for this one, you can follow the exact same steps as the first one. Multiply both sides by 2 to get rid of the 1/2 divide both sides by f subtract k from both sides what do you get?

  40. KE8717504
    • 3 years ago
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    one sec

  41. KE8717504
    • 3 years ago
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    i even wrote the wrong problem

  42. etemplin
    • 3 years ago
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    its ok, it happens

  43. KE8717504
    • 3 years ago
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    b=1/2f(r+k)

  44. KE8717504
    • 3 years ago
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    And I'm solving for r

  45. etemplin
    • 3 years ago
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    thats the exact same

  46. KE8717504
    • 3 years ago
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    i got 2c/f-k but that's not right

  47. etemplin
    • 3 years ago
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    where did you get a c theres no c in b=1/2f(r+k)

  48. KE8717504
    • 3 years ago
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    i have no idea

  49. KE8717504
    • 3 years ago
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    i got that one may i ask another onee

  50. KE8717504
    • 3 years ago
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    ?

  51. etemplin
    • 3 years ago
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    yes

  52. KE8717504
    • 3 years ago
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    |dw:1351649553338:dw|

  53. KE8717504
    • 3 years ago
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    that is my next one

  54. etemplin
    • 3 years ago
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    hmmm give me a sec

  55. KE8717504
    • 3 years ago
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    k

  56. etemplin
    • 3 years ago
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    ok start by simplifying sqrt((xy)^2)=xy yes?

  57. KE8717504
    • 3 years ago
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    I guess i don't know lol

  58. etemplin
    • 3 years ago
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    it is. trust me anyway, on the top, you can simplify sqrt(x^5y^9) to sqrt(x^4y^8)*sqrt(xy) sqrt(x^4y^8)=x^2*y^4 so 7x^2y^4*sqrt(xy) ---------------- 3xy one x and 1 y on the top and bottom cancel answer: (7xy^3*sqrt(xy))/3

  59. etemplin
    • 3 years ago
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    |dw:1351649829505:dw|

  60. KE8717504
    • 3 years ago
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    can u draw that for me because it's hard for me to read and understand it

  61. KE8717504
    • 3 years ago
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    is that 3 suppose to be with the root?

  62. KE8717504
    • 3 years ago
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    or is it with the Y?

  63. KE8717504
    • 3 years ago
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    @etemplin

  64. etemplin
    • 3 years ago
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    its with the y

  65. etemplin
    • 3 years ago
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    were the original numbers with the root or were they outside?

  66. KE8717504
    • 3 years ago
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    because i put it with the y and it said it was wrong.

  67. KE8717504
    • 3 years ago
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    they were on the outside of the root but they were part of the root

  68. etemplin
    • 3 years ago
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    were the original numbers (the 3 and 7) with the root or in front of it?

  69. KE8717504
    • 3 years ago
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    they were in front

  70. KE8717504
    • 3 years ago
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    |dw:1351650274050:dw|

  71. etemplin
    • 3 years ago
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    that changes the answer then i assumed they were not part of the root; i assumed they were multiplied by it my mistake

  72. KE8717504
    • 3 years ago
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    like that

  73. KE8717504
    • 3 years ago
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    it's fine

  74. etemplin
    • 3 years ago
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    \[\sqrt[7]{x ^{5}y ^{9}} or 7\sqrt{x ^{5}y ^{9}}\]

  75. KE8717504
    • 3 years ago
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    ok what is that

  76. etemplin
    • 3 years ago
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    which is it?

  77. KE8717504
    • 3 years ago
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    is that the answer?

  78. KE8717504
    • 3 years ago
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    the first one

  79. etemplin
    • 3 years ago
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    yea. when i solved it the first time i did it like the 2nd choice the new answer is|dw:1351650440270:dw|

  80. KE8717504
    • 3 years ago
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    so my y is on the outside??

  81. etemplin
    • 3 years ago
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    yes

  82. etemplin
    • 3 years ago
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    you cant simplify the bottom at all, and the top can be just a little bit

  83. KE8717504
    • 3 years ago
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    what u mean

  84. etemplin
    • 3 years ago
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    im just saying the answer is the one i drew the reason is given in the post after that

  85. KE8717504
    • 3 years ago
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    ok

  86. KE8717504
    • 3 years ago
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    that wasn't correct

  87. etemplin
    • 3 years ago
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    is the 3 part of the root like the 7 is?

  88. etemplin
    • 3 years ago
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    another answer is |dw:1351651467504:dw|

  89. etemplin
    • 3 years ago
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    i apologize for any incorrect answers

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