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anonymous

  • 5 years ago

Cylinder help..if the volume of a cylinder is 340 how do i find the perimeter, surface aria, and the hight

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  1. anonymous
    • 5 years ago
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    You know that the Surface Area equation of a cylinder right? which is:\[SA = 2\pi r^2 + 2\pi r H\] and the volume is : \[V = \pi r^2 H\] Perimeter is : \[P = 2\pi R\] all you have to do is find a relationship between all of them and you'll be able to figure out the height, perimeter and SA , so you'll get : Notice that the volume is = pi r ^2 h and the perimeter is = 2pi r so the SA can be written as : \[SA = 2(340)H + PH\]

  2. anonymous
    • 5 years ago
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    are you sure that that's the only given?

  3. anonymous
    • 5 years ago
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    yep i am givin x 1-a2 i think that is the radious and i have to find all of those for numbers 1-12

  4. anonymous
    • 5 years ago
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    x = 1- a2?

  5. anonymous
    • 5 years ago
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    you need more given, I doubt that the only given is the volume lol

  6. anonymous
    • 5 years ago
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    ok so the whole thing is a project we have to figure out what is best for making a box..we are given x values 1-12 and the volume of 340 cubic inches 1-12 i believe are the radices

  7. anonymous
    • 5 years ago
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    i need to find the hight. the surface aria..and the perimeter of a cylinder

  8. anonymous
    • 5 years ago
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    with that info

  9. anonymous
    • 5 years ago
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    Alright, pick one of the radiuses and plug it in the Volume's equation to find H :) see if the number makes sense.

  10. anonymous
    • 5 years ago
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    ok so i take piXr squared and get the hight?

  11. anonymous
    • 5 years ago
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    for example .3.14X1 squared?

  12. anonymous
    • 5 years ago
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    V = pi r^2 H , you can plug in any value or R, then find H, and after finding H you can plug it in the perimeter, and when you find the perimeter you'll be able to find the SA, the new equation that I have wrote up there ^_^

  13. anonymous
    • 5 years ago
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    as far as I can go lol, since I don't have enough given

  14. anonymous
    • 5 years ago
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    im not getting how you find h when you need it in the equation

  15. anonymous
    • 5 years ago
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    alright, we have the volume right? :)

  16. anonymous
    • 5 years ago
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    \[V = 340 \]

  17. anonymous
    • 5 years ago
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    you have values of x, which are considered as the radius right?

  18. anonymous
    • 5 years ago
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    so you can take R = 4 for example and plug it in the following equation:\[V = \pi r^2h\] so we have R and V, we can solve for H, right?

  19. anonymous
    • 5 years ago
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    lol, are you getting the idea now? :)

  20. anonymous
    • 5 years ago
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    \[340=\pi(4) squared ?\]

  21. anonymous
    • 5 years ago
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    actually it's :\[H = (340)/(8\pi)\] and calculate it, that ofcourse if you choose your radius to be 4 ^_^ clear?

  22. anonymous
    • 5 years ago
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    wait it's 32 pi instead of 8 pi I forgot to square it lol!

  23. anonymous
    • 5 years ago
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    ok so i take x=340/r is this correct...if not i give up for the night and am going to bed..i am sure i am making this 100X harder than it actually is

  24. anonymous
    • 5 years ago
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    and that will give me the hight ?

  25. anonymous
    • 5 years ago
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    don't give up lol, alright, let's try once more! WE have x values, right?

  26. anonymous
    • 5 years ago
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    yes those values= the radius...so if the virst x value is 1 how do i get the hight?

  27. anonymous
    • 5 years ago
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    it's simple, just take your time and concentrate with me

  28. anonymous
    • 5 years ago
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    Okay so x = r = 1 we have given: R = 1 V = 340 find H= ?

  29. anonymous
    • 5 years ago
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    yep thats what i need to do...now how?

  30. anonymous
    • 5 years ago
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    The equation is : \[V = \pi r^2 h\]

  31. anonymous
    • 5 years ago
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    plug in the given in the equation and solve for H : 340 = pi (1)^2 H H = 340/pi H = 108.23 in ^_^

  32. anonymous
    • 5 years ago
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    we have found the height now :)

  33. anonymous
    • 5 years ago
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    now we need to find the perimeter right?

  34. anonymous
    • 5 years ago
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    are you following me lol? =P wakey wakey~

  35. anonymous
    • 5 years ago
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    hold on

  36. anonymous
    • 5 years ago
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    okay ^_^

  37. anonymous
    • 5 years ago
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    ok so the hight for a radios of 1 is 108.2 correct

  38. anonymous
    • 5 years ago
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    excellent!

  39. anonymous
    • 5 years ago
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    ok so now perimeter

  40. anonymous
    • 5 years ago
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    now, we can find the perimeter right? The equation is : P = 2pi R we have R = 1, solve :)

  41. anonymous
    • 5 years ago
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    ok so i take \[2piX(1)squared\]

  42. anonymous
    • 5 years ago
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    i got 6.28

  43. anonymous
    • 5 years ago
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    lol no, the equation is :\[P = 2 \pi R\] substitute R in the equation and you'll get P ^_^

  44. anonymous
    • 5 years ago
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    lol yeah , that's right :)

  45. anonymous
    • 5 years ago
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    now, we can find the SA right?

  46. anonymous
    • 5 years ago
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    ok so now the SA

  47. anonymous
    • 5 years ago
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    The equation is :\[SA = 2\pi r^2 + 2\pi r h\] we have R = 1 and H = 108.2 substitute the given and solve ^_^

  48. anonymous
    • 5 years ago
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    i got 686.1

  49. anonymous
    • 5 years ago
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    very good !

  50. anonymous
    • 5 years ago
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    you have solved the question using R = 1 ^_^ did you get the idea now? :)

  51. anonymous
    • 5 years ago
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    ok let me make sure i can get this for 2 now and then ill let you go..if you dont mind

  52. anonymous
    • 5 years ago
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    good morning

  53. anonymous
    • 5 years ago
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    hi

  54. anonymous
    • 5 years ago
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    alright, it's the same idea blackford, but different values for R ^_^, good morning andy

  55. anonymous
    • 5 years ago
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    same way of solving , good luck! ^_^

  56. anonymous
    • 5 years ago
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    i got 120.8

  57. anonymous
    • 5 years ago
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    for?

  58. anonymous
    • 5 years ago
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    the hight with a radios of 2

  59. anonymous
    • 5 years ago
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    hmm it should be 27.06 (340)/(4pi) = H H = 27.06 in

  60. anonymous
    • 5 years ago
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    and andy going quite lol

  61. anonymous
    • 5 years ago
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    oops i see i think..i just realized you want h by itself

  62. anonymous
    • 5 years ago
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    yep ^_^ since we want to find it

  63. anonymous
    • 5 years ago
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    ok for a R of 3 i got 12.02 for the hight..correct?

  64. anonymous
    • 5 years ago
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    excellent ^_^

  65. anonymous
    • 5 years ago
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    ok i think i am getting it now

  66. anonymous
    • 5 years ago
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    you got the idea now :)

  67. anonymous
    • 5 years ago
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    you can proceed on your own right?

  68. anonymous
    • 5 years ago
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    for now thank you :)

  69. anonymous
    • 5 years ago
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    np ^_^

  70. anonymous
    • 5 years ago
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    i got a new one

  71. anonymous
    • 5 years ago
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    alright

  72. anonymous
    • 5 years ago
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    now for a square pyramid. this time i need the hight,perimeter, SA, and lateral aria

  73. anonymous
    • 5 years ago
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    volume still 340 and same x values 1-12

  74. anonymous
    • 5 years ago
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    All you have to do is write down the new equations for the square pyramid

  75. anonymous
    • 5 years ago
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    i think that x= base length now

  76. anonymous
    • 5 years ago
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    Volume is : \[V = (b^2h)/3\] Surface Area is : \[SA = 2bs + b^2\] and perimeter : \[P = (\S-B)(1/2)(l)\] find the relationship and solve ^_^ same idea as the previous one :)

  77. anonymous
    • 5 years ago
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    P = (S-B)(1/2)(L) *

  78. anonymous
    • 5 years ago
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    oh and the lateral area is = PL/2 where P = perimeter and L = slant height ^_^

  79. anonymous
    • 5 years ago
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    ok so 340=1squared(h)/3

  80. anonymous
    • 5 years ago
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    oh yeah i want the slant hight not lateral aria

  81. anonymous
    • 5 years ago
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    so \[340=1^2(h).....340=1/3 =.33 340/.33..correct or no\]

  82. anonymous
    • 5 years ago
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    340(3) = h h = 1020

  83. anonymous
    • 5 years ago
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    lol, I see what's your problem, you got the idea, but have trouble with calculation, take your time answering the questions :) step at a time ^_^

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