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anonymous

  • one year ago

Hey guys it would be awesome if someone could help... Thank You A candy bar box is in the shape of a triangular prism. The volume of the box is 2,400 cubic centimeters. http://learn.flvs.net/webdav/assessment_images/educator_mjmath2_v14/mjmath2_%20test2_m5_g4_p.jpg Part A: What is the height of the box? Part B: What is the approximate amount of cardboard used to make the sides of the candy box? Explain how you got your answer. (5 points)

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  1. anonymous
    • one year ago
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    @DanJS

  2. anonymous
    • one year ago
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    @Deeezzzz

  3. DanJS
    • one year ago
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    Links to signed in things dont work.. but

  4. DanJS
    • one year ago
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    |dw:1437756105426:dw|

  5. anonymous
    • one year ago
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    A triangular prism is shown with base of triangle labeled 16 cm, sides of triangles labeled 17 cm, and length of the box equal to 20 cm.

  6. DanJS
    • one year ago
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    Cant link pages that you signed into , im guessing, does not work

  7. anonymous
    • one year ago
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    does that help

  8. anonymous
    • one year ago
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    the descri[tion

  9. DanJS
    • one year ago
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    yes

  10. anonymous
    • one year ago
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    great

  11. DanJS
    • one year ago
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    |dw:1437756241528:dw|

  12. anonymous
    • one year ago
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    perfect

  13. DanJS
    • one year ago
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    The volume of that shape , is just the area of the triangular face times the extruded length of the box

  14. DanJS
    • one year ago
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    \[V = 1/2*base*height*depth\]

  15. DanJS
    • one year ago
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    To find the height, need to work with the triangle like this...

  16. anonymous
    • one year ago
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    proceed

  17. DanJS
    • one year ago
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    |dw:1437756382708:dw|

  18. DanJS
    • one year ago
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    So the height, is found with the pythagorean theorem 17^2 = 8^2 + h^2

  19. anonymous
    • one year ago
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    can i help

  20. DanJS
    • one year ago
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    \[V = 1/2*base*height*depth \] \[V = 1/2 * 16 * \sqrt{17^2-8^2} * 20\]

  21. anonymous
    • one year ago
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    yes of course @kittylover1234

  22. anonymous
    • one year ago
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    that seem right i agree

  23. DanJS
    • one year ago
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    oh, they gave you the volume at the start, so you could have figured h, without all the triangle things... V = 2400 = 1/2 * 16 * h * 20

  24. anonymous
    • one year ago
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    yep so good so far

  25. anonymous
    • one year ago
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    give me a sec im trying to understand

  26. DanJS
    • one year ago
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    either way, the height comes out the same...k

  27. anonymous
    • one year ago
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    ummm ook...

  28. DanJS
    • one year ago
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    For the Surface Area, you have to add up the area of all the sides, all the dimensions are given

  29. DanJS
    • one year ago
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    2 triangles, 3 rectangles

  30. anonymous
    • one year ago
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    so the hieght is 15 right?

  31. DanJS
    • one year ago
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    right

  32. anonymous
    • one year ago
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    ok then for the surfcae area i would have to add up all the areas

  33. anonymous
    • one year ago
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    i got the answer

  34. DanJS
    • one year ago
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    |dw:1437756997046:dw|

  35. anonymous
    • one year ago
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    what did you guys get

  36. anonymous
    • one year ago
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    what is it kitty ;)

  37. DanJS
    • one year ago
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    didnt calcualte

  38. anonymous
    • one year ago
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    me niether

  39. anonymous
    • one year ago
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    mine was wrong dang

  40. anonymous
    • one year ago
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    calculated wrong

  41. DanJS
    • one year ago
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    1240

  42. anonymous
    • one year ago
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    that what you got

  43. DanJS
    • one year ago
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    \[S.A. = 2[\frac{ 1 }{ 2 }]*16*15 + [20*17]+[20*17]+[20*16]\]

  44. anonymous
    • one year ago
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    I got 1,320

  45. DanJS
    • one year ago
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    2 triangles, 3 rectangles

  46. anonymous
    • one year ago
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    does any one agree

  47. anonymous
    • one year ago
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    oh nvm your rigt Dan

  48. anonymous
    • one year ago
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    right**

  49. DanJS
    • one year ago
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    |dw:1437757284740:dw|

  50. anonymous
    • one year ago
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    Yup your def Right Dan Thanks so much both of you...!! @DanJS @kitty_lover1234

  51. DanJS
    • one year ago
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    welcome

  52. anonymous
    • one year ago
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    welcome

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