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Katieann16

  • 2 years ago

y varies jointly as x and z, and y = 32 m, x = 6 m, and z = 8 m. What is the value of y when x = 5 m and z = 12 m?

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  1. Mertsj
    • 2 years ago
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    Can you write the joint variation equation for x y and z? Use k for the constant.

  2. Katieann16
    • 2 years ago
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    @jim_thompson5910

  3. Mertsj
    • 2 years ago
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    y=kxz

  4. Mertsj
    • 2 years ago
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    Now plug in the given values and find k.

  5. jim_thompson5910
    • 2 years ago
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    does Mertsj's response make sense?

  6. Katieann16
    • 2 years ago
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    a little, im still a little confused ..

  7. Katieann16
    • 2 years ago
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    so it would be 32m= k *5m*12m

  8. jim_thompson5910
    • 2 years ago
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    y = kxz we're given y = 32 m, x = 6 m, and z = 8 m plug all this in to get 32 = k*6*8 now solve for k Note: I think we can ignore the m's because I think they refer to meters (eg: x = 6 m means x is 6 meters long)

  9. Katieann16
    • 2 years ago
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    4?

  10. jim_thompson5910
    • 2 years ago
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    no

  11. jim_thompson5910
    • 2 years ago
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    32 = k*6*8 32 = k*48 k = ??

  12. Katieann16
    • 2 years ago
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    its 4?

  13. jim_thompson5910
    • 2 years ago
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    no it's not 4 divide both sides by 48 to solve for k

  14. Katieann16
    • 2 years ago
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    0.6?

  15. jim_thompson5910
    • 2 years ago
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    more like 0.667 but I would keep it as a fraction so k = 2/3

  16. jim_thompson5910
    • 2 years ago
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    y = kxz y = (2/3)xz y = (2/3)*5*12 plug in x = 5 and z = 12 y = ???

  17. Katieann16
    • 2 years ago
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    my choices are 90m, 4m, 40m, 9m

  18. jim_thompson5910
    • 2 years ago
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    5*12 is what

  19. Katieann16
    • 2 years ago
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    60

  20. jim_thompson5910
    • 2 years ago
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    now multiply that by 2 then divide that result by 3

  21. Katieann16
    • 2 years ago
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    4

  22. jim_thompson5910
    • 2 years ago
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    close, but not quite

  23. Katieann16
    • 2 years ago
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    40*

  24. jim_thompson5910
    • 2 years ago
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    yep

  25. Katieann16
    • 2 years ago
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    thank you(:

  26. jim_thompson5910
    • 2 years ago
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    you're welcome

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