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anonymous

  • one year ago

Find the x value for point L such that JL and KL form a 3:1 ratio. A. 5.25 B. 2.33 C. 2.25 D. 4.33

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  1. anonymous
    • one year ago
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  2. mathstudent55
    • one year ago
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    |dw:1439400000835:dw|

  3. mathstudent55
    • one year ago
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    We place L on the segment, between J and K, in a way that JL = 3KL Ok so far?

  4. anonymous
    • one year ago
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    yes, perfect.

  5. mathstudent55
    • one year ago
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    Now we need to find the x-coordinate of point L.

  6. mathstudent55
    • one year ago
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    Here is our original graph again. |dw:1439400224309:dw|

  7. mathstudent55
    • one year ago
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    Since point L creates a ratio of 3:1 for JL to KL, and 3 + 1 = 4, then point L is 1/4 of the way from K to J. Ok with this?

  8. anonymous
    • one year ago
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    Ok!

  9. mathstudent55
    • one year ago
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    The x-coordinate of point L is the same as the x-coordinate of the point below it on a horizontal line going from K to the x-coordinate of J. See figure below. |dw:1439400776719:dw|

  10. mathstudent55
    • one year ago
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    Just like KL is 1/4 of KJ, so KL' is also 1/4 of KJ'.

  11. mathstudent55
    • one year ago
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    |dw:1439400914014:dw|

  12. mathstudent55
    • one year ago
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    We need to find the x-coordinate of point L'. The x-coordinate of point L is the same.

  13. mathstudent55
    • one year ago
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    |dw:1439400979255:dw|

  14. mathstudent55
    • one year ago
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    |dw:1439401009408:dw|

  15. mathstudent55
    • one year ago
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    Along the horizontal line J'K, it's easy to see that the distance from J' to K is 7. Also, we know that from J to L it's 3/4 of the way from J to K. So from J' to L' it is 3/4 of the distance of J' to K. Since the distance from J' to K is 7, the distance from J' to L' is 3/4 * 7 = 5.25

  16. anonymous
    • one year ago
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    thank you so much! you helped me a lot!

  17. mathstudent55
    • one year ago
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    |dw:1439401248977:dw|

  18. mathstudent55
    • one year ago
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    Finally, we add 5.25 to the x-coordinate of J' to get: 5.25 + -3 = 2.25

  19. mathstudent55
    • one year ago
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    You're welcome.

  20. anonymous
    • one year ago
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    can you help me with another one please?

  21. mathstudent55
    • one year ago
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    Sure

  22. mathstudent55
    • one year ago
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    pls start a new post

  23. anonymous
    • one year ago
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    okay i will

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