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anonymous

  • one year ago

someone please help rise/run medal !!!!!

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    What do you think the answer is? (What is your understanding?) :)

  3. anonymous
    • one year ago
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    im pretty sure that the answer is 1/40 any ideas ?

  4. anonymous
    • one year ago
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    Rise/Run is also known as slope. Slope's may also be calculated like this: y2 - y1 / x2 - x1

  5. anonymous
    • one year ago
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    ok

  6. anonymous
    • one year ago
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    |dw:1433863318239:dw|

  7. anonymous
    • one year ago
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    We need two points to plug in to these values. Remember: (x, y) We will label these two values.

  8. anonymous
    • one year ago
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    So the graph shows these points: (3, 120) (2, 80)

  9. anonymous
    • one year ago
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    Is everything clear so far?

  10. anonymous
    • one year ago
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    yes sir ^~^

  11. anonymous
    • one year ago
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    i know all this lol but its nice to see what i did wrong

  12. anonymous
    • one year ago
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    |dw:1433863432536:dw|

  13. anonymous
    • one year ago
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    Oh okay, then plug in these values into the formula I provided you :)

  14. anonymous
    • one year ago
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    y2 - y1 / x2 - x1

  15. anonymous
    • one year ago
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    Need any help?

  16. anonymous
    • one year ago
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    es therefore the avnswer i gave was correct right ?

  17. anonymous
    • one year ago
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    if not then im still confused

  18. anonymous
    • one year ago
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    No...

  19. anonymous
    • one year ago
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    Plug in your numbers to understand where I'm coming from.

  20. anonymous
    • one year ago
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    |dw:1433864274794:dw|

  21. anonymous
    • one year ago
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    I think you accidentally switched the top & bottom :)

  22. anonymous
    • one year ago
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    Understand?

  23. anonymous
    • one year ago
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    Please don't forget to tip those who help you out with a medal // fan :)

  24. anonymous
    • one year ago
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    ok so its 40 after u plug in the numbers thanks so much ! took me awhile to understand

  25. anonymous
    • one year ago
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    :) You're welcome.

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spraguer (Moderator)
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