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anonymous

  • one year ago

I need help with this question, The graph below shows the value of Edna's profits f(t), in dollars, after t months: graph of quadratic function f of t having x intercepts at 6, 0 and 18, 0, vertex at 12, negative 36, and passes through point 21, 41.25 What is the closest approximate average rate of change for Edna's profits from the 18th month to the 21st month?

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  1. anonymous
    • one year ago
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    whees the graph??

  2. anonymous
    • one year ago
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    I don't know how to put it on here but the part that says graph of quadratic function f of x that is what is on the graph.

  3. anonymous
    • one year ago
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    click on attach file i can't understand u

  4. anonymous
    • one year ago
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    Oh ok sorry on my computer the screen is small so i didn't see that okay one sec and i will put the graph.

  5. anonymous
    • one year ago
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    ok :3

  6. anonymous
    • one year ago
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  7. anonymous
    • one year ago
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    Here is the graph

  8. anonymous
    • one year ago
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    ok wait up

  9. anonymous
    • one year ago
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    ok

  10. anonymous
    • one year ago
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    rate of change= slope

  11. anonymous
    • one year ago
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    do you know how to find the slope?

  12. anonymous
    • one year ago
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    since you are not answering i will show ya

  13. anonymous
    • one year ago
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    slope=\[\frac{ y2-y1 }{ x2-x1 }\]

  14. anonymous
    • one year ago
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    so your two points from 18 months to 21 months is: (0,18) and (21,41.25).

  15. anonymous
    • one year ago
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    Hey sorry it wouldn't let me type lol

  16. anonymous
    • one year ago
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    plug the points into the slope formula and you get:\[\frac{ 41.25-18 }{ 21-0 } =\frac{ 23.25 }{ 21 }\]

  17. anonymous
    • one year ago
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    thats the rate o change :3

  18. anonymous
    • one year ago
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    g'night btw

  19. anonymous
    • one year ago
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    Oh okay thank you! Lol

  20. anonymous
    • one year ago
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    :)

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