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anonymous

  • one year ago

Algebra!!!

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  1. anonymous
    • one year ago
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    The function below shows the number of car owners f(t), in thousands, in a city in different years t: f(t) = 0.25t2 − 0.5t + 3.5 The average rate of change of f(t) from t = 2 to t = 6 is ______ thousand owners per year.

  2. anonymous
    • one year ago
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    @Vocaloid @futurelegend @thomaster @LynFran

  3. Vocaloid
    • one year ago
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    [f(6)-f(2)]/(6-2)

  4. Vocaloid
    • one year ago
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    find f(6) and f(2), then plug them into the equation

  5. anonymous
    • one year ago
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    You need to take the derivatives. \[f'(t) = 0.5t -5\]

  6. anonymous
    • one year ago
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    Oh lord, I have no idea.

  7. anonymous
    • one year ago
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    Wait, is it 2?

  8. anonymous
    • one year ago
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    @saseal ? @Vocaloid ?

  9. Vocaloid
    • one year ago
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    f(t) = 0.25t^2 − 0.5t + 3.5 f(6) = ?

  10. anonymous
    • one year ago
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    Yea, I suppose.

  11. anonymous
    • one year ago
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    OK.

  12. anonymous
    • one year ago
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    @DaBest21

  13. anonymous
    • one year ago
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    @ganeshie8 @Hero @ParthKohli @thomaster @triciaal @Michele_Laino @alekos

  14. Michele_Laino
    • one year ago
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    please see the procedure of @Vocaloid

  15. triciaal
    • one year ago
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    the average rate of change is like the slope of a best fit line slope of a line is change in y / corresponding change in x = y2 - y1 find the y value when x = 6 this is y2 find the y value when x = 2 this is y1 the change in x from 6 to 2 this is the same thing done above by @Vocaloid

  16. anonymous
    • one year ago
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    The line isn't straight, its a quadratic curve. \[f(t) = 0.25t ^{2} + 0.5t +3.5\]

  17. anonymous
    • one year ago
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    oops a typo in the function should be \[f(t) = 0.25t^2-0.5t+3.5\]

  18. anonymous
    • one year ago
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    What vocaloid found was the total change during the total period from t=2 to t=6, but the question asked for average rate of change from period t=2 to t=6.

  19. triciaal
    • one year ago
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    |dw:1438098854324:dw|

  20. triciaal
    • one year ago
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    |dw:1438100759868:dw|

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