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anonymous

  • 5 years ago

The temperature outside dropped from 42 degrees at 5:00 am to 24 degrees at 5:00 pm. Find the rate of change in the temperature

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  1. anonymous
    • 5 years ago
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    do you have the general formula?

  2. anonymous
    • 5 years ago
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    No I do not

  3. anonymous
    • 5 years ago
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    use:( y" -y')?(x"-x1)

  4. anonymous
    • 5 years ago
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    The temperature dropped 18 degrees (42-24) in 12 hours, so the rate of change was a decrease of 1.5 degrees per hour. (18/12) or Newton's Law of Cooling? T(t) = Ts + (To - Ts)*e^(-k*t) ; where: t is the time in the preferred units (seconds, minutes, hours, etc.) T(t) is the temperature of the object at time t Ts is the sorrounding constant temperature To is the initial temperature of the object k is a constant to be found

  5. anonymous
    • 5 years ago
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    Is the y2= 42 or 24? Is the y1 = 42 or 24?

  6. anonymous
    • 5 years ago
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    y2=5pm,y1=5am subtrac you get 12 x2=24, x1= 42 subtrac you det negative =-18 slope= 12/-18=-3/4 the rate -3/4

  7. anonymous
    • 5 years ago
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    sorry -3/2 not -3/4

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