anonymous
  • anonymous
Let f(t)=t^2+4t+2. Find a value of t such that the average rate of change of f(t) from 0 to t equals 10. I know how to find the average rate of change but I am little lost on how to get it to equal 10.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1378831912340:dw| is this what ur looking for?
anonymous
  • anonymous
No not really, I don't see how you got 5 as the answer and it wasn't right.
tkhunny
  • tkhunny
How do you find the average rate of change? Please demonstrate an example.

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anonymous
  • anonymous
A(x)=f(x)-f(a)/x-a do you find the rate of change and then set it to equal to 10 and solve it?
tkhunny
  • tkhunny
Well, it is hoped that you mean something other than what you wrote. I'm sure you meant A(x) = (f(x) - f(a))/(x - a). Let's not forget our Order of Operations. Otherwise, that's very good. You are looking in the right place. So, what's stopping you from solving \(10 = \dfrac{f(t_{0}) - f(0)}{t_{0} - 0}\) for \(t_{0}\)
anonymous
  • anonymous
@tkhunny I think the change in t is throwing me off. Do you times t and t^2 and become t^3? I have no idea what I am doing here... \[\frac{ t^2+4t+2-2 }{ t^2+4t+2-0 }=10 \] before doing any simplying
tkhunny
  • tkhunny
Why do you have f(t) in the denominator? That should be just "t".
anonymous
  • anonymous
Oh I'm dumb.....oops
tkhunny
  • tkhunny
I am somewhat concerned by what I see, here. 1) You should not be saying "Do I times...". This is not good mathematical language. You can say "3 times 4" to mean 3*4, but "Do I times 3 by 4?" No. The right word is "multiply". 2) You should not be contemplating multiplication where addition is clearly indicated. Conclusion: You will have a very hard time in this course if you spend a lot of your time relearning algebra. You simply must get up to speed in these fundamental areas.
tkhunny
  • tkhunny
Did you get it, yet? \(\dfrac{t^{2} + 4t + 2 - 2}{t - 0} = \dfrac{t^{2} + 4t}{t} = t+4 = 10\) Oh, that simplified nicely!

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