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Posted by **Melinda** on Sunday, December 4, 2011 at 11:35am.

it says to find the instantaneous rate at distance t = 6s with the given graph

First of all, it is a linear line,

do i need to find the equation of the graph?

After finding the equation, i plug in 5.99 and 6.01? and solve ?

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**MathMate**, Sunday, December 4, 2011 at 12:49pmIf it is truly linear, the rate (slope of the line) is the same as the instantaneous rate at all times.

Try 5.99 and 6.01, it should give the same answer as 5 and 7.

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**Melinda**, Sunday, December 4, 2011 at 12:51pmi got 0.25 answer, I wish I could show you the graph... T.T

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**MathMate**, Sunday, December 4, 2011 at 1:06pmI do not know the domain of the graph (i.e. the max min values of t), but if you try different intervals, and if they always give the same answer of 0.25, you should be OK.

If you can scan the graph and post it at one of the image sites and post the link, then we can all see it. However, I believe it is not necessary if the graph is truly linear.

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**Melinda**, Sunday, December 4, 2011 at 1:18pmHere is the link

imgur dot com/3V1AX

The question asks: A distance versus time graph is shown below. What is the instantaneous rate of change in the distance at t=6s

sorry, it won't rotate, you can save the pic and rotate >.<

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**Melinda**, Sunday, December 4, 2011 at 1:19pmI can't post exact link due to errors :S

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**MathMate**, Sunday, December 4, 2011 at 1:47pmYes, it looks like the graph between (4,3) and (8,4) is a straight line, so if you apply the formula for slope:

m=(4-3)/(8-4)=1/4 as you had it.

Since t=6 is within the above interval of [4,8], the calculated value of slope (m=1/4) applies.

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**Melinda**, Sunday, December 4, 2011 at 1:47pmSo I am right? =D

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**Melinda**, Sunday, December 4, 2011 at 1:50pmBtw, may I post another question here with image, I just want you to verify if I am correct :|

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**MathMate**, Sunday, December 4, 2011 at 2:00pmYes, your answer is correct.

Feel free to post another question with image, but with a new thread. Posts tagged onto old threads tend to get lost.

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**Melinda**, Sunday, December 4, 2011 at 2:00pmOk, thanks~

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**MathMate**, Sunday, December 4, 2011 at 2:05pmYou're welcome!

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