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Mathematics
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Rate of change
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@princeharryyy
a) i) (0-25)/(5-2.5) = -10, ii) 0 to 1s => 16/1 = 16 iii) 1s tp 2s => (24 - 16)/(2-1) = 8
b) no, the rate of change in each case is different. The rate is given as final value of height - initial value of height / final time - initial time Which comes out to be different in each case. Moreover graph is not constant over the period of intervals.
Average rate over 5 seconds = > (0-0)/(5-1) = 0
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are you sure?
of course! y ?
oh nvm, your correct :)
ok
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