h = 0.2 t2 + 6.9 t + 16
h(10) = 0.2(10)2 + 6.9(10) + 16 = 105 in.
h(2) = 0.2(2)2 + 6.9(2) + 16 = 30.6 in.
Average rate of change = (f(10) - f(2))/ (10 - 2) = (105 - 30.6)/(10 - 2) = 74.4/8 = 9.3 in./yr.
Ela J.
asked 04/20/23Your friend's height, for the first 10 years of their life, can be modeled with the function h`=-0.2t^2+6.9t+16 , where t is the time in years since your friend was born and h is the height in inches. . What was the average rate of change, in inches per year, from year 2 to year 10?
h = 0.2 t2 + 6.9 t + 16
h(10) = 0.2(10)2 + 6.9(10) + 16 = 105 in.
h(2) = 0.2(2)2 + 6.9(2) + 16 = 30.6 in.
Average rate of change = (f(10) - f(2))/ (10 - 2) = (105 - 30.6)/(10 - 2) = 74.4/8 = 9.3 in./yr.
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