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Answer:

All figures rounded to 4 decimal places

[tex][10, 11]: \quad \boxed{ 10.57 \; m/s}[/tex]

[tex][10, 10.5]: \quad \boxed{ 10.985\; m/s}[/tex]

[tex][10, 10.1]: \quad \boxed{ 11.317\;m/s}[/tex]

[tex][10, 10.01]: \quad \boxed{ 11.3917\;m/s}[/tex]

[tex][10, 10.001]: \quad \boxed{ 11.3992\;m/s}[/tex]

Step-by-step explanation:

The given function for height h in meters in terms of time t in seconds is:

[tex]h(t) = 28t- 0.83t^2[/tex]

To compute the average velocity for each of the intervals provided in the question

  • Find the height of the arrow at the lower and upper limits of the interval
  • Find the difference between these two values
  • Divide the difference by the time difference (the interval length)

Interval [10, 11]

[tex]h(10) = 28(10) - 0.83\cdot 10^2 = 197 $ meters\\\\h(11) = 28(11) - 0.83 \cdot 11^2 = 207.57 $ meters\\\\Change in height = 207.57 - 197 = 10.57 $ meters\\\\\text{Elapsed time} = 11 - 10 = 1 $ second\\\\\text{Average Velocity} = \dfrac{10.57 \; meters}{1 \;second} = 10.57 $ m/s[/tex]

Interval [10, 10.5]

[tex]h(10) = 28(10) - 0.83\cdot 10^2 = 197 $ meters\\\\h(10.5) = 28(10.5) - 0.83 \cdot (10.5)^2 = 202.4925 $ meters\\\\Change in height = 202.4925 - 197 = 5.4925 $ meters\\\text{Elapsed time} = 10.5 - 10 = 0.5 $ second\\\\\text{Average Velocity} = \dfrac{ 5.4925 \; meters}{0.5 \;second} = 10.985$ m/s[/tex]

Interval [10, 10.1]

[tex]h(10) = 28(10) - 0.83\cdot 10^2 = 197 $ meters\\\\h(10.1) = 28(10.1) - 0.83\cdot (10.1)^2 = 198.1317 $ meters\\\\\text{Change in height} = 198.1317 - 197 = 1.1317 $meters\\\\\text{Elapsed time} = 10.1 - 10 = 0.1 $ second\\\\\text{Average Velocity} = \dfrac{ 1.1317 \; meters}{0.1 \;second} = 11.317 $ m/s[/tex]

Interval [10, 10.01]

[tex]h(10) = 28(10) - 0.83\cdot 10^2 = 197 $ meters\\[/tex]

[tex]h(10.01) = 28(10.01) - 0.83 \cdot (10.01)^2 = 197.113917 $ meters\\[/tex]

[tex]\text{Change in height} = 197.113917 - 197 = 0.113917 $ meters[/tex]

[tex]\text{Elapsed time} = 10.01 - 10 = 0.01 $ second\\[/tex]

[tex]\text{Average Velocity} = \dfrac{ 0.113917\; meters}{0.01 \;second} = 11.3917 $ m/s[/tex]

Interval [10, 10.001]

[tex]h(10) = 28(10) - 0.83\cdot 10^2 = 197 $ meters\\\\[/tex]

[tex]h(10.001) = 28(10.001) - 0.83 \cdot (10.001)^2 = 197.01139917 $ meters\\[/tex]

[tex]\text{Change in height} = 197.01139917 - 197 = 0.01139917 $ meters\\[/tex]

[tex]\text{Elapsed time} = 10.001 - 10 = 0.001 $ second\\[/tex]

[tex]\text{Average Velocity} = \dfrac{ 0.01139917 \; meters}{0.001 \;second} = 11.39917 $ m/s[/tex]

Just a note:
You really don't have to compute h(10) each time, since the lower limit on the interval is always 10. So you can compute h(10) once and this is equal to 197

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