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Svet_ta [14]
3 years ago
11

Find the average value of the function f(t)=(t-2)^2 on [0,6]

Mathematics
1 answer:
pentagon [3]3 years ago
4 0

Answer:

2

Step-by-step explanation:

The "average value of function f(x) on interval [a, b] is given by:

                      f(b) - f(a)

ave. value = ---------------

                         b - a

Here f(t)=(t-2)^2.

Thus, f(b) = (b - 2)^2.  For b = 6, we get:

          f(6) = 6^2 - 4(6) + 4, or f(6) = 36 - 24 + 4 = 16

For a = 0, we get:

           f(0) = (0 - 2)^2 = 4

Plugging these results into the ave. value function shown above, we get:

                      16 - 4

ave. value = ------------ = 12/6 = 2

                         6 - 0

The average value of the function f(t)=(t-2)^2 on [0,6] is 2.


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Answer:

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Step-by-step explanation:

1. This pair of intersecting lines form four pairs of supplementary angles, which may be one approach to this problem.

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7. The sum of the angles are known to 360 degrees, as provided previously, such that m∠1 + m∠2 + m∠3 + m∠4 = 360°, ⇒ 35 + 145 + 35 + 145 = 360, ⇒ <em>360 = 360</em>

8. <em>This proves that the m∠1 = 35 degrees ( ° )</em>

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