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3 years ago

I WILL MARK BRAINLIEST

Mathematics

2 answers:

Elina [12.6K]3 years ago

8 0

The answer to this question is D. 364

neonofarm [45]3 years ago

3 0

A^2 +b^2 =c^2
a^2 +27^2=365^2
a^2+ 729=133,225
a^2=133,225- 729
a^2=132,496
a^2= The square root of 132,496
a=364

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Answer:

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3 years ago

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3 years ago

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tankabanditka [31]

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The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha

Xelga [282]

Answer:

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

Step-by-step explanation:

Let suppose that function $f$ is continuous and integrable in the given intervals, by integral definition of average we have that:

$\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6$ (1)

$\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20$ (2)

By Fundamental Theorems of Calculus we expand both expressions:

$\frac{F(3)-F(-2)}{3-(-2)} = -6$

$F(3) - F(-2) = -30$ (1b)

$\frac{F(5)-F(3)}{5-3} = 20$

$F(5) - F(3) = 40$ (2b)

We obtain the average value of $f$ over the interval $[-2, 5]$ by algebraic handling:

$F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)$

$F(5) - F(-2) = 10$

$\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}$

$\bar f = \frac{10}{7}$

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

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3 years ago