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poizon [28]
3 years ago
6

Help me out with this one guys (10 points)

Mathematics
1 answer:
Anna007 [38]3 years ago
6 0
M is 3 and -3

p is all real number

r has no solution
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Find the range of the function F(x) = the integral from 0 to x of the square root of 4-t^2 dt
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Note that √(4 - t²) is defined only as long as 4 - t² ≥ 0, or -2 ≤ t ≤ 2. Then the real integral exists only if -2 ≤ x ≤ 2. (Otherwise we deal with complex numbers.)

If x = 2, then the integral corresponds to the area of a quarter-circle with radius 2. This means that the integral has a maximum value of 1/4 • π • 2² = π.

On the opposite end, if x = -2, then the integral has the same value, but the integral from 0 to -2 is equal to the negative integral from -2 to 0. So the minimum value is -π.

For all x in between, we observe that the integrand is continuous over the rest of its domain, so F(x) is continuous.

Then the range of F(x) is the interval [-π, π].

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2 years ago
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6.0 x 10^2

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= 24000073

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= 6.0 x 10^2

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3 years ago
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Write a number that is greater than 714,587
TEA [102]
A number that is greater than 714,587 is 1,000,000
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3 years ago
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Answer:

f(5) = 3200

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Using the recursive formula to generate the terms, that is

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3 years ago
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