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Viefleur [7K]
3 years ago
15

How do I do number 16?

Mathematics
1 answer:
PolarNik [594]3 years ago
8 0

Answer:  x = 121 degrees

Step-by-step explanation:

(n-2) x 180

(6-2) x 180

4 x 180=720

599 +x = 720

599 - 599 = x  +x = 720 - 599 = 121

and then add up all the degrees

88 + 105 + 142 + 140 + 124 + 121 = 720

so there for you answer for x is 121 degrees

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Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]
Zina [86]

Answer: Absolute minimum: f(-1) = -2\sqrt{6}

              Absolute maximum: f(\sqrt{12.5}) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = t\sqrt{25-t^{2}}

f'(t) = 1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)

f'(t) = \sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }

f'(t) = \frac{25-2t^{2}}{\sqrt{25-t^{2}} } = 0

For this function to be zero, only denominator must be zero:

25-2t^{2} = 0

t = ±\sqrt{2.5}

\sqrt{25-t^{2}} ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = -\sqrt{2.5} and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = t\sqrt{25-t^{2}}

f(-1) = -1\sqrt{25-(-1)^{2}}

f(-1) = -\sqrt{24}

f(-1) = -2\sqrt{6}

f(\sqrt{12.5}) = \sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}

f(\sqrt{12.5}) = 12.5

f(5) = 5\sqrt{25-5^{2}}

f(5) = 0

Therefore, absolute maximum is f(\sqrt{12.5}) = 12.5 and absolute minimum is

f(-1) = -2\sqrt{6}.

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3 years ago
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Abner receives a toy catalog in the mail.The catalog has 25 pages and shows 11 toys on each page.Abner wants to know what is the
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