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Zolol [24]
2 years ago
12

5(6p-3)-2 = 2(15p - 11)+5 one solution, no solution, infinite solution ​

Mathematics
1 answer:
pantera1 [17]2 years ago
3 0

Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

-2x=3+7

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Consider the following function. f(x) = 16 − x2/3 Find f(−64) and f(64). f(−64) = f(64) = Find all values c in (−64, 64) such th
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Answer:

This does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64).

Step-by-step explanation:

The given function is

f(x)=16-\frac{x^2}{3}

To find f(-64), we substitute x=-64 into the function.

f(-64)=16-\frac{(-64)^2}{3}

f(-64)=16-\frac{4096}{3}

f(-64)=-\frac{4048}{3}

To find f(64), we substitute x=64 into the function.

f(64)=16-\frac{(64)^2}{3}

f(64)=16-\frac{4096}{3}

f(64)=-\frac{4048}{3}

To find f'(c), we must first find f'(x).

f'(x)=-\frac{2x}{3}

This implies that;

f'(c)=-\frac{2c}{3}

f'(c)=0

\Rightarrow -\frac{2c}{3}=0

\Rightarrow -\frac{2c}{3}\times -\frac{3}{2}=0\times -\frac{3}{2}

c=0

For this function to satisfy the Rolle's Theorem;

It must be continuous on [-64,64].

It must be differentiable  on (-64,64).

and

f(-64)=f(64).

All the hypotheses are met, hence this does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64) is the correct choice.

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Hope this helped!!!!!
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