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Valentin [98]
4 years ago
8

Explain why rolle's theorem does not apply to the function even though there exist a and b such that f(a) = f(b). (select all th

at apply.) f(x) = cot x 2 , [π, 9π] there are points on the interval [a, b] where f is not continuous. none of these. there are points on the interval (a,
b.where f is not differentiable. f(a) does not equal f(b) for all possible values of a and b in the interval [π, 9π]. f '(a) does not equal f '(b) for any values in the interval [π, 9π].
Mathematics
1 answer:
frozen [14]4 years ago
5 0
The reason Rolle's theorem does not apply to the function cot x^2 on the given interval is that on the interval there are points where f is not continuous, and as such not differentiable either.
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3n + 8 = 53<br> I don’t understand how to find the n
olganol [36]

Answer:

n=15

Step-by-step explanation:

subtract 8 from both sides to get 3n=45

divide both sides by 3 to solve for n to get n=15

3 0
3 years ago
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Find the first , fourth , and eighth terms of the sequence A(n) = -3 •2 ^x - 1
Usimov [2.4K]

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5 0
4 years ago
A store paid $12 for a sweatshirt and sold the sweatshirt of $19.20. What was the percent markup (increase) of the sweatshirt? *
Nookie1986 [14]

The percent markup of the sweatshirt is 60%

Here, we want to calculate the percenatge of profit made by the store

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From the question, we can deduce the following;

cost price is the price the store bought the sweatshirt = $12

sales price is the amount in which they sold the sweatshirt = $19.20

Thus, we proceed to substitute these values into the markup equation above;

\begin{gathered} \text{Perecentage markup = }\frac{(19.20-12)}{12}\text{ }\times\text{ 100\%} \\  \\ \text{Percentage markup = }\frac{7.20}{12}\text{ }\times\text{ 100\%} \\  \\ =\text{ 60\%} \end{gathered}

8 0
1 year ago
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Marta_Voda [28]

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c = 2\pi \: r

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r =  \frac{d}{2}

So,

c = \pi \: d \\ d =  \frac{c}{\pi}  \\ d =  \frac{604}{\pi} \:  cm

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3 years ago
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