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Citrus2011 [14]
3 years ago
15

How to flip a desk 5 feet long and then flip another desk and it was 10 feet long what would be the answer

Mathematics
1 answer:
-BARSIC- [3]3 years ago
6 0
I think this may be multiplication so 5 times 10 equls50
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She subtracted 7x instead of dividing it. Personally, i wouldn't have done that. I would subtract the 14 first, then divide.

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A mini-cone has a diameter of 3 inches and holds 10 in3 of ice cream. A mega-cone holds 49.13 in3 of ice cream. If the cones hav
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The answer would be D
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Select the correct answer.
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Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
aliya0001 [1]

Answer:

f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...

We first compute the n-th derivative of f(x)=\ln(1+2x), note that

f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\

Now, if we compute the n-th derivative at 0 we get

f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!

and so the Maclaurin series for f(x)=ln(1+2x) is given by

f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2  \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}

3 0
3 years ago
"Suppose that you are a city planner who obtains a sample of 20 randomly selected members of a mid-sized town in order to determ
frosja888 [35]

Answer: 2.093

Step-by-step explanation:

As per give , we have

Sample size : n= 20

Degree of freedom : df= n-1=19

Significance level : \alpha: 1-0.95=0.05

Since , the sample size is small (n<30) so we use t-test.

For confidence interval , we find two-tailed test value.

Using students's t-critical value table,

Critical t-value : t_{\alpha/2, df}=t_{0.025,19}=2.093

Thus, the critical value for the 95% confidence interval = 2.093

4 0
3 years ago
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