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4 years ago

Brainliest and 20 points!!!!

Mathematics

1 answer:

denis23 [38]4 years ago

6 0

Answer:

Step-by-step explanation:

The n th term ( explicit formula) for a geometric sequence is

f(n) = a $(r)^{n-1}$

where a is the first term and r the common ratio

Here a = 2 and r = - 8 ÷ 2 = - 4, thus

f(n) = 2 $(-4)^{n-1}$

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Derivative of g(x) = 3x^2(x^2-(2/x))

Elza [17]

When you are doing derivatives, you don't want to do it right away because there is still ( ) we need to remove that first get,

g(x)=3x^4-6x

Now we can do the derivatives!

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I hope this helps!

4 0

3 years ago

A circle has a circumference of 25.12 inches what is the diameter of the circle

BigorU [14]

Answer:

Step-by-step explanation:

Recall these two equations for circumference and diameter.

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3 years ago

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3 years ago

a pumpkin pie is made in a pie pan measuring 9 inches in diameter. it is cut into 8 equal slices. what is the length of the crus

lukranit [14]

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7 0

3 years ago

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