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3 years ago

What is meant by the term financial planning?

Mathematics

1 answer:

lidiya [134]3 years ago

5 0

A. Long term goals are reached over an extended period of time, so your current income does not affect them.

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The volume of a cylindrical container is 226 cubic centimeters. The radius of the container is 6 centimeters. Find the height of

coldgirl [10]

Here you go i hope i could help you

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3 years ago

-2x-9y=-25 4x+9y=23 Help me

Tatiana [17]

X=25/11 and y=17/11 if that’s what your asking for

6 0

2 years ago

What can you infer about the population from each data set represented below? Please help

Softa [21]

Answer:

Well around 5-8 miles were jogged daily.

Step-by-step explanation:

Thats all i know

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3 years ago

A diver begins at sea level and descends vertically at a rate of 2 1/2 ft. per second. How long does the diver take to reach -15

galina1969 [7]

2 1/2 feet can be written as 2.5 feet.

Divide the total distance by the distance per second:

15.6 feet / 2.5 feet per second = 6.24 seconds.

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