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2 years ago

PLEASE HELP, What is the annual interest rate she is earning on her deposit?

Mathematics

1 answer:

uysha [10]2 years ago

5 0

i have no ideaaaaa HAHAHAHA

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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

2 years ago

Anyone know how to find the constant of proportionality???!!!

dmitriy555 [2]

Answer:

You find the constant of proportionality by how much u need to add or subtract the same number to get the same answer and in a graph you need to see if the line goes through the origin (0,0).

Step-by-step explanation:

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2 years ago

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Is x=15 a function?

Mama L [17]

Answer:

yes i do believe so

Step-by-step explanation:

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3 years ago

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Four times the lesser of two consecutive even integers is 12 less than twice the greater number. Find the integers.

valkas [14]

The numbers are: 2n and (2n+2)
we can suggest this equation:
4(2n)=2(2n+2)-12

We compute this equation:
8n=4n+4-12
8n-4n=-8
4n=-8
n=-8/4
n=-2

we compute the numbers:
2n=2(-2)=-4
2n+2=-4+2=-2

answer: the numbers are -4 and -2.

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3 years ago

Which expression is equivalent to StartFraction 4 a + 4 Over 2 a EndFraction times StartFraction a squared Over a + 1 EndFractio

Cloud [144]

Answer:

Its 2a or C on EDGE 2020

Step-by-step explanation:

Took the test

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3 years ago

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