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2 years ago

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Sam buys a rare stamp for his stamp collection at an auction. he buys the stamp for 920£ the stamp increased in value by 7% in e

ach of the first 2 years decreased in value by 4% in the third year

1 answer:

VladimirAG [237]2 years ago

5 0

Answer:

anong grade po yan baka alam kopo

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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

What is the equation for the hyperbola shown?

amid [387]

Answer:

D. y²/5² - x²/8² = 1

Step-by-step explanation:

A and B are both incorrectly oriented, and D is the only hyperbola that contains the points (0,5) and (0,-5).

Verification (0,5) and (0,-5) are in the hyperbola:

First replace x and y with corresponding x and y values (We will start with x=0 and y=5)

$\frac{5^{2}}{5^{2}}-\frac{0^{2}}{8^{2}}=1$

Then simplify.

$\frac{25}{25}-\frac{0}{16}=1$

$1-0=1$

$1=1$

If the result is an equation (where both sides are equal to each other) then the original x and y values inputted are valid. The same is true with x and y inputs x=0 and y=-5, or any other point along the hyperbola.

6 0

3 years ago

Read 2 more answers

help it is due today and not links or files of downloads of anything like that just type in your answer plz and thank u :)

pantera1 [17]

The answer should be x=4

4 0

3 years ago

Read 2 more answers

Yes please help me on this

otez555 [7]

15 abbabababababababababbababa

3 0

2 years ago

An accounting firm purchased a new digital camera that cost $2,500. The life of the camera is estimated to be 5 years. The total

lozanna [386]

2500-700 = 1800
1800/5 = 360

The annual depreciation of the camera is $360.

5 0

3 years ago