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2 years ago

Pls help (pic above) 2 answers

Mathematics

1 answer:

True [87]2 years ago

3 0

Answer:

Commutative property of multiplication/associative property of multiplication

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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

F(x)=x^2 -5x+1<br><br><br> f(x+h)-f(x)/h =

Sergeu [11.5K]

The answer is

F = x - 5 + 1/5

8 0

3 years ago

Which expressions is equivolent to 6[3+2x4] ?

nordsb [41]

Im not exactly sure what you are asking but 18 + 48 is equivalent and simplified!

4 0

3 years ago

Select the correct expressions. 1 Identify each expression that represents the slope of a tangent to the curve y= * +1 at any po

Olegator [25]

The expressions which represents the slope of a tangent to the curve $y=\frac{1}{x+1}$ at any point (x, y) are:

$f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}$

<h3>The slope of a tangent to the curve.</h3>

Mathematically, the slope of a tangent line to the curve is given by this equation:

$f'(x) =limh \rightarrow 0\frac{f(x+h)-f(x)}{h}$

Given the function:

$f(x)=y=\frac{1}{x+1}$

When (x + h), we have:

$f(x+h)=y=\frac{1}{x+h+1}$

Next, we would find the derivative of f(x):

$f'(x) =limh \rightarrow 0\frac{\frac{1}{x+h+1} -\frac{1}{x+1}}{h}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -(x+h+1)}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -x-h-1}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}\\\\f'(x) = \frac{-1}{(x+1)(x+1)} \\\\f'(x) = \frac{-1}{(x+1)^2}$

Read more on slope of a tangent here: brainly.com/question/26015157

#SPJ1

5 0

2 years ago

Fine the area of this triangle. I don’t know how to do it so it would help if you show me how too as well. Thanks

saw5 [17]

Answer:

gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter. ... On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth's mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects.

3 0

2 years ago