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3 years ago

Line ab contains the points a(-2,3) and b(4,5). Line ab has a slope that is

Mathematics

1 answer:

gayaneshka [121]3 years ago

5 0

Given two points $A = (A_x, A_y),\ B = (B_x,B_y)$ , the slope of the line passing through the two points is

$m = \dfrac{A_y-B_y}{A_x-B_x}$

So, in your case, you have

$m = \dfrac{3-5}{-2-4} = \dfrac{-2}{-6} = \dfrac{1}{3}$

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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]

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$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

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

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

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3 years ago

If you run a marathon of 26 1/5 miles in a time of 5 4/5 how far do you run in 1 hour<br><br>pls

alisha [4.7K]

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So, to find our miles per hour, we have to divide the miles by the hours to get: $\frac{\frac{131}{5}\text{miles}}{\frac{29}{5}\text{hours}}=\frac{131\text{miles}}{29\text{hours}} \\ \text{ Now we have to divide the top and the bottom by 29 to get our miles per hour so: }\\ \frac{\frac{131}{29}}{\text{hour}} \approx \frac{4.52 \text{miles}}{\text{hour}}$

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