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3 years ago

What is the equation of the line that passes through the points (-9,-6) and (-9, -8)?

Mathematics

1 answer:

andrey2020 [161]3 years ago

7 0

Answer:

x=-9

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-8-(-6))/(-9-(-9))

m=(-8+6)/(-9+9)

m=-2/0

undefined

x=-9

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$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

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

What is the equation of the line that passes through the points (-9,-6) and (-9, -8)?​

What is the equation of the line that passes through the points (-9,-6) and (-9, -8)?