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3 years ago

5

A grid shows the positions of a subway stop and your house. The subway stop is located at (-1, 9) and your house is located at (

2, -4). What is the distance, to the nearest unit, between your house and the subway stop? A.) 5 B.)6 C.)12 D.)13

1 answer:

sammy [17]3 years ago

4 0

The distance between the house and the subway is given by:
d=√[(x-x1)²+(y-y1)²]
Thus given the coordinates:
(-1,9) and (2,-4)
thus
d=√[(2-(-1))²+(-4-9)²]
d=√[(3)²+(-13)²]
d=√[9+169]
d=√178
d=13.34 ~13 units
Answer: D) 13

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Given the polynomial function f(x) = log(1+x) …...(1) centered at x=0

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$f^{l} (0) = \frac{1}{1+0} (1)= 1$

using formula $\frac{d}{dx} x^{n-1} =nx^{n-1}$

Again Differentiating Equation(2) with respective to 'x' we get

$f^{l} (x) = \frac{-1}{(1+x)^2} (\frac{d}{dx}((1+x))$

$f^{ll} (x) = \frac{-1}{(1+x)^2} (1)$ ….(3)

At x=0

$f^{ll} (0) = \frac{-1}{(1+0)^2} (1)= -1$

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$f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (1)= \frac{(-1)^2 (2)!}{(1+x)^3}$ ….(4)

At x=0

$f^{lll} (0) = \frac{(-1)(-2)}{(1+0)^3} (1)$

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and so on...

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Step :-2

Taylors theorem expansion of f(x) is

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At x=a =0

$f(x) = f(0) + \frac{x}{1!} f^{l}(0) +\frac{(x)^2}{2!}f^{ll}(0)+\frac{(x)^3}{3!}f^{lll}(0)+\frac{(x)^4}{4!}f^{lV}(0)+\frac{(x)^5}{5!}f^{V}(0)+\frac{(x)^6}{6!}f^{VI}(0)+...….. \frac{(x-0)^n}{n!}f^{n}(0)$

Substitute all values , we get

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On simplification we get

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