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3 years ago

Find the distance between (-2,4) and (3,-2):

Mathematics

2 answers:

Sedbober [7]3 years ago

8 0

Answer:

7.8 units

Step-by-step explanation:

sqrt[ (-2-4)² + (3--2)² ]

sqrt[ 36 + 25 ]

sqrt(61)

7.8102496759

Serjik [45]3 years ago

7 0

Option C: The distance between the two points is 7.8 units

Explanation:

Given that the two points are (-2,4) and (3,-2)

Distance between the two points:

The distance between the two points can be determined using the formula,

$c=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

Substituting the coordinates (-2,4) and (3,-2), we get,

$c=\sqrt{(3+2)^2+(-2-4)^2}$

Simplifying the terms, we get,

$c=\sqrt{(5)^2+(-6)^2}$

Squaring the terms, we have,

$c=\sqrt{25+36}$

Adding, we get,

$c=\sqrt{61}$

$c=7.8$

Thus, the distance between the two points is 7.8 units

Therefore, Option C is the correct answer.

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3 years ago

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FromTheMoon [43]

Answer:

The Taylor series is $\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}$ .

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Step-by-step explanation:

The Taylor expansion.

Recall that as we want the Taylor series centered at $a=3$ its expression is given in powers of $(x-3)$ . With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of $\ln(1+x)$ .

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$\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).$

Now, in order to make a more compact notation write $\frac{x-3}{3}=y$ . Thus, the above expression becomes

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$\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.$

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$\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.$

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We find the radius of convergence with the Cauchy-Hadamard formula:

$R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},$

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$\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}$

Applying the properties of roots

$\sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.$

Hence,

$R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}$

Recall that

$\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.$

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