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2 years ago

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The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th

e midpoint of the hypotenuse and passing through the vertex opposite the hypotenuse is:

1 answer:

ElenaW [278]2 years ago

4 0

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

$|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}$

By doing so, we find:

$|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}$

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

$|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1$

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2 years ago

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Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is $f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, $x=0$ is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

$f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

$f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-\frac{5}{x^2}+\frac{8}{x^3}$

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$x=\frac{8}{5}$

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

$f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}$

$f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3$

Therefore, the function increases on the interval $(0,\frac{8}{5})$ and decreases on the interval $(-\infty,0),(\frac{8}{5},\infty)$ .

C) Since we determined that the slope is 0 when $x=\frac{8}{5}$ from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, $f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}$ , meaning there's an extreme at the point $(\frac{8}{5},\frac{41}{16})$ , but is it a maximum or minimum? To answer that, we will plug in $x=\frac{8}{5}$ into the second derivative which is $f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$ . If $f''(x)>0$ , then it's a minimum. If $f''(x)$ , then it's a maximum. If $f''(x)=0$ , the test fails. So, $f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}$ , which means $(\frac{8}{5},\frac{41}{16})$ is a local maximum.

D) Now set the second derivative equal to 0 and solve:

$f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$

$0=\frac{10}{x^3}-\frac{24}{x^4}$

$0=10x-24$

$-10x=-24$

$x=\frac{24}{10}$

$x=\frac{12}{5}$

We then test where $f''(x)$ is negative or positive by plugging in test values. I will use -1 and 3 to test this:

$f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34$ , so the function is concave down on the interval $(-\infty,0)\cup(0,\frac{12}{5})$

$f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0$ , so the function is concave up on the interval $(\frac{12}{5},\infty)$

The inflection point is where concavity changes, which can be determined by plugging in $x=\frac{12}{5}$ into the original function, which would be $f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}$ , or $(\frac{12}{5},\frac{43}{18})$ .

E) See attached graph

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2 years ago

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ryzh [129]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

What is 1/3 add 1/4

Harrizon [31]

$\frac{1}{3} + \frac{1}{4} = \frac{1\times 4}{3\times 4} + \frac{1\times 3}{4\times 3} = \frac{4}{12} + \frac{3}{12} = \frac{4+3}{12} = \boxed{\frac{7}{12}}$

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2 years ago

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lorasvet [3.4K]

<h2>Answer:</h2>

B. 12/5

<h2>Step-by-step explanation:</h2>

In a division, we should swap the second fraction and change the equation to times. After that, we can multiply each fraction easily.

4/5 : 1/3

= 4/5 x 3/1

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In conclusion, the simplified quotient of 4/5 : 1/3 is 12/5

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1 year ago