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3 years ago

What is the value of x?

Mathematics

1 answer:

Eduardwww [97]3 years ago

3 0

Answer: 15

Step-by-step explanation:

x-6/(x-6)+3 = x/x+5

The x in the equation is 15

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I am Lyosha [343]

Answer:

I agree with the other answer

Step-by-step explanation:

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3 years ago

Read 2 more answers

A box contains 10 cards numbered 1 through 10. Which events are mutually exclusive? A. A card is drawn from the box and the numb

coldgirl [10]

The answer is A. 5 is not an even number, so both events can't take place together.

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3 years ago

Find the average value of f(x)=e2x over the interval [2, 4].

shusha [124]

<h2>Option B is the correct answer.</h2>

Step-by-step explanation:

We need to find average value of $e^{2x}$ in [2,4]

Area of $e^{2x}$ in [2,4] is given by

$\int_{2}^{4}e^{2x}dx=\frac{1}{2}\times \left [ e^{2x}\right ]^4_2\\\\\int_{2}^{4}e^{2x}dx=\frac{1}{2}\times(e^8-e^4)=1463.18$

Area of $e^{2x}$ in [2,4] = 1463.18

Difference = 4 - 2 = 2

Average value = Area of $e^{2x}$ in [2,4] ÷ Difference

Average value = 1463.18 ÷ 2

Average value = 731.59

Option B is the correct answer.

5 0

3 years ago

Helppppp I have until 11:30 and it’s 10:36

forsale [732]

Answer:

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Step-by-step explanation:

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3 years ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

1 year ago