If x^2+2=3^2/3+3^-2/3 then prove that 3x^3+9x-8=0

Ill give brainliest - please help ASAP WITH THE WORK THO PLEASE?

ohaa [14]

Answer:

5.

we have.

11x=½(16x+16+50)

22x=16x+66

22x-16x=66

6x=66

x=66/6

x=11

6.

118°=½(6x-11+181)

236=6x+170

6x=236-170

x=66/6

x=11°

7.

34°=½(arcAK-arc LN)

34×2=(110-arc LN)

arc LN =110-68

arc LN=42°

8.

<L=½(arc EN-arc KM)

<L=½(139°-73°)

<L=33

4 0

3 years ago

Find the indefinite integral using the substitution provided.

Nady [450]

Answer: $7\text{Ln}\left(e^{2x}+10\right)+C$

This is the same as writing 7*Ln( e^(2x) + 10) + C

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Explanation:

Start with the equation $u = e^{2x}+10$

Apply the derivative and multiply both sides by 7 like so

$u = e^{2x}+10\\\\\frac{du}{dx} = 2e^{2x}\\\\7\frac{du}{dx} = 7*2e^{2x}\\\\7\frac{du}{dx} = 14e^{2x}\\\\7du = 14e^{2x}dx\\\\$

The "multiply both sides by 7" operation was done to turn the 2e^(2x) into 14e^(2x)

This way we can do the following substitutions:

$\displaystyle \int \frac{14e^{2x}}{e^{2x}+10}dx\\\\\\\displaystyle \int \frac{1}{e^{2x}+10}14e^{2x}dx\\\\\\\displaystyle \int \frac{1}{u}7du\\\\\\\displaystyle 7\int \frac{1}{u}du\\\\\\$

Integrating leads to

$\displaystyle 7\int \frac{1}{u}du\\\\\\7\text{Ln}\left(u\right)+C\\\\\\7\text{Ln}\left(e^{2x}+10\right)+C\\\\\\$

Be sure to replace 'u' with e^(2x)+10 since it's likely your teacher wants a function in terms of x. Also, do not forget to have the plus C at the end. This is a common mistake many students forget to do.

To verify the answer, you can apply the derivative to it and you should get back to the original integrand of $\frac{14e^{2x}}{e^{2x}+10}$