All categories

2 years ago

My question is down there

Mathematics

1 answer:

slavikrds [6]2 years ago

7 0

I don’t really see the question ?

You might be interested in

den301095 [7]

Answer:

Step-by-step explanation:

m∠JXM + m∠NXP = 180° - m∠MXN = 90°

7 0

3 years ago

How do i Write in Digits the fifty five hundredths

grandymaker [24]

fifty five hundredths is equal to 0.55

6 0

2 years ago

Read 2 more answers

C=100+25h solve for h

aivan3 [116]

Answer:

h=(c/25)-4

Step-by-step explanation:

c-100 = 25h. "/25"

(c/25)-4 = h

7 0

3 years ago

Read 2 more answers

How do you make the colored lines short enough to make a "K"?

earnstyle [38]

Hmmm wait let me try to do it

8 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

=======================================================

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 $\frac{14e^{2x}}{e^{2x}+10}$

4 0

2 years ago