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pickupchik [31]
3 years ago
14

What would you assume at the beginning of an indirect proof for the following?

Mathematics
1 answer:
stich3 [128]3 years ago
5 0
B. When the sky is cloudy it rains.

I hope this helps:)
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Please help ! I’m so confused
Naily [24]

Answer:

2/7 is the scale factor.

Step-by-step explanation:

All you have to do is get two sides that are equal to each other. Let's say Side SR and YX. 3 and 10.5. Do 3/10.5, which is 30/105, which simplified is 6/21, which again simplified is 2/7. So, the scale factor is 2/7.

7 0
3 years ago
How many applications of integration by parts are required to evaluate integral (x^7)(e^x) dx?
charle [14.2K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2860229

_______________


Make it more general, and see what happens when you try to reduce the exponent of  x  in the following integral:

\mathsf{\mathtt{I}_0=\displaystyle\int\! x^k\cdot e^x\,dx\qquad\qquad (k\ge 1,~~k\in\mathbb{N})}


Now, integrate it by parts:

\begin{array}{lcl}
\mathsf{u=x^k}&\quad\Rightarrow\quad&\mathsf{du=k\cdot x^{k-1}\,dx}\\\\
\mathsf{dv=e^x\,dx}&\quad\Leftarrow\quad&\mathsf{v=e^x}
\end{array}


\mathsf{\displaystyle\int\! u\,dv=u\cdot v-\int\! v\,du}\\\\\\
\mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-\int\! e^x\cdot k\cdot x^{k-1}\,dx}\\\\\\
\mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-k\int\! x^{k-1}\cdot e^x\,dx}\\\\\\
\mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-k\cdot \mathtt{I}_1}

where \mathsf{\mathtt{I}_1=\displaystyle\int\! x^{k-1}\cdot e^x\,dx.}


So after one iteration, the exponent of  x  was decreased by one unit.

The question is:  after how many iterations will the exponent of  x  equals zero?

     After exactly  k  iterations, of course.

Therefore, for  k = 7, you have to apply integration by parts  7  times, to get rid of that polynomial factor. Then, there will be one last integral left to evaluate:

\mathsf{\displaystyle\int\! e^x\,dx}

But this one doesn't need to be evaluated by parts. You can directly write the result:

\mathsf{\displaystyle\int\! e^x\,dx=e^x+C}


Shortly, for the integral

\mathsf{\mathtt{I}_0=\displaystyle\int\! x^7\cdot e^x\,dx}

you have to apply integration by parts  7  times (not  8  times).


I hope this helps. =)


Tags:  <em>indefinite integral integration by parts reduction formula product polynomial exponential differential integral calculus</em>

8 0
4 years ago
your teacher ask you to find a recipe that includes two ingredients with a ratio of 1/2 cup and 1/8 cups
klemol [59]
The answer is 0.625 I am not so sure
8 0
3 years ago
Se compraron dos sacos
RoseWind [281]

Answer:

10 es la respuesta amigo buenas tardes

4 0
3 years ago
Solve for x
Ivan

9514 1404 393

Answer:

  x = 4

Step-by-step explanation:

The measures of all angles in this geometry are 90 degrees.

  21x +6 = 90

  21x = 84 . . . . . subtract 6

  x = 4 . . . . . . . divide by 21

5 0
3 years ago
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