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Marrrta [24]
3 years ago
11

Charles can do a job in 1 hour, bill can do the same job in 2 hours, and bob can do the job in 3 hours. How long does it take th

em to do the job working together?
Mathematics
2 answers:
KiRa [710]3 years ago
5 0
Charles can do the whole job in an hour, Bill can do 1/2 of the job in an hour, and Bob does 1/3 of the job in an hour.  Therefore, they do a total of 1+1/2+1/3=11/6 of the job in an hour, meaning it takes them 6/11 of an hour.  This is 6/11*60=360/11 minutes, or approximately 33 minutes.
Papessa [141]3 years ago
3 0

Charles does a complete job in 1 hour.
Bill does 1/2 of a job in an hour.
Bob does 1/3 of a job in an hour.

If they're all working on the same job, or copies of the same one,
then working together, they do (1 + 1/2 + 1/3)  job per hour.

   (1) + (1/2) + (1/3)  =  (6/6) + (3/6) + (2/6) = 11/6  job/hour

Working at the rate of  11/6 job per hour, the team needs <em>6/11 hou</em><em>r</em> per job.

6/11 hour is 32 minutes 43-21/33 seconds .

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Find \(\int \dfrac{x}{\sqrt{1-x^4}}\) Please, help
ki77a [65]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2867785

_______________


Evaluate the indefinite integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}


Make a trigonometric substitution:

\begin{array}{lcl}&#10;\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\&#10;&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}


so the integral (i) becomes

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\&#10;\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\&#10;\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\&#10;\mathsf{=\displaystyle\frac{1}{2}\,t+C}


Now, substitute back for t = arcsin(x²), and you finally get the result:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}          ✔

________


You could also make

x² = cos t

and you would get this expression for the integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}          ✔


which is fine, because those two functions have the same derivative, as the difference between them is a constant:

\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\&#10;=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\&#10;=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\&#10;=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}

\mathsf{=\dfrac{\pi}{4}}         ✔


and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.


I hope this helps. =)

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3 years ago
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Answer:

y = - 2x - 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

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m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

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bagirrra123 [75]
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3 years ago
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