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# Tutor profile: Pavan S.

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Pavan S.
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## Questions

### Subject:Spanish

TutorMe
Question:

Elige la opción con el conector mas adecuado: "Hola Pedro, no puedo ir al trabajo hoy, _________, estoy enfermo." a)pero b)también c)dado que d)sino que e)además

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Pavan S.

Dado que se intenta explicar la razón de la falta al trabajo, se requiere un conector causal, o sea, la opción correcta es c) dado que.

### Subject:Pre-Calculus

TutorMe
Question:

Given the forces $$F_1=-300 [N] \hat{i}$$ and $$F_2=(200 \hat{r}+ \dfrac{\pi}{3} \hat{\theta}) [N]$$, write the magnitude and direction of the sum of the forces.

Inactive
Pavan S.

Let's start by writing out $$F_2$$ in cartesian form, using that $$F_2 \hat{i} = \cos(\dfrac{\pi}{3}) ||F_2||$$ and that $$F_2 \hat{j} = \sin(\dfrac{\pi}{3}) ||F_2||$$ we obtain that $$F_2 = (100 \hat{i} + 100 \sqrt{3} \hat{j})[N]$$, so summing $$F_1$$ and $$F_2$$ we obtain $$F_1+F_2= (-200 \hat{i} + 100 \sqrt{3} \hat{j})[N]$$, that means that the magnitude of the vector we are interested in is $$\sqrt{(-200)^2+( 100 \sqrt{3})^2} = 100 \sqrt{7}$$. Remembering that the $$\tan(angle)$$=$$\hat{j}$$ component of vector / $$\hat{i}$$ component of vector we obtain that the angle is $$\arctan{\dfrac{-\sqrt{3}}{2}}$$ rad $$\approx$$-0.714, but we know that our vector should actually be between $$\dfrac{\pi}2$$ rad and $$\pi$$ rad, meaning we should add $$\pi$$ to the obtained angle. Finally, we obtain that $$||F_1+F_2||$$=$$100 \sqrt{7}$$ and the angle it forms with the x-axis is 2.43 rad.

### Subject:Calculus

TutorMe
Question:

Given the function: $$f(x)=|x|\cos(x)$$ find the average value in the interval $$[-\pi,\pi]$$

Inactive
Pavan S.

Given that the average value of a function $$g(y)$$ in the interval $$[a,b]$$ is obtained by $$\dfrac1 {b-a} \int_a^b g(y)dy$$, the average value of $$f(x)$$ in $$[-\pi,\pi]$$ is $$\dfrac1 {2\pi} \int_{-\pi}^\pi|x| \cos(x) dx$$. So, let's solve $$\int_{-\pi}^\pi |x| \cos(x) dx$$. Remembering that $$\cos(x)$$ and $$|x|$$ are even functions we obtain that their product is even, and also noting that we are integrating in a symmetric interval around 0, we can reduce the integral to: $$2 \int_{0}^\pi |x| \cos(x) dx$$ and given that $$|x|=x$$ for $$x$$ in $$[0,\pi]$$, we are left with $$2 \int_{0}^\pi x \cos(x) dx$$. Integrating $$\int_{0}^\pi x \cos(x) dx$$, by parts we have: $$\int_{0}^\pi x \cos(x) dx = x \sin(x) |_0^\pi - \int_{0}^\pi \sin(x) dx$$$$=\cos(x)|_0^\pi=-2$$. That implies that, $$\int_{-\pi}^\pi |x| \cos(x) dx=2 \int_{0}^\pi |x| \cos(x) dx=-4$$, so the average value of $$f(x)$$ in $$[-\pi,\pi]$$ is $$\dfrac{-2}{\pi}$$

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