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# Tutor profile: Rohan B.

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Rohan B.
B.E. chemical & M.Sc Mathematics student at BITS PILANI
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## Questions

### Subject:Trigonometry

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Question:

Prove the identity $$4 sin(x) cos(x) = sin(4x) / cos(2x)$$.

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Rohan B.

L.H.S -> $$2 * 2sin(x) cos(x)$$ = $$2 * sin(2x)$$ .{ Identity :- $$2sin(a) cos(a)$$ = $$2 * sin(2a)$$ } Multiply and divide by cos(2x) $$\frac{(2 * sin(2x) * cos(2x)}{cos(2x)}$$ $$\frac{sin(4x)}{cos(2x)}$$ Hence Proved.

### Subject:Basic Math

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Question:

For all x in the domain of the function $$\frac{x + 1}{x^{3}-x}$$ , simply this function.

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Rohan B.

Take common out from denominator, it will look like $${x(x^{2}-1)}.$$ we know $$x^{2}-1 = (x + 1)(x - 1)$$ $$\frac{x + 1}{x (x + 1)(x - 1)}$$ = $$\frac{1}{x (x - 1)}$$ $$\frac{1}{x (x - 1)}$$ = $$\frac{1}{(x^{2}-x)}$$ $$\frac{x + 1}{x (x + 1)(x - 1)}$$ = $$\frac{1} {(x^2-x)}$$ Answer.

### Subject:Chemical Engineering

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Question:

The heat capacity of sulphuric acid has the units J/(g mol)($$^{\circ}$$ C) and is given by the relation heat capacity = 139.1 + 1.56 * $$10^(-1)$$T where T is expressed in $$^{\circ}$$C. Modify the formula so that the resulting expressions has the associated units pf Btu/(lb mol)($$^{\circ}$$R ) and T as in $$^{\circ}$$R.

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Rohan B.

The symbol $$^{\circ}$$C in the denominator of the heat capacity stands for the unit temperature difference, $$\Delta$$$$^{\circ}$$ C, not the temperature, where the units of T in the equation are in $$^{\circ}$$C. First, we have to substitute the proper equation in the formula to convert T in $$^{\circ}$$C to T in $$^{\circ}$$R, and then by multiplication by conversion factors convert the units on the righthand side of the equation to Btu/(lb mol)($$^{\circ}$$R ) as requested. T($$^{\circ}$$C) = ((T$$^{\circ}$$R) - 460 - 32)/1.8). heat capacity = (139.1 + 156 * $$10^(-1)$$T($$^{\circ}$$C) * by conversion factor. Units need to convert units: 1 Btu = 1055J 1 lb = 454g Note the suppression of the $$\Delta$$ symbol in the conversion between $$\Delta$$$$^{\circ}$$C and $$\Delta$$$$^{\circ}$$R.

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