Alan

The resistive force is given by μ_s*W, where W is the weight.  The resistance is overcome a force of 659.2N is applied, so 

μ_s*87.9*9.81 = 659.2

μ_s = 659.2/(87.9*9.81)

$${\frac{{\mathtt{659.2}}}{\left({\mathtt{87.9}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}} = {\mathtt{0.764\: \!468\: \!009\: \!356\: \!383\: \!3}}$$

or μ_s ≈ 0.764

Note that there are two solutions to x² = 13.9, namely x = +√13.9 and x = -√13.9.

To solve ln(2x-1)² = 4 first use one of the properties of logarithms (see the Formulary) to get

2ln(2x-1) = 4 so that

ln(2x-1) = 2

Now exponentiate:

e^ln(2x-1) = e², so

2x-1 = e², because e^ln(2x-1) is just 2x-1

2x = 1+e²

x = (1+e²)/2

Melody, the second one is essentially the same as the first and the fourth is the same as the third.  The minus sign would be part of the constant that you need to turn the proportionality into an equality.

You have used 150 instead of 1500 Melody.

If the number in question is 39^{(46252+73)*4733} then, since the unit value of powers of 39 alternate between 1, if it's an even power, and 9, if it's an odd power we have: (46252+73)*4733 =  219256225 is odd, so the value in the units position of 39^{(46252+73)*4733} is 9.

If the number in question is 39⁴⁶²⁵²+73⁴⁷³³ then the unit digit of the first term will be 1 (because of the even power) and we need to find the unit digit of the second term.

The unit digit of powers of 73 run through the sequence 1, 3, 9, 7 depending on the remainder of the power after dividing by 4 (0,1,2, 3 respectively).  Since the remainder of 4733 when divided by 4 is 1, the unit digit place of 73⁴⁷³³ is 3.  

When the two terms are added together therefore, the value in the units position of 39⁴⁶²⁵²+73⁴⁷³³  is 1+3 = 4.

Just type this, exactly as written, into the calculator on the Home page here.

I take it C is the angle, and you are talking about a triangle in which case you can use the Cosine rule in the form

$$c^2=a^2+b^2-2abcos(C)$$  to find c.

$${\mathtt{c}} = {\sqrt{{{\mathtt{192}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{291}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{192}}{\mathtt{\,\times\,}}{\mathtt{291}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{41.3}}^\circ\right)}}} \Rightarrow {\mathtt{c}} = {\mathtt{193.896\: \!211\: \!065\: \!943\: \!792\: \!5}}$$

or c ≈193.9

I'll leave you to use the sine rule $$\frac{sin(A)}{a}=\frac{sin(C)}{c}$$ to find A, and similarly for B (or just use the fact that the three angles must add to 90° to get the last one). 

Good try Anonymous, but, unfortunately, in your answer you forgot the warning about being careful with the units!  P is in millions, so we have:

1103.4*e^0.0149t = 1500

Divide both sides by 1103.4

e^0.0149t = 1500/1103.4

Take log to base e of both sides

0.0149t = ln(1500/1103.4)

Divide both sides by 0.0149

t = ln(1500/1103.4)/0.0149

$${\mathtt{t}} = {\frac{{ln}{\left({\frac{{\mathtt{1\,500}}}{{\mathtt{1\,103.4}}}}\right)}}{{\mathtt{0.014\: \!9}}}} \Rightarrow {\mathtt{t}} = {\mathtt{20.608\: \!643\: \!372\: \!615\: \!506}}$$  years

This is 21 years to the nearest year, so it will be 2005+21 = 2026 when India's population reaches 1500million.

$$We have $$\sqrt{(1-cos(x))(1+cos(x))} = \sqrt{1-cos^2(x)}=\sqrt{sin^2(x)}=sin(x)$$