Alan

$$\frac{12-x}{1+x}=0.3$$

Multiply both sides by (1 + x)

$$12-x = 0.3(1+x)$$

Multpliy each term in brackets on the right-hand side by the 0.3

$$12-x=0.3+0.3x$$

Subtract 0.3 from both sides and add x to both sides

$$12-0.3=x+0.3x$$

Collect like terms

$$11.7=1.3x$$

Divide both sides by 1.3

$$x=\frac{11.7}{1.3}$$

or $$x=9$$

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How about the following:

Perhaps I should clarify a step:  Note that n*rⁿ = r*n*r^n-1= r*d(rⁿ)/dr  (or perhaps this just muddies the waters - in which case ignore it!)

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Anonymous is correct, but just to flesh it out a little:

Momentum is conserved (assuming no friction etc.) so:

Initial momentum = 0

Final momentum of wagon + child after bricks are thrown = (62 + 150 - 6)*velocity

Final momentum of bricks = -6*2  (assuming velocity is negative to the west and positive to the east)

(62+150-6)*velocity - 6*2 = 0

velocity = 6*2/(62+150-6) m/s

$${\mathtt{velocity}} = {\frac{{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{2}}}{\left({\mathtt{62}}{\mathtt{\,\small\textbf+\,}}{\mathtt{150}}{\mathtt{\,-\,}}{\mathtt{6}}\right)}} \Rightarrow {\mathtt{velocity}} = {\mathtt{0.058\: \!252\: \!427\: \!184\: \!466}}$$ 

or velocity ≈ 0.058 m/s

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 This could also be solved by putting p(t) = 50 and solving for t.  There will be two values for t (one for the ball on its way up and one for it on its way down).  The difference between them gives the time you are looking for.

The equation can be written as : 50 = -16t² + 50t + 20 or:

16t² - 50t + 30 = 0 

$${\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{50}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,\small\textbf+\,}}{\mathtt{30}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{145}}}}{\mathtt{\,-\,}}{\mathtt{25}}\right)}{{\mathtt{16}}}}\\
{\mathtt{t}} = {\frac{\left({\sqrt{{\mathtt{145}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{25}}\right)}{{\mathtt{16}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{0.809\: \!900\: \!338\: \!825\: \!481\: \!5}}\\
{\mathtt{t}} = {\mathtt{2.315\: \!099\: \!661\: \!174\: \!518\: \!5}}\\
\end{array} \right\}$$

Giving the same result as found graphically by Chris

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pi is defined as the ratio of the circumference (c) of a circle to its diameter (d).   

pi = c/d

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It's possible this means there are "x" and "y" values given for years (?) 1995 and 1996, such that the following holds:

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If tan(α/2) =  3 then α/2 = tan^-1(3) and α = 2tan^-1(3).  Most calculators will give the 1st quadrant value for tan^-1(3), so just finding sin(2tan^-1(3)) will give the correct result:

$${\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{3}}\right)} = {\mathtt{143.130\: \!102\: \!354\: \!156}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{3}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.6}}$$

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You need to put a 3 in front of the NaOH and the NaCl terms to make this balance.

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Yes, that works too.  I always find it easier to remember just the sine and cosine relationships though, and work out the other trig identities from those if I need them.

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