                                                                                                                                
 \begin{problembox}                                                                                                             
     \subsection{Formulae}                                                                                                      
     \begin{questions}                                                                                                          
         \Question[3] Temperatures can be measured in either degrees Fahrenheit or degrees Celsius. To convert from one scale t 
 the other, the following formula is used: \(F=\frac{9}{5} C+32\).                                                              
         \begin{parts}                                                                                                          
             \part Rearrange the formula to make \(C\) the subject.                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = \frac{5}{9}(F - 32)\)                                                                                    
             \end{solutionordottedlines}                                                                                        
             \part On a particular day in Melbourne, the temperature was \(28^{\circ} \mathrm{C}\). What is this temperature    
 measured in Fahrenheit?                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(F = \frac{9}{5} \times 28 + 32 = 82.4^{\circ} \mathrm{F}\)                                                   
             \end{solutionordottedlines}                                                                                        
             \part In Boston, USA, the minimum overnight temperature was \(4^{\circ} \mathrm{F}\). What is this temperature     
 measured in Celsius?                                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = \frac{5}{9}(4 - 32) = -15.56^{\circ} \mathrm{C}\)                                                        
             \end{solutionordottedlines}                                                                                        
             \part What number represents the same temperature in \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\) ?   
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = F\) \\                                                                                                   
                 \(\frac{5}{9}(F - 32) = F\) \\                                                                                 
                 \(5F - 160 = 9F\) \\                                                                                           
                 \(4F = 160\) \\                                                                                                
                 \(F = 40^{\circ}\)                                                                                             
             \end{solutionordottedlines}                                                                                        
             \part An approximate conversion formula, used frequently when converting oven temperatures, is \(F=2 C+30\). Use   
 this to convert these temperatures:                                                                                            
             \begin{subparts}                                                                                                   
                 \subpart an oven temperature of \(180^{\circ} \mathrm{C}\)                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(F = 2 \times 180 + 30 = 390^{\circ} \mathrm{F}\)                                                             
             \end{solutionordottedlines}                                                                                        
                 \subpart an oven temperature of \(530^{\circ} \mathrm{F}\)                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(C = \frac{530 - 30}{2} = 250^{\circ} \mathrm{C}\)                                                            
             \end{solutionordottedlines}                                                                                        
             \end{subparts}                                                                                                     
         \end{parts}                                                                                                            
         \Question[2] Gareth the gardener has a large rectangular vegetable patch and he wishes to put in a path around it usin 
 concrete pavers that measure \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\). The path is to be 1 paver wide. Let \(n\) be the numb 
 of pavers required. If the vegetable patch measures \(x\) metres by \(y\) metres, find a formula for \(n\) in terms of \(x\) a 
 \(y\).                                                                                                                         
         \begin{solutionordottedlines}[2cm]                                                                                     
             \(n = 2(x + y) \times 2\) because each side needs \(x \times 2\) and \(y \times 2\) pavers, and there are two      
 lengths and two widths.                                                                                                        
         \end{solutionordottedlines}                                                                                            
         \Question[3] If \(s=\frac{n}{2}(2 a+(n-1) d)\) :                                                                       
         \begin{parts}                                                                                                          
             \part find the value of \(s\) when \(n=10, a=6\) and \(d=3\)                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(s = \frac{10}{2}(2 \times 6 + (10 - 1) \times 3) = 5(12 + 27) = 5 \times 39 = 195\)                          
             \end{solutionordottedlines}                                                                                        
             \part find the value of \(a\) when \(s=350, n=20\) and \(d=4\)                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(350 = \frac{20}{2}(2a + (20 - 1) \times 4)\) \\                                                              
                 \(350 = 10(2a + 76)\) \\                                                                                       
                 \(35 = 2a + 76\) \\                                                                                            
                 \(a = \frac{35 - 76}{2} = -20.5\)                                                                              
             \end{solutionordottedlines}                                                                                        
             \part find the value of \(d\) when \(s=460, n=10\) and \(a=10\)                                                    
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(460 = \frac{10}{2}(2 \times 10 + (10 - 1)d)\) \\                                                             
                 \(460 = 5(20 + 9d)\) \\                                                                                        
                 \(92 = 20 + 9d\) \\                                                                                            
                 \(d = \frac{92 - 20}{9} = 8\)                                                                                  
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] The formula for the geometric mean \(m\) of two positive numbers \(a\) and \(b\) is \(m=\sqrt{a b}\).     
         \begin{parts}                                                                                                          
             \part Find \(m\) if \(a=16\) and \(b=25\).                                                                         
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(m = \sqrt{16 \times 25} = \sqrt{400} = 20\)                                                                  
             \end{solutionordottedlines}                                                                                        
             \part Find \(a\) if \(m=7\) and \(b=16\).                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(7 = \sqrt{a \times 16}\) \\                                                                                  
                 \(49 = a \times 16\) \\                                                                                        
                 \(a = \frac{49}{16} = 3.0625\)                                                                                 
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] If \(x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) :                                                              
         \begin{parts}                                                                                                          
             \part find \(x\) if \(b=4, a=1\) and \(c=-24\)                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(x = \frac{-4 + \sqrt{4^2 - 4 \times 1 \times (-24)}}{2 \times 1}\) \\                                        
                 \(x = \frac{-4 + \sqrt{16 + 96}}{2}\) \\                                                                       
                 \(x = \frac{-4 + \sqrt{112}}{2}\) \\                                                                           
                 \(x = \frac{-4 + 10.58}{2}\) \\                                                                                
                 \(x = 3.29\)                                                                                                   
             \end{solutionordottedlines}                                                                                        
             \part find \(c\) if \(a=1, x=6\) and \(b=2\)                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(6 = \frac{-2 + \sqrt{2^2 - 4 \times 1 \times c}}{2 \times 1}\) \\                                            
                 \(12 = -2 + \sqrt{4 - 4c}\) \\                                                                                 
                 \(14 = \sqrt{4 - 4c}\) \\                                                                                      
                 \(196 = 4 - 4c\) \\                                                                                            
                 \(c = \frac{4 - 196}{4} = -48\)                                                                                
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] A pillar is in the shape of a cylinder with a hemispherical top. If \(r\) metres is the radius of the bas 
 and \(h\) metres is the total height, the volume \(V\) cubic metres is given by the formula \(V=\frac{1}{3} \pi r^{2}(3 h-r)\) 
         \begin{parts}                                                                                                          
             \part Rearrange the formula to make \(h\) the subject.                                                             
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(V = \frac{1}{3} \pi r^2 (3h - r)\) \\                                                                        
                 \(3V = \pi r^2 (3h - r)\) \\                                                                                   
                 \(3V = 3\pi r^2 h - \pi r^3\) \\                                                                               
                 \(3V + \pi r^3 = 3\pi r^2 h\) \\                                                                               
                 \(h = \frac{3V + \pi r^3}{3\pi r^2}\)                                                                          
             \end{solutionordottedlines}                                                                                        
             \part Find the height of the pillar, correct to the nearest centimetre, if the radius of the pillar is \(0.5       
 \mathrm{~m}\) and the volume is \(10 \mathrm{~m}^{3}\).                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(h = \frac{3 \times 10 + \pi \times 0.5^3}{3\pi \times 0.5^2}\) \\                                            
                 \(h = \frac{30 + \frac{1}{8}\pi}{\frac{3}{4}\pi}\) \\                                                          
                 \(h = \frac{30 + 0.3927}{2.3562}\) \\                                                                          
                 \(h = \frac{30.3927}{2.3562}\) \\                                                                              
                 \(h = 12.90 \mathrm{~m}\) \\                                                                                   
                 \(h \approx 1290 \mathrm{~cm}\)                                                                                
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[3] Rearrange each of these formulas to make the pronumeral in brackets the subject. (All of the pronumerals  
 represent positive numbers.)                                                                                                   
         \begin{parts}                                                                                                          
             \part \(A=\ell \times w\hfill{}(\ell)\)                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(\ell = \frac{A}{w}\)                                                                                         
             \end{solutionordottedlines}                                                                                        
             \part \(V=\pi r^{2} h\hfill{}(r)\)                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(r = \sqrt{\frac{V}{\pi h}}\)                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part \(\frac{1}{x}+\frac{1}{y}=\frac{2}{z}\hfill{}(z)\)                                                           
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(\frac{1}{x}+\frac{1}{y}=\frac{2}{z}\) \\                                                                     
                 \(\frac{y+x}{xy}=\frac{2}{z}\) \\                                                                              
                 \(z = \frac{2xy}{x+y}\)                                                                                        
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] If a stone is dropped off a cliff, the number of metres it has fallen after a certain number of seconds i 
 found by multiplying the square of the number of seconds by 4.9.                                                               
         \begin{parts}                                                                                                          
             \part Find the formula for the distance \(d\) metres fallen by the stone in \(t\) seconds.                         
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(d = 4.9t^2\)                                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part Find the distance fallen in 1.5 seconds.                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(d = 4.9 \times 1.5^2 = 4.9 \times 2.25 = 11.025 \mathrm{~m}\)                                                
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[3] If \(t=\sqrt{\frac{M}{M-m}}\) :                                                                           
         \begin{parts}                                                                                                          
             \part express the formula with \(m\) as the subject                                                                
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(t^2 = \frac{M}{M-m}\) \\                                                                                     
                 \(t^2(M-m) = M\) \\                                                                                            
                 \(t^2M - t^2m = M\) \\                                                                                         
                 \(t^2m = t^2M - M\) \\                                                                                         
                 \(m = \frac{t^2M - M}{t^2}\)                                                                                   
             \end{solutionordottedlines}                                                                                        
             \part express the formula with \(M\) as the subject                                                                
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(t^2 = \frac{M}{M-m}\) \\                                                                                     
                 \(t^2(M-m) = M\) \\                                                                                            
                 \(M - t^2m = M/t^2\) \\                                                                                        
                 \(M(1 - 1/t^2) = t^2m\) \\                                                                                     
                 \(M = \frac{t^2m}{1 - 1/t^2}\)                                                                                 
             \end{solutionordottedlines}                                                                                        
             \part find the value of \(M\) if \(m=3\) and \(t=\sqrt{2}\).                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(M = \frac{(\sqrt{2})^2 \times 3}{1 - 1/(\sqrt{2})^2}\) \\                                                    
                 \(M = \frac{2 \times 3}{1 - 1/2}\) \\                                                                          
                 \(M = \frac{6}{1/2}\) \\                                                                                       
                 \(M = 12\)                                                                                                     
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] The total surface area \(S \mathrm{~cm}^{2}\) of a cylinder is given in terms of its radius \(r           
 \mathrm{~cm}\) and height \(h \mathrm{~cm}\) by the formula \(S=2 \pi r(r+h)\).                                                
         \begin{parts}                                                                                                          
             \part Express this formula with \(h\) as the subject.                                                              
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(S = 2 \pi r^2 + 2 \pi r h\) \\                                                                               
                 \(h = \frac{S - 2 \pi r^2}{2 \pi r}\)                                                                          
             \end{solutionordottedlines}                                                                                        
             \part What is the height of such a cylinder if the radius is \(7 \mathrm{~cm}\) and the total surface area is \(50 
 \mathrm{~cm}^{2}\) ? Calculate your answer in centimetres, correct to 2 decimal places.                                        
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(h = \frac{500 - 2 \pi \times 7^2}{2 \pi \times 7}\) \\                                                       
                 \(h = \frac{500 - 2 \pi \times 49}{2 \pi \times 7}\) \\                                                        
                 \(h = \frac{500 - 307.88}{43.98}\) \\                                                                          
                 \(h = \frac{192.12}{43.98}\) \\                                                                                
                 \(h = 4.37 \mathrm{~cm}\)                                                                                      
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] The sum \(S\) of the squares of the first \(n\) whole numbers is given by the formula \(S=\frac{n(n+1)(2  
 n+1)}{6}\). Find the sum of the squares of:                                                                                    
         \begin{parts}                                                                                                          
             \part the first 20 whole numbers                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(S = \frac{20(20+1)(2 \times 20+1)}{6}\) \\                                                                   
                 \(S = \frac{20 \times 21 \times 41}{6}\) \\                                                                    
                 \(S = \frac{17220}{6}\) \\                                                                                     
                 \(S = 2870\)                                                                                                   
             \end{solutionordottedlines}                                                                                        
             \part all the numbers from 5 to 21 inclusive                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(S_{21} = \frac{21(21+1)(2 \times 21+1)}{6}\) \\                                                              
                 \(S_{21} = \frac{21 \times 22 \times 43}{6}\) \\                                                               
                 \(S_{21} = 3311\) \\                                                                                           
                 \(S_{4} = \frac{4(4+1)(2 \times 4+1)}{6}\) \\                                                                  
                 \(S_{4} = \frac{4 \times 5 \times 9}{6}\) \\                                                                   
                 \(S_{4} = 30\) \\                                                                                              
                 \(S = S_{21} - S_{4} = 3311 - 30 = 3281\)                                                                      
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] For the formula \(D=\sqrt{\frac{f+x}{f-x}}\), make \(x\) the subject.                                     
         \begin{solutionordottedlines}[2cm]                                                                                     
             \(D^2 = \frac{f+x}{f-x}\) \\                                                                                       
             \(D^2(f-x) = f+x\) \\                                                                                              
             \(D^2f - D^2x = f+x\) \\                                                                                           
             \(D^2x + x = D^2f - f\) \\                                                                                         
             \(x(D^2 + 1) = D^2f - f\) \\                                                                                       
             \(x = \frac{D^2f - f}{D^2 + 1}\)                                                                                   
         \end{solutionordottedlines}                                                                                            
         \Question[2] Cans in a supermarket are displayed in a triangular stack with one can at the top, two cans in the second 
 row from the top, three cans in the third row from the top, and so on. What is the number of cans in the display if the number 
 of rows is:                                                                                                                    
         \begin{parts}                                                                                                          
             \part \(n\)                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(n = \frac{n(n+1)}{2}\)                                                                                       
             \end{solutionordottedlines}                                                                                        
             \part 35                                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(n = \frac{35(35+1)}{2} = \frac{35 \times 36}{2} = 630\)                                                      
             \end{solutionordottedlines}                                                                                        
         \end{parts}                                                                                                            
         \Question[2] Rearrange this formula to make the pronumeral in brackets the subject. \[v^{2}=u^{2}+2 a s\hfill{}(u)\]   
             \begin{solutionordottedlines}[2cm]                                                                                 
                 \(u^2 = v^2 - 2as\) \\                                                                                         
                 \(u = \sqrt{v^2 -                                                                                              
                                                                                                                                