 \subsection{Substitution into formulas}                                                                    
 \begin{questions}                                                                                          
     \Question[3] For each part, find the value of the subject when the other pronumerals have the value    
 indicated.                                                                                                 
     \begin{parts}                                                                                          
         \part \(A=\frac{1}{2}(a+b) h\), where \(a=4, b=6, h=10\)                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(A=\frac{1}{2}(4+6) \cdot 10 = \frac{1}{2} \cdot 10 \cdot 10 = 50\)                       
             \end{solutionordottedlines}                                                                    
         \part \(t=a+(n-1) d\), where \(a=30, n=8, d=4\)                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(t=30+(8-1) \cdot 4 = 30+7 \cdot 4 = 30+28 = 58\)                                         
             \end{solutionordottedlines}                                                                    
         \part \(E=\frac{1}{2} m v^{2}\), where \(m=8, v=4\)                                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(E=\frac{1}{2} \cdot 8 \cdot 4^{2} = 4 \cdot 16 = 64\)                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] For each part, find the value of the subject when the other pronumerals have the value    
 indicated. Calculate \(\mathbf{a}-\mathbf{c}\) correct to 3 decimal places and \(\mathbf{d}\) correct to 2 
     \begin{parts}                                                                                          
         \part \(x=\sqrt{a b}\), where \(a=40, b=50\)                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(x=\sqrt{40 \cdot 50} = \sqrt{2000} \approx 44.721\)                                      
             \end{solutionordottedlines}                                                                    
         \part \(V=\pi r^{2} h\), where \(r=12, h=20\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(V=\pi \cdot 12^{2} \cdot 20 = 144 \pi \cdot 20 = 2880 \pi \approx 9047.78\)              
             \end{solutionordottedlines}                                                                    
         \part \(T=2 \pi \sqrt{\frac{\ell}{g}}\), where \(\ell=88.2, g=9.8\)                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(T=2 \pi \sqrt{\frac{88.2}{9.8}} \approx 2 \pi \sqrt{9} = 2 \pi \cdot 3 \approx 18.850\)  
             \end{solutionordottedlines}                                                                    
         \part \(A=P(1+R)^{n}\), where \(P=10000, R=0.065, n=10\)                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(A=10000(1+0.065)^{10} \approx 10000 \cdot 1.8194 \approx 18194.00\)                      
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[1] For the formula \(S=2(\ell w+\ell h+h w)\), find \(h\) if \(S=592, \ell=10\) and \(w=8\). 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(592=2(10 \cdot 8+10h+8h)\)                                                               
                 \(296=80+18h\)                                                                             
                 \(216=18h\)                                                                                
                 \(h=12\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[1] For the formula \(s=u t+\frac{1}{2} a t^{2}\), find \(a\) if \(s=1000, u=20\) and \(t=5\) 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(1000=20 \cdot 5+\frac{1}{2} a \cdot 5^{2}\)                                              
                 \(1000=100+\frac{1}{2} a \cdot 25\)                                                        
                 \(900=\frac{1}{2} a \cdot 25\)                                                             
                 \(36=a\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[1] For the formula \(t=a+(n-1) d\), find \(n\) if \(t=58, d=3\) and \(a=7\).                 
             \begin{solutionordottedlines}[2cm]                                                             
                 \(58=7+(n-1) \cdot 3\)                                                                     
                 \(51=(n-1) \cdot 3\)                                                                       
                 \(17=n-1\)                                                                                 
                 \(n=18\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \Question[2] Given \(v^{2}=u^{2}+2 a x\) and \(v>0\), find the value of \(v\) (correct to 1 decimal    
 place) when:                                                                                               
     \begin{parts}\begin{multicols}{2}                                                                      
         \part \(u=0, a=5\) and \(x=10\)                                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{0^{2}+2 \cdot 5 \cdot 10} = \sqrt{100} = 10.0\)                                  
             \end{solutionordottedlines}                                                                    
         \part \(u=2, a=9.8\) and \(x=22\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{2^{2}+2 \cdot 9.8 \cdot 22} \approx \sqrt{4+431.2} \approx \sqrt{435.2} \approx  
 20.9\)                                                                                                     
             \end{solutionordottedlines}                                                                    
     \end{multicols}\end{parts}                                                                             
     \Question[2] Given \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), find the value of:                         
     \begin{parts}\begin{multicols}{2}                                                                      
         \part \(u\) when \(f=2\) and \(v=4\)                                                               
             \begin{solutionordottedlines}[2cm]                                                             
                 \(\frac{1}{2}=\frac{1}{u}+\frac{1}{4}\)                                                    
                 \(\frac{1}{2}-\frac{1}{4}=\frac{1}{u}\)                                                    
                 \(\frac{1}{4}=\frac{1}{u}\)                                                                
                 \(u=4\)                                                                                    
             \end{solutionordottedlines}                                                                    
         \part \(u\) when \(f=3\) and \(v=4\)                                                               
             \begin{solutionordottedlines}[2cm]                                                             
                 \(\frac{1}{3}=\frac{1}{u}+\frac{1}{4}\)                                                    
                 \(\frac{1}{3}-\frac{1}{4}=\frac{1}{u}\)                                                    
                 \(\frac{1}{12}=\frac{1}{u}\)                                                               
                 \(u=12\)                                                                                   
             \end{solutionordottedlines}                                                                    
     \end{multicols}\end{parts}                                                                             
     \Question[6] The formula for finding the number of degrees Fahrenheit \((F)\) for a temperature given  
 a number of degrees Celsius \((C)\) is \(F=\frac{9}{5} C+32\).                                             
                                                                                                            
     Fahrenheit temperatures are still used in the USA, but in Australia we commonly use Celsius.           
                                                                                                            
     Calculate the Fahrenheit temperatures which people in the USA would recognise for:                     
     \begin{parts}                                                                                          
         \part the freezing point of water, \(0^{\circ} \mathrm{C}\)                                        
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 0+32 = 32\)                                                          
             \end{solutionordottedlines}                                                                    
         \part the boiling point of water, \(100^{\circ} \mathrm{C}\)                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 100+32 = 180+32 = 212\)                                              
             \end{solutionordottedlines}                                                                    
         \part a nice summer temperature of \(25^{\circ} \mathrm{C}\)                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(F=\frac{9}{5} \cdot 25+32 = 45+32 = 77\)                                                 
             \end{solutionordottedlines}                                                                    
         \hspace{-1cm}\item[] Now calculate the Celsius temperatures which people in Australia would        
 recognise for:                                                                                             
         \part \(50^{\circ} \mathrm{F}\)                                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(C=\frac{5}{9}(50-32) = \frac{5}{9} \cdot 18 \approx 10\)                                 
             \end{solutionordottedlines}                                                                    
         \part \(104^{\circ} \mathrm{F}\)                                                                   
             \begin{solutionordottedlines}[2cm]                                                             
                 \(C=\frac{5}{9}(104-32) = \frac{5}{9} \cdot 72 \approx 40\)                                
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] Sam throws a stone down to the ground from the top of a cliff \(s\) metres high, with an  
 initial speed of \(u \mathrm{~m} / \mathrm{s}\). It accelerates at \(a \mathrm{~m} / \mathrm{s}^{2}\). The 
 stone hits the ground with a speed of \(v \mathrm{~m} / \mathrm{s}\) given by the formula \(v^{2}=u^{2}+2  
 s\). Find the speed at which the stone hits the ground, correct to 2 decimal places, if:                   
     \begin{parts}                                                                                          
         \part \(u=0, a=9.8\) and \(s=50\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{0^{2}+2 \cdot 9.8 \cdot 50} = \sqrt{980} \approx 31.30\)                         
             \end{solutionordottedlines}                                                                    
         \part \(u=5, a=9.8\) and \(s=35\)                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(v=\sqrt{5^{2}+2 \cdot 9.8 \cdot 35} \approx \sqrt{25+686} \approx \sqrt{711} \approx     
 26.67\)                                                                                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
 \end{questions}                                                                                            
                                                                                                            
 \subsection{Changing the subject of a formula}                                                             
                                                                                                            
 \begin{questions}                                                                                          
     \Question[] Given the formula \(v=u+a t\) :                                                            
     \begin{parts}                                                                                          
         \Part[1] rearrange the formula to make \(u\) the subject                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=v-a t\)                                                                                
             \end{solutionordottedlines}                                                                    
         \Part[2] find the value of \(u\) when:                                                             
             \begin{solutionordottedlines}[2cm]                                                             
             \end{solutionordottedlines}                                                                    
         \begin{subparts}                                                                                   
             \subpart \(v=20, a=2\) and \(t=5 \quad\)                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=20-2 \cdot 5 = 20-10 = 10\)                                                            
             \end{solutionordottedlines}                                                                    
             \subpart \(v=40, a=-6\) and \(t=4\)                                                            
             \begin{solutionordottedlines}[2cm]                                                             
                 \(u=40-(-6) \cdot 4 = 40+24 = 64\)                                                         
             \end{solutionordottedlines}                                                                    
         \end{subparts}                                                                                     
         \Part[1] rearrange the formula to make \(a\) the subject                                           
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{v-u}{t}\)                                                                        
             \end{solutionordottedlines}                                                                    
         \Part[2] find the value of \(a\) when:                                                             
             \begin{solutionordottedlines}[2cm]                                                             
             \end{solutionordottedlines}                                                                    
         \begin{subparts}                                                                                   
             \subpart \(v=20, u=15\) and \(t=2 \quad\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{20-15}{2} = \frac{5}{2} = 2.5\)                                                  
             \end{solutionordottedlines}                                                                    
             \subpart \(v=-26.8, u=-14.4\) and \(t=2\)                                                      
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{-26.8-(-14.4)}{2} = \frac{-26.8+14.4}{2} = \frac{-12.4}{2} = -6.2\)              
             \end{solutionordottedlines}                                                                    
             \subpart \(v=\frac{1}{2}, u=\frac{2}{3}\) and \(t=\frac{5}{6}\)                                
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{\frac{1}{2}-\frac{2}{3}}{\frac{5}{6}} =                                          
 \frac{\frac{3}{6}-\frac{4}{6}}{\frac{5}{6}} = \frac{-\frac{1}{6}}{\frac{5}{6}} = -\frac{1}{5}\)            
             \end{solutionordottedlines}                                                                    
         \end{subparts}                                                                                     
         \Part[2] rearrange the formula to make \(t\) the subject and find \(t\) when \(v=6, u=7\) and      
 \(a=-3\).                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(t=\frac{v-u}{a}\)                                                                        
                 \(t=\frac{6-7}{-3} = \frac{-1}{-3} = \frac{1}{3}\)                                         
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[6] Rearrange each of these formulas to make the pronumeral in brackets the subject.          
     \begin{parts}                                                                                          
         \part \(y=m x+c\hfill{}(x)\)                                                                       
             \begin{solutionordottedlines}[2cm]                                                             
                 \(x=\frac{y-c}{m}\)                                                                        
             \end{solutionordottedlines}                                                                    
         \part \(C=2 \pi r\hfill{}(r)\)                                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(r=\frac{C}{2 \pi}\)                                                                      
             \end{solutionordottedlines}                                                                    
         \part \(s=u t+\frac{1}{2} a t^{2}\hfill{}(a)\)                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(a=\frac{2(s-u t)}{t^{2}}\)                                                               
             \end{solutionordottedlines}                                                                    
         \part \(V=\frac{1}{3} \pi r^{2} h\hfill{}(h)\)                                                     
             \begin{solutionordottedlines}[2cm]                                                             
                 \(h=\frac{3 V}{\pi r^{2}}\)                                                                
             \end{solutionordottedlines}                                                                    
         \part \(S=\frac{n}{2}(a+\ell)\hfill{}(n)\)                                                         
             \begin{solutionordottedlines}[2cm]                                                             
                 \(n=\frac{2 S}{a+\ell}\)                                                                   
             \end{solutionordottedlines}                                                                    
         \part \(E=m g h+\frac{1}{2} m v^{2}\hfill{}(h)\)                                                   
             \begin{solutionordottedlines}[2cm]                                                             
                 \(h=\frac{E-\frac{1}{2} m v^{2}}{m g}\)                                                    
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
     \Question[4] The kinetic energy \(E\) joules of a moving object is given by \(E=\frac{1}{2} m v^{2}\), 
 where \(m \mathrm{~kg}\) is the mass of the object and \(v \mathrm{~m} / \mathrm{s}\) is its speed.        
                                                                                                            
     Rearrange the formula to make \(m\) the subject and use this to find the mass of the object when its   
 energy and speed are, respectively:                                                                        
     \begin{parts}                                                                                          
         \part 400 joules, \(10 \mathrm{~m} / \mathrm{s}\)                                                  
             \begin{solutionordottedlines}[2cm]                                                             
                 \(m=\frac{2 E}{v^{2}}\)                                                                    
                 \(m=\frac{2 \cdot 400}{10^{2}} = \frac{800}{100} = 8\)                                     
             \end{solutionordottedlines}                                                                    
         \part 28 joules, \(4 \mathrm{~m} / \mathrm{s}\)                                                    
             \begin{solutionordottedlines}[2cm]                                                             
                 \(m=\frac{2 \cdot 28}{4^{2}} = \frac{56}{16} = 3.5\)                                       
             \end{solutionordottedlines}                                                                    
     \end{parts}                                                                                            
 \end{questions}                                                                                            


 \subsection{Constructing Formulas}                                                                                                                                             
 \begin{questions}                                                                                                                                                              
     \Question[8] Find a formula for:                                                                                                                                           
     \begin{parts}                                                                                                                                                              
         \part the number of cents \(z\) in \(x\) dollars and \(y\) cents                                                                                                       
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(z = 100x + y\)                                                                                                                                               
             \end{solutionordottedlines}                                                                                                                                        
         \part the number of minutes \(x\) in \(y\) minutes and \(z\) seconds                                                                                                   
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(x = y + \frac{z}{60}\)                                                                                                                                       
             \end{solutionordottedlines}                                                                                                                                        
         \part the number of hours \(x\) in \(y\) minutes and \(z\) seconds                                                                                                     
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(x = \frac{y}{60} + \frac{z}{3600}\)                                                                                                                          
             \end{solutionordottedlines}                                                                                                                                        
         \part the cost \(\$ m\) of 1 book if 20 books cost \(\$ c\)                                                                                                            
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(m = \frac{c}{20}\)                                                                                                                                           
             \end{solutionordottedlines}                                                                                                                                        
         \part the cost \(\$ n\) of 1 suit if 5 suits cost \(\$ m\)                                                                                                             
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(n = \frac{m}{5}\)                                                                                                                                            
             \end{solutionordottedlines}                                                                                                                                        
         \part the cost \(\$ m\) of 1 tyre if \(x\) tyres cost \(\$ y\)                                                                                                         
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(m = \frac{y}{x}\)                                                                                                                                            
             \end{solutionordottedlines}                                                                                                                                        
         \part the cost \(\$ p\) of \(n\) suits if 4 suits cost \(\$ k\)                                                                                                        
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(p = \frac{k}{4}n\)                                                                                                                                           
             \end{solutionordottedlines}                                                                                                                                        
         \part the cost \(\$ q\) of \(x\) cars if 8 cars cost \(\$ b\)                                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(q = \frac{b}{8}x\)                                                                                                                                           
             \end{solutionordottedlines}                                                                                                                                        
     \end{parts}                                                                                                                                                                
     \Question[7] In each part, find a formula from the information given.                                                                                                      
     \begin{parts}                                                                                                                                                              
         \part A hire car firm charges \(\$ 20\) per day plus 40 cents per \(\mathrm{km}\). What is the total cost \(\$ C\) for a day in which \(x \mathrm{~km}\) was travelled 
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(C = 20 + 0.40x\)                                                                                                                                             
             \end{solutionordottedlines}                                                                                                                                        
         \part If there are 50 litres of petrol in the tank of a car and petrol is used at the rate of 4 litres per day, what is the number of litres \(y\) that remains after  
 \(x\) days?                                                                                                                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(y = 50 - 4x\)                                                                                                                                                
             \end{solutionordottedlines}                                                                                                                                        
         \part Cooking instructions for a forequarter of lamb are as follows: preheat oven to \(220^{\circ} \mathrm{C}\) and cook for \(45 \mathrm{~min}\) per kg plus an       
 additional \(20 \mathrm{~min}\). What is the formula relating the cooking time \(T\) minutes and weight \(w \mathrm{~kg}\) ?                                                   
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(T = 45w + 20\)                                                                                                                                               
             \end{solutionordottedlines}                                                                                                                                        
         \part In a sequence of numbers the first number is 2 , the second number is 4 , the third is 8 , the fourth is 16, etc. Assuming the doubling pattern continues, what  
 the formula you would use to calculate \(t\), the \(n\)th number?                                                                                                              
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(t = 2^n\)                                                                                                                                                    
             \end{solutionordottedlines}                                                                                                                                        
         \part A piece of wire of length \(x \mathrm{~cm}\) is bent into a circle of area \(A \mathrm{~cm}^{2}\). What is the formula relating \(A\) and \(x\) ?                
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(A = \frac{\pi}{4\pi^2}x^2 = \frac{x^2}{4\pi}\)                                                                                                               
             \end{solutionordottedlines}                                                                                                                                        
     \end{parts}                                                                                                                                                                
     \Question[10] A cyclic quadrilateral has all its vertices on a circle. Its area \(A\) is given by Brahmagupta's formula \[ A^{2}=(s-a)(s-b)(s-c)(s-d) \] where \(a, b, c\) 
 and \(d\) are the side lengths of the quadrilateral and \(s=\frac{a+b+c+d}{2}\) is the 'semi-perimeter'. Find the exact area of a cyclic quadrilateral with side lengths:      
     \begin{parts}                                                                                                                                                              
         \part \(4,5,6,7\)                                                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(s = \frac{4+5+6+7}{2} = 11\) \\                                                                                                                              
                 \(A^{2} = (11-4)(11-5)(11-6)(11-7) = 7 \cdot 6 \cdot 5 \cdot 4\) \\                                                                                            
                 \(A = \sqrt{7 \cdot 6 \cdot 5 \cdot 4} = \sqrt{840} = 2\sqrt{210}\)                                                                                            
             \end{solutionordottedlines}                                                                                                                                        
         \part \(7,4,4,3\)                                                                                                                                                      
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(s = \frac{7+4+4+3}{2} = 9\) \\                                                                                                                               
                 \(A^{2} = (9-7)(9-4)(9-4)(9-3) = 2 \cdot 5 \cdot 5 \cdot 6\) \\                                                                                                
                 \(A = \sqrt{2 \cdot 5 \cdot 5 \cdot 6} = \sqrt{300} = 10\sqrt{3}\)                                                                                             
             \end{solutionordottedlines}                                                                                                                                        
         \part \(8,9,10,13\)                                                                                                                                                    
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(s = \frac{8+9+10+13}{2} = 20\) \\                                                                                                                            
                 \(A^{2} = (20-8)(20-9)(20-10)(20-13) = 12 \cdot 11 \cdot 10 \cdot 7\) \\                                                                                       
                 \(A = \sqrt{12 \cdot 11 \cdot 10 \cdot 7} = \sqrt{9240} = 2\sqrt{2310}\)                                                                                       
             \end{solutionordottedlines}                                                                                                                                        
         \part \(39,52,25,60\)                                                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(s = \frac{39+52+25+60}{2} = 88\) \\                                                                                                                          
                 \(A^{2} = (88-39)(88-52)(88-25)(88-60) = 49 \cdot 36 \cdot 63 \cdot 28\) \\                                                                                    
                 \(A = \sqrt{49 \cdot 36 \cdot 63 \cdot 28} = \sqrt{3895584} = 1974\)                                                                                           
             \end{solutionordottedlines}                                                                                                                                        
         \part \(51,40,68,75\)                                                                                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                                                                                 
                 \(s = \frac{51+40+68+75}{2} = 117\) \\                                                                                                                         
                 \(A^{2} = (117-51)(117-40)(117-68)(117-75) = 66 \cdot 77 \cdot 49 \cdot 42\) \\                                                                                
                 \(A = \sqrt{66 \cdot 77 \cdot 49 \cdot 42} = \sqrt{11388348} = 3372\)                                                                                          
             \end{solutionordottedlines}                                                                                                                                        
     \end{parts}                                                                                                                                                                
 \end{questions}                                                                                                                                                                
                                                                                                                                                                                
