Question

V varies partly as D and partly as the square of D when Vequals to 5 ,D equals to 2, and when V equal to 9 and D equals to 3

write this law connecting V and D ​

Answer 1

`\stackrel{ \textit{Partial Variation} }{V=aD+bD^2}\qquad \impliedby \begin{array}{llll} \textit{V\textit{ varies partly}}\\ \textit{with D and partly with }D^2 \end{array} \\\\[-0.35em] ~\dotfill\\\\ \textit{we also know that} \begin{cases} V=5\\ D=2\\[-0.5em] \hrulefill\\ V=9\\ D=3 \end{cases}\implies \begin{array}{llll} 5=a2+b2^2&\qquad &5=2a+4b\\\\ 9=a3+b3^2&&9=3a+9b \end{array} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{5=2a+4b}\implies 5-2a=4b\implies \cfrac{5-2a}{4}=b \\\\\\ \stackrel{\textit{using the 2nd equation}}{9=3a+9b}\implies \stackrel{\textit{substituting from above}}{9=3a+9\left( \cfrac{5-2a}{4} \right)}\implies 9=3a+\cfrac{45-18a}{4} \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{4}}{4(9)=4\left( 3a+\cfrac{45-18a}{4} \right)}\implies 36=12a+45-18a\implies -9=-6a`

`\cfrac{-9}{-6}=a\implies \boxed{\cfrac{3}{2}=a}\hspace{5em}b=\cfrac{5-2\left( (3)/(2) \right)}{4}\implies \boxed{b=\cfrac{1}{2}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill V=\cfrac{3}{2}D+\cfrac{1}{2}D^2~\hfill`