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\(1\dfrac {7}{10} = \dfrac {17}{10}\\ 8\dfrac {1}{2} = \dfrac {17}{2}\)

\(\dfrac {\dfrac {17}{10}}{\dfrac {17}{2} } = \dfrac {17}{10}\times \dfrac {2}{17}=\dfrac{2}{10}=\dfrac 1 5\)

\(6(y-1)+8=32\\ 6(y-1)=24 \\ (y-1)=4 \\ y=5\)

\(h(t)=-9.8 \dfrac {t^2}{2}+v_0 t + h_0 \\ \mbox{In this case we have }v_0=0 \mbox{ and we'll let }h_0=0.03 \mbox{ and solve for when }h=0\)

\(0=-4.9 t^2 + 0.03 \\ 4.9t^2 = 0.03 \\ t^2 = \dfrac{0.03}{4.9} \\ t=\sqrt{ \dfrac{0.03}{4.9}} = 0.078s\)

\(822\%=\dfrac {822}{100}=8\dfrac {22}{100}=8 \dfrac{11}{50}\)

-f(x) is a reflection of f(x) with respect to the x axis.

If you're in a drawing program and you have some shape and you select it and "flip vertical" this is the same sort of reflection that -f(x) does to f(x).

\(4^x - 4^{-x} \\ \mbox{I'm not sure what sort of answer you are looking for here. We can do this..} \\ 4^x = e^{\ln(4) x} \mbox{ and } 4^{-x} = e^{-\ln(4)x} \mbox{ so we have }\\ 4^x - 4^{-x} = e^{\ln(4) x} - e^{-\ln(4)x} = \\ 2\sinh(\ln(4)x)\)

You can't be 3 3/8 full.

\(PV=nRT\)

\(V_1=\dfrac 4 3 \pi \left(\dfrac {50}{2}\right)^3 \\ V_2 = \dfrac 4 3 \pi \left(\dfrac {51}{2}\right)^3 \\ \mbox{The rest of the quantities other than temperature stay constant.} \)

\(\dfrac{P V_1}{P V_2} = \dfrac{n R T_1}{n R T_2} \Rightarrow \dfrac {V_1}{V_2}=\dfrac {T_1}{T_2}\)

\(\dfrac {V_1}{V_2}=\dfrac{50^3}{51^3} = \dfrac {125000}{132651} \\ \mbox{so }\dfrac {T_1}{T_2} = \dfrac {125000}{132651} \\ T_2 = T_1 \dfrac{132651}{125000} = 19 \dfrac{132651}{125000} = 20.163 ^\circ C\)

\(\mbox{20% of 4.2 }=\dfrac {20}{100} \times 4.2 = 0.2 \times 4.2 = 0.84\)

\(10C1 = \begin{pmatrix}10 \\ 1\end{pmatrix} = \dfrac {10!}{1!(10-1)!} = \dfrac {10!}{9!} = 10\)