Verified Answer :

naǫ

A:
\(F=\frac{9}{5}C+32 \\ \\ F-32=\frac{9}{5}C \\ \\ \boxed{C=\frac{5}{9}(F-32)}\)

B:
\(F=10^\circ F \\ \\ C=\frac{5}{9}(F-32)=\frac{5}{9}(10-32)=\frac{5}{9} \times (-22)=\frac{-110}{9}=-12\frac{2}{9} \\ \\ \boxed{C=-12 \frac{2}{9} ^\circ C}\)

C:
\(-np-70\ \textgreater \ 40 \ \ \ \ \ \ |\hbox{add 70} \\ \\ -np\ \textgreater \ 110 \ \ \ \ \ \ \ \ \ \ \ |\hbox{divide by (-p), flip the sign} \\ \\ \boxed{n\ \textless \ -\frac{110}{p}}\)

Related Questions