[tex]\displaystyle\bf\\ 7.~c)\\-\frac{3}{10}\cdot\left(-\frac{5}{9}\right)+\left(-\frac{1}{6}\right)=\\\\\\=\frac{3}{10}\cdot\frac{5}{9}+\left(-\frac{1}{6}\right)=\\\\\\=\frac{3\cdot5}{10\cdot9}-\frac{1}{6}=\\\\\\=\frac{1\cdot1}{2\cdot3}-\frac{1}{6}=\frac{1}{6}-\frac{1}{6}=\boxed{\bf0}\\\\\\\implies~\boxed{\bf~-\frac{3}{10}\cdot\left(-\frac{5}{9}\right)+\left(-\frac{1}{6}\right)=0}[/tex]
