Answer:
[tex]\large\boxed{h\leq-\dfrac{19}{6}\to h\in\left(-\infty,\ -\dfrac{19}{6}\right]}[/tex]
Step-by-step explanation:
[tex]\text{First convert the mixed numbers to the improper fractions:}\\\\10\dfrac{2}{3}=\dfrac{10\cdot3+2}{3}=\dfrac{32}{3}\\\\4\dfrac{1}{3}=\dfrac{4\cdot3+1}{3}=\dfrac{13}{3}\\\\\text{Solve the inequality}\\\\10\dfrac{2}{3}+2h\leq4\dfrac{1}{3}\\\\\dfrac{32}{3}+2h\leq\dfrac{13}{3}\qquad\text{multiply both sides by 3}\\\\3\!\!\!\!\diagup^1\cdot\dfrac{32}{3\!\!\!\!\diagup_1}+3\cdot2h\leq3\!\!\!\!\diagup^1\cdot\dfrac{13}{3\!\!\!\!\diagup_1}\\\\32+6h\leq13\qquad\text{subtract 32 from both sides}\\\\6h\leq-19\qquad\text{divide both sides by 6}\\\\h\leq-\dfrac{19}{6}[/tex]
