Question

15 и 16 номер пжж сделайте​

Answer 1

$15)\; \; 4\frac{1}{3}<|x|<9,8\; \; \to \; \; \left \{ {{|x|<9,8} \atop {|x|>4\frac{1}{3}}} \right.\\\\a)\; \; |x|<9,8\; \; \to \; \; -9,8<x<9,8\\\\b)\; \; |x|>4\frac{1}{3}\; \; \to \; \; \left [ {{x>4\frac{1}{3}} \atop {x<-4\frac{1}{3}}} \right.\; \; \to \; \; x\in (-\infty ,-4\frac{1}{3}\; )\cup (4\frac{1}{3}\; ,+\infty )\\\\c)\; \; \left \{ {{x\in (-9,8\; ;\; 9,8)} \atop {x\in (-\infty ,-4\frac{1}{3})\cup (4\frac{1}{3},+\infty )}} \right.\; \; \to \; \; \underline {x\in (-9,8\; ;\; -4\frac{1}{3})\cup (4\frac{1}{3}\; ;\; 9,8)}$

$16)\; \; A=\frac{1}{4\cdot 6}+\frac{1}{6\cdot 8}+\frac{1}{8\cdot 10}=\frac{10+5+3}{240}=\frac{18}{240}=\frac{3}{40}\\\\120A=120\cdot \frac{3}{40}=\frac{120\cdot 3}{40}=3\cdot 3=9$