Question

Помогите срочно!!!! ​

Answer 1

$9<x<12\; \; ,\; \; 3<y<4\\\\a)\; \; 3\cdot 9+4\cdot 3<3x+4y<3\cdot 12+4\cdot 4\\\\39<3x+4y<91\\\\b)\; \; \frac{1}{12}<\frac{1}{x}<\frac{1}{9}\; \; ,\; \; \frac{1}{4}<y<\frac{1}{3}\\\\-\frac{1}{9}<-\frac{1}{x}<-\frac{1}{12}\\\\\frac{1}{4}-\frac{1}{9}<\frac{1}{y}+(-\frac{1}{x})<\frac{1}{3}-\frac{1}{12}\\\\\frac{5}{36}<\frac{1}{y}-\frac{1}{x}<\frac{1}{3}$

$c)\; \; 9<y^2<16\; ,\; \; 81<x^2<144\; ,\; \; -144<-x^2<-81\\\\9-144<y^2+(-x^2)<16-81\\\\-135<y^2-x^2<-65\\\\d)\; \; \frac{9}{12}<\frac{y^2}{x}<\frac{16}{9}\; \; ,\; \; -\frac{16}{9}<-\frac{y^2}{x}<-\frac{3}{4}$

$2)\; \; \alpha =\frac{180^\circ -\beta }{2}=90^\circ -\frac{\beta }{2}\; \; ,\; \; 50^\circ <\beta <52^\circ \; ,\\\\\frac{50^\circ }{2}<\frac{\beta }{2}<\frac{52^\circ }{2}\; \; ,\; \; 25^\circ <\frac{\beta }{2}<26^\circ \; \; ,\; \; -26^\circ <-\frac{\beta }{2}<-25^\circ \\\\64^\circ <90^\circ -\frac{\beta }{2}<65^\circ \\\\64^\circ <\alpha <65^\circ$