Question

Cat Mai Repede Va Rog!!! ​

Answer 1

Răspuns:

Explicație pas cu pas:

$f(x)=\frac{2}{3} x-2$

∩OX, y=0 $\frac{2}{3} x-2=0$

2x-6=0

x=3 A(3,0)

∩OY, x=0 y=0-2=-2 B(0,-2)

b. Fie OD⊥AB

ΔAOB dr in O

$OD=\frac{AO\cdot OB}{AB}$

Aflam AB din Pitagora

AB²=OB²+AO²

AB²=4+9=13

AB=√13

Deci

$OD=\frac{2\cdot 3}{\sqrt{13} } =\frac{6\sqrt{13} }{13}$

c. f(x)+f(x+1)≤f(x-2)+4

$\frac{2}{3} x-2+\frac{2}{3}(x+1)-2\leq \frac{2}{3}(x-2)-2+4 \ \ |\cdot 3\\\\2x-6+2x+2-6\leq 2x-4-6+12\\\\2x\leq 12\\\\x\leq 6$

x∈(-∞,6]