Question

AM NEVOIE DE REZOLVARE URGENT !!! RASPUNDETI DOAR DACA STITI !!!

Answer 1

Răspuns:

$f:R->R, f(x)=2x+3$ $A(-1,a) ; B(b,5)$

$a)$

$=> f(-1)=a <=> 2*(-1)+3=a <=>-2+3=a => a = 1\\=> f(b)=5 <=> 2*(b)+3=5 <=>2b+3=5 => 2b=2 => b = 1$

$b) A(-1,1) ; B(1,5)$

$|AB|^2 = (xb-xa)^2+(yb-ya)^2 <=> (1-(-1))^2+(5-1)^2 <=> (2)^2+(4)^2 <=> 4+16 = 20 => |AB| = \sqrt{20} = 2\sqrt{5}$

Explicație pas cu pas: