The domain of g alone is {x | x ≠ 0}, and the domain of f is all reals. So the domain of (f ◦ g) is the domain of g
{x | x ≠ 0}.
(f ◦ g)(x) = 1/x + 3.
The range of g(x) = 1/x is actually the same as its domain {y | y ≠ 0}. Adding three, the range of f ◦ g is all reals except for 3,
{y | y ≠ 3}
The line y = 3 is actually an asymptote (horizontal) to the graph of f ◦ g.
Option a, c,d are correct.
step-by-step explanation:
from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have
tz=sz(given)
bz=zb(common)
therefore, by rhs rule,δtzb ≅δszb
by cpctc, sz≅tz
also, from δctz and δasz,
tz=sz(given)
∠tcz=∠saz(90°)
by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz
also,from δasz and δzsb,
zs=sz(common)
∠zbs=∠saz=90°
by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb
hence, option a, c,d are correct.
