F(x) = 2x g(x) = 2x + 1
Find
4) (a). Include any restrictions on the domain.
V22
20+1;
22+
O A. (1) (2) = 9,*+-2
,
O B. () (-) = 4,16-1
c. (1) (z) = 21,0 + 0
O D. (4) (x) = ***, > 0
2x+
20+1
2

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.

answer: 10 meter

step-by-step explanation:

need details

the answer is 11.2 (i know i’m late)

b