Suppose that the functions r and s are defined for all real numbers x as follows. r(x) = 2x+4

s(x) = 6x

Write the expressions for (r-s)(x) and (r:s) (x) and evaluate (r+s)(-2).

(r - s) (x) = 1

x

?

(r+s) (-2)

Help me please

`1234567890-=/*-qwertyuiop[]asdfghjkl; '\\zxcvbnm,./+ /

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.

the answer is a

step-by-step explanation: