What is the general equation of a cosine function with an amplitude of 3, a period of 4x, and a horizontal shift of
?
© y = 4500S(3(x-1))
y-3cos(4=(x+))
y=3cos(0.5(x+5))
y=4COS(0.2(x+ ))

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.

tangent

step-by-step explanation:

i think asking google will you get the answer to that question