If a point is chosen inside the large circle,
what is the probability that it will also be
inside the small circle?

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6

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.

sorry can't

step-by-step explanation:

i googled though and something came up but it was in science langueage not common english sooo

answer: the anser is 3

step-by-step explanation:

c- 2o is equal to 9