Are my choices correct?

pls check math geniuses!!

X would be 10, and 2 because:

(10-10) = 0 and (2-2) = 0

The extraneous solution is correct because the equation does not hold true when x is 2 and only hold trues when it is 10.

Edit: the extraneous solution is 2 is because it doesn't hold true to the equation.

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.

don't know what that says sorry i can't you

step-by-step explanation:

idk what it says