Complete the square to solve the quadratic equation. 3x^2+30-12=0

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.

f(2x−x2)=−3x2+6x−1

step-by-step explanation:

set up the composite function and evaluate.

f(2x−x2)=−3x2+6x−1

my answer is$ 12

step-by-step explanation:

because 15 -2=12

i belive those are the ones that are under 90 degreese.