Find the coordinates of the point 1/4 of the way from b to a

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.

1. a^4+8a^3+24a^2+32a+16

2.y^8+8y^7+28y^6+56y^5+70y^4+56y^3+28y^2+8y+1

3.x^6-3x^4y^2+3x^2y^4-y^6

your mom

step-by-step explanation: