Evaluate \dfrac32y-3+\dfrac53z
2
3

y−3+
3
5

zstart fraction, 3, divided by, 2, end fraction, y, minus, 3, plus, start fraction, 5, divided by, 3, end fraction, z when y=4y=4y, equals, 4 and z=3z=3z, equals, 3.

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.

answer: i guess 5 units?

step-by-step explanation: