Choose the student who solved the equation correctly. 12a = 6

Maria's work. 12 a = 6. StartFraction 12 a Over 12 EndFraction = StartFraction 6 Over 12 EndFraction. A = 2.
Jessi's work. 12 a = 6. StartFraction 12 a Over 12 EndFraction = StartFraction 6 Over 12 EndFraction. A = one-half.
Marcus's work. 12 a = 6. StartFraction 12 a Over 6 EndFraction = StartFraction 6 Over 6 EndFraction. A = 1.
Trevor's work. 12 a = 6. StartFraction 12 a Over 6 EndFraction = StartFraction 6 Over 6 EndFraction. A = 2.

jessi!! i took the test <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.

congruent segments

step-by-step explanation: