Quadrilateral ABCD is inscribed in a circle with angle measures m
*PLEASE ANSWER IMMEDIATELY*

B and D

Step-by-step explanation:

First method :

All angles of a quadrilateral sums to 360 degrees

A + B + C + D = 360

=> 5x^2 + 36x - 336 = 0

=> x = 6 or x = - 56/5 [It gives negative value of angles]

So, A = 58, B =  109, C = 122, D = 71

Second method :

Sum of opposite angles of a quadrilateral inscribed in a circle is 180 degrees

A + C = 180

=> x = 6

So, A = 58, B =  109, C = 122, D = 71

[If there's any mistake, please inform me]

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.

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation: