The constraints of a problem are listed below. What are the vertices of the feasible region? 2x+3y>12
5x+2y >=15
x>=0 y>=0

A) (0,0) --- (0,4) --- (21/11, 30/11)--- (3,0)
B) (0,0) ---(0,15/2) ---(21/11,30/11),--- (6,0)
C) (0,4) ---(21/11,30/11)--- (3,0)
D) (0,15/2)---(21/11,30/11)---(6,0)​

A IS THE ANSWER

Step-by-step explanation:

I USED DESMOS APP try it really helps

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)