Set X is made up of the possible ways five students, represented by A, B, C, D, and E, can be formed into groups of three. Set Y is made up of the possible ways five students can be formed into groups of three if student A must be in all possible groups. Which statements about the situation are true? Select three options.

Set X has 10 possible groupings.
X Y
Set Y = {ABC, ABD, ABE, ACD, ACE, ADE}
If person E must be in each group, then there can be only one group.
There are three ways to form a group if persons A and C must be in it.

answer: ok

step-by-step explanation:

ghjmutmum

step-by-step explanation:

graph b

step-by-step explanation:

one solution- graph a (clear intercepting point.)

no solution- graph b (they're parallel, there will never be a point of interception.)

infinite solutions- graph c (all points intercept.)

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)