Answer: bruh nobody needs to know this I won’t use this irl
Step-by-step explanation:
Answer:What are the equivalence classes of the equivalence relations in Exercise 3? A binary relation defined on a set S is said to be equivalence relation if it is reflexive, symmetric and transitive. An equivalence relation defined on a set S, partition the set into disjoint equivalence classes
Answer:
x= -3 x = 1/2 x=-2
Step-by-step explanation:
f(x)=(x+3) (2x-1)(x+2)
Set equal to zero
0 =(x+3) (2x-1)(x+2)
Using the zero product property
x+3 =0 2x-1 =0 x+2 =0
x= -3 2x =1 x = -2
x= -3 x = 1/2 x=-2
They are basically like groups. Take -3x+2x , Just do -3 +2 = -1 and then just add the x. So -1x. Next you have -5y+10y . Just do -5 + 10 = 5 and remember to add the x on so your final answer should be -1x + 5y
