Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
Answer:
There are two points of intersection
(-4,1) and (1,-4)
Step-by-step explanation:
Answer:
the answer is 45 over 6
Step-by-step explanation:
A shortcut is "2 x _____ is 6", which is 3.
Multiply 15 by 3 also, to get 45,
45/6
Answer:
A is the answer I go with although Im not that sure
Answer:
8
Step-by-step explanation:
