Answer:
2/6561
Step-by-step explanation:
Geometric sequence formula : 
where an = nth term, a1 = first term , r = common ratio and n = term position
given ratio : 1/3 , first term : 2 , given this we want to find the 9th term
to do so we simply plug in what we are given into the formula
recall formula :
define variables : a1 = 2 , r = 1/3 , n = 9
plug in values
a9 = 2(1/3)^(9-1)
subtract exponents
a9 = 2(1/3)^8
evaluate exponent
a9 = 2 (1/6561)
multiply 2 and 1/6561
a9 = 2/6561
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:
∠GAC ≅ ∠HFD by the Property of Congruence.
Step-by-step explanation:
I'm going to be honest the question is a little confusing cause of the beginning, but if you're looking for which angles are actually congruent it's ∠GAC ≅ ∠HFD
Answer:
m=3
b=(0,5)
Step-by-step explanation:
