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
Step-by-step explanation:
Pythagoras' theorem for the smallest one :


= 52
Pythagoras' theorem for the middle one :
=
+ 
Pythagoras' theorem for the biggest one :


Using the formula before (for
) it becomes :



16 + 8a = 52 + 36
16+8a = 88
8a = 88-16
8a = 72
a = 9
Verifying :



= 117
The biggest one :



True
Answer:
theres no questions
Step-by-step explanation:
Answer: 31
Step-by-step explanation:
add the 2 lengths of the given parts.