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:
446mm
Step-by-step explanation:
If we box off parts of the area, it makes it easier to solve. I personally broke it into tiny bits:
Upper left box: 16mm
Bigger box (excluding little box): 90mm
Big rectangle: 340mm
Now, add them all together.
Equals 446mm
Answer:
x=12
Step-by-step explanation:
（X-8)/7=2
X-8=14
X=22
A = L x W = 8 x 17 = 136
answer
C. 136 sq. units
Step-by-step explanation:
Taking the first coordinate point (3,16.5)
where x= 3 and y= 16.5
optionB
