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
X+4x = 180
5x = 180
x =180/5
x = 36
answer
D. 36
Intersection of the first two lines:
Multiply the first equation by 4 and the second by 5:
Subtract the two equations:
Plug this value for y in one of the equation, for example the first:
So, the first point of intersection is 
We can find the intersection of the other two lines in the same way: we start with
Use the fact that x and y are the same to rewrite the second equation as
And since x and y are the same, the second point is 
So, we're looking for a line passing through and . We may use the formula to find the equation of a line knowing two of its points, but in this case it is very clear that both points have the same coordinates, so the line must be 
In the attached figure, line 
is light green, line 
is dark green, and their intersection is point A.
Simiarly, line 
is red, line 
is orange, and their intersection is B.
As you can see, the line connecting A and B is the red line itself.
Answer:
Length = 8 feet
Step-by-step explanation:
Perimeter = 2 x ( length + width )
20 = 2 x (length + 2)
10 = Length + 2
10 - 2 = Length
Therefore, Length = 8 feet
Answer:
h=3.5
Step-by-step explanation:
divide 28 by 8
hope I helped
