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:
A two-digit number can be written as:
a*10 + b*1
Where a and b are single-digit numbers, and a ≠ 0.
We know that:
"The sum of a two-digit number and the number obtained by interchanging the digits is 132."
then:
a*10 + b*1 + (b*10 + a*1) = 132
And we also know that the digits differ by 2.
then:
a = b + 2
or
a = b - 2
So let's solve this:
We start with the equation:
a*10 + b*1 + (b*10 + a*1) = 132
(a*10 + a) + (b*10 + b) = 132
a*11 + b*11 = 132
(a + b)*11 = 132
(a + b) = 132/11 = 12
Then:
a + b = 12
And remember that:
a = b + 2
or
a = b - 2
Then if we select the first one, we get:
a + b = 12
(b + 2) + b = 12
2*b + 2 = 12
2*b = 12 -2 = 10
b = 10/2 = 5
b = 5
then a = b + 2= 5 + 2 = 7
The number is 75.
And if we selected:
a = b - 2, we would get the number 57.
Both are valid solutions because we are changing the order of the digits, so is the same:
75 + 57
than
57 + 75.
Answer:
Answer Choice D
Step-by-step explanation:
Does not pass the vertical line test but the rest of the answer choices do which makes D not a function and your solution.
Can you please make my answer the brainliest I would very much appreciate it. Thanks!
GOING ACROSS
First: 2/4
Second: 5/6
Third: 5/6
Fourth: 2/4
Fifth: other
