∠ACB is an inscribed angle, so
m∠ACB= (1/2)mAB =(1/2)*50=25⁰
m∠ACB= 25⁰
-4(2y)+11y=15, -8y+11y=3y=15: y=5
X=2(5)=10
To check your work -4(10)+11(5)=15
15=15
Answer:
The probability is 
Step-by-step explanation:
We can divide the amount of favourable cases by the total amount of cases.
The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8,
For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function
with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.
Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.
We can conclude that the probability for 8 rooks not being able to capture themselves is

Answer:
D
Step-by-step explanation:
i think it's correct if not I'm sorry