Answer:
x=-18
Step-by-step explanation:
multiply the invisible 1 outside the parenthesis with the numbers inside. next combine like terms on the left side of the equation. then simplify the equation. then divide -2x by -36. then multiply -1 to your answer and the outcome gives -18
Answer:
70 and 110
Step-by-step explanation:
First angle -- 2x
Second angle -- 3x + 5
Therefore,
2x + 3x + 5 = 180
5x = 175
x = 35
Therefore, 2x -- 70
3x + 5 -------------110
Thenks and mark me brainliest :)
I am sorry but this cannot be put as a mixed number. The answer is 11. To dot his do 11/3 divided by 2/6 u would first do same change flip and do 11/3 times 6/2 which then u would multiply straight across and u would get 66/6 which equals 11 because 66 divided by 6 equals 11. I hope this helped u! ;)
Answer: (0,3)
Step-by-step explanation: To find the x-value of the vertex for any quadratic, you can use the formula -b/2a. In this case, there is no clear b, but you can think of it like y=-x^2 +0x+3 so in the form ax^2 + bx +c, b=0. Therefore the x-value is 0/ -2, or 0. So we know the vertex is (0, y). To find y, just plug 0 into the original equation for x, getting you y=0+3, so y=3, so the vertex is (0,3)
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
