The answers to the question have been written below
<h3>The changes to a polygon</h3>
1. If the number of sides of the polygon should increase then the measure of the interior angle would increase by 180.
2. A regular polygon has the sum of its external angles as 360. This is solved as 360/n = 0
Hence the value of the exterior would move towards 0 as its sides increases.
4. As the total sum of the of the polygon changes, it would tend towards infinity.
5. The total sum of the exterior angles as the number of sides changes would tends towards 360 degrees.
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Answer:
it is more intense by 3.0
Step-by-step explanation:
Answer:
-3/8= -6/16=-9/24=-12/32=-15/40....
3/8=6/16=9/24=12/32=15/40=18/48=21/56.....
Answer:
x = - 4 y = - 4
Step-by-step explanation:
x+y= - 8
-9x-6y=60
First, solve for x in the first equation:
x+y = - 8 Subtract y from both sides
x + y - y = -8 - y y cancels on the left
x = - 8 - y
Now plug in what you found for x into the 2nd equation and solve for y.
- 9x - 6y = 60
-9(- 8 - y) - 6y = 60 Multiply out
72 + 9y - 6y = 60
72 + 3y = 60 Subtract 72 from both sides
72 - 72 + 3y = 60 - 72 72 cancels on the left
3y = - 12 Divie both sides by 3
3y/3 = -12/3 3 cancels on the left because 3/3 = 1
y = -4
Now plug your answer for y back into the first equation to get x.
x + y = -8
x + (-4) = - 8 Add 4 to each side
x - 4 + 4 = - 8 + 4 4 cancels on the left
x = -4
x = - 4 and y = - 4
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
