For both questions can someone please fill in the missing numbers in the tables thannks!

Mathematics

1 answer:

Serggg [28]2 years ago

8 0

Step-by-step explanation:

a.
if x=-1 then,
y=2*(-1)-1
y=-2-1
y=-3

if y=-1 then,
-1=2x-1
-2x=-1+1
-2x=0
x=0

if x=1 then,
y=2*1-1
y=2-1
y=1

if y=3 then,
3=2x-1
-2x=-1-3
-2x=-4
-2/-2x=-4/-2
x=2

b.
if x=0 then,
0+y=-2
y=-2

if x=1 then,
y+1=-2
y=-1-1
y=-3

if x=2 then,
y+2=-2
y=-2-2
y=-4

if x=3 then,
y+3=-2
y=-2-3
y=-5

You might be interested in

Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r

erastova [34]

Answer:

The probability is $\frac{56!}{64!}$

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, $64 \choose 8 .$

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 $f : A \rightarrow A ,$ 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

$\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}$