Answer:
9.11*10^-4 %
Step-by-step explanation:
To find the probability, you simply need to find the possible outcomes that allows no rooks to be in danger, and the possible amount of ways to place the rooks.
For the first outcome, you start by putting 1 rook in the first columns, you have 8 possible rows to do this. The next rook in the next column will only have 7 possible rows, as you have to exclude the one where the previous rook is located. The next rook, 6 possibilities, the next 5, and so on. So we conclude that the total amount of ways so that none of the rooks can capture any of the other rooks is 8*7*6*5*4*3*2*1 = 8! = 40320
In order to find the total amount of ways to place the rooks, you can just use a combinatoric:
![\left[\begin{array}{ccc}64\\8\end{array}\right]= \frac{64!}{8!(64-8)!} = 4.43*10^9](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D64%5C%5C8%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cfrac%7B64%21%7D%7B8%21%2864-8%29%21%7D%20%3D%204.43%2A10%5E9)
Then:
P = 
Since it is a parallelogram cente at point T,
then the measure of PT is equal to the measure of TR. And the measure of QT is
equal to the measure of TS.
PT = TR
a + 4 = 2a
4 = 2a -a
a = 4
PT = TR = 8 units
QT = TS
b = 2b -3
3 = 2b – b
b = 3
<span>QT = TS = 3 units</span>
Answer:
1
Step-by-step explanation:
First, convert all the secants and cosecants to cosine and sine, respectively. Recall that 
and 
.
Thus:
Let's do the first part first: (Recall how to divide fractions)
For the second term:
So, all together: (same denominator; combine terms)
Note the numerator; it can be derived from the Pythagorean Identity:
Thus, we can substitute the numerator:
Everything simplifies to 1.
Answer:
Find the question mark using the model I've provided and that'll be the answer!
Answer:
Yes its correct the unhighlighted is complement
