Answer:
Total derangement = 4
Step-by-step explanation:
To find - How many derangement of {1, 2, 3, 4, 5, 6} begin with the integers 1, 2, and 3, in some order?
Solution -
Given that,
The first three numbers are 1, 2, 3
The only derangement possible are - 231 and 312
Now,
The remaining elements 4, 5, 6 are the 4th, 5th and 6th number in the derangement
Now,
In order for the permutation of the derangement, the only possibility for 123 are 564 and 645
∴ we get
The Total derangement are -
231564
231645
312564
312645
So,
Total derangement = 4
cot(<em>θ</em>) = cos(<em>θ</em>)/sin(<em>θ</em>)
So if both cot(<em>θ</em>) and cos(<em>θ</em>) are negative, that means sin(<em>θ</em>) must be positive.
Recall that
cot²(<em>θ</em>) + 1 = csc²(<em>θ</em>) = 1/sin²(<em>θ</em>)
so that
sin²(<em>θ</em>) = 1/(cot²(<em>θ</em>) + 1)
sin(<em>θ</em>) = 1 / √(cot²(<em>θ</em>) + 1)
Plug in cot(<em>θ</em>) = -2 and solve for sin(<em>θ</em>) :
sin(<em>θ</em>) = 1 / √((-2)² + 1)
sin(<em>θ</em>) = 1/√(5)
Answer:
a. 2x + 4
Step-by-step explanation:
Four more than twice a number
Let x = unknown number
Four more than twice of x
Twice of x can be written as 2x
Four more than 2x
Four more can be written as + 4
We get 2x + 4
X^2 + y^2 = 11^2 or x^2 + y^2 = 121 is the equation.
