Answer:
5040,56
Step-by-step explanation:
We have to construct pass words of 4 digits
a) None of the digits can be repeated
We have total digits as 0 to 9.
4 digits can be selected form these 10 in 10P4 ways (since order matters in numbers)
No of passwords = 10P4
= 
b) start with 5 and end in even digit
Here we restrain the choices by putting conditions
I digit is compulsorily 5 and hence only one way
Last digit can be any one of 0,2,4,6,8 hence 5 ways
Once first and last selected remaining 2 digits can be selected from remaining 8 digits in 8P2 ways (order counts here)
=56
Answer: Abby- Marvel
Step-by-step explanation:
Just way better and deeper plotline in my personal opinion.
Sin(θ - 180)
sin(θ)cos(180) - cos(θ)sin(180)
sin(θ)[-1] - cos(θ)[0]
-sin(θ) - 0
-sin(θ)
The next larger hundred is 5,100 .
The next smaller hundred is 4,900 .
5,000 is exactly midway between 4,900 and 5,100.
Neither of them is closer to 5,000 than the other one is.
So 5,000 is already rounded to the nearest hundred.
The LCM is (x+4)(x-3)(x+5)