Lol... I don't know if this is a joke, but it's basically just restating the precious problem, but switching numbers. So the answer would be 2,592.
Answer:
see explanation
Step-by-step explanation:
Using the sum/ difference → product formula
cos x - cos y = - 2sin(
)sin (
)
sin x - sin y = 2cos (
)sin (
)
Given
(cosA - cosB)² + (sinA - sinB )²
= [ - 2sin(
)sin(
) ]² + [ 2cos(
)sin(
) ]²
= 4sin² (
)sin² (
) + 4cos² (
)sin² (
)
= 4sin² (
)[ sin² (
) + cos² (
) ← sin²x + cos²x = 1
= 4sin² (
) × 1
= 4sin² (
) = right side ⇒ proven
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
Subtract negative = add
88 + 35 = 123
The solution is 123
Answer:
Third option
Step-by-step explanation:
Please help me by marking me brainliest. I'm one away :)