Answer:
a and d
Step-by-step explanation:
brainliest if you want :)
Answer:
- 3x + 1
Step-by-step explanation:
Answer:
Check the explanation
Step-by-step explanation:
(a)Let p be the smallest prime divisor of (n!)^2+1 if p<=n then p|n! Hence p can not divide (n!)^2+1. Hence p>n
(b) (n!)^2=-1 mod p now by format theorem (n!)^(p-1)= 1 mod p ( as p doesn't divide (n!)^2)
Hence (-1)^(p-1)/2= 1 mod p hence [ as p-1/2 is an integer] and hence( p-1)/2 is even number hence p is of the form 4k+1
(C) now let p be the largest prime of the form 4k+1 consider x= (p!)^2+1 . Let q be the smallest prime dividing x . By the previous exercises q> p and q is also of the form 4k+1 hence contradiction. Hence P_1 is infinite
Answer:
0.2146
Step-by-step explanation:
From the picture:
The radius of the circle = r. This means that the area of the circle = πr²
Also For the square, the length of the square = 2r, Therefore the area of the square = Length × length = 2r × 2r = 4r²
The area inside the square but outside the circle = Area of square - Area of circle = 4r² - πr² = r²(4 - π) = 0.8584r²
The ratio of the area inside the square but outside the circle to the area of the square = r²(4 - π) / 4r² = (4 - π) / 4 = 1 - π/4 = 0.2146