Answer:CA, BC, BA
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: YOU MUST USE PEMDAS
Step-by-step explanation:
pls give me brainliest im almost lvled up
Answer:
A. square root of 3 times 5
Step-by-step explanation:
The midsegment theorem says that the midsegment is half the third side and parallel as well. 20*2 is 40 so x = 35
