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
Which is it
2x times y times 3 = 6xy
or,
2xxyx3= 6x^3y
HD = 10.5
Step-by-step explanation:
Given BH = 3, GH = 2, BF = 10
Step 1: To find HF:
HF = BF – BH
HF = 10 – 3
HF = 7
Step 2: To find HD:
We know that if two chords intersects inside a circle, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.
⇒ GH × HD = BH × HF
⇒ 2 × HD = 3 × 7
⇒ HD = 10.5
Hence, the value of HD = 10.5.
2x
so...
2x (4t - 2 - 9)
2x (4t - 11)
Hope this helped!
~Just a girl in love with Shawn Mendes
