Let p be an odd prime and let a be an integer with p - a.
(a) prove that a(p−1)/2 is congruent to either 1 or −1 modulo p.
(b) prove that a(p−1)/2 is congruent to 1 modulo p if and only if a is a quadratic residue modulo p. (hint. let g be a primitive root for p and usethefact,3.60,thatgm is a quadratic residue if and only if m is even.)
(c) prove that a(p−1)/2 ≡¡a p¢(mod p). (this holds even if p|a.)
(d) use (c) to prove theorem 3.61(a), that is, prove that

don't know what that says sorry i can't you

step-by-step explanation:

idk what it says

answer: .64

step-by-step explanation

set up the formula for standard deviation for this set of numbers, then simplify

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.

wait this depends wether emily and zach how they are thinking.

yes they are the same.