Answer:
The prove is as given below
Step-by-step explanation:
Suppose there are only finitely many primes of the form 4k + 3, say {p1, . . . , pk}. Let P denote their product.
Suppose k is even. Then P ≅ 3^k (mod 4) = 9^k/2 (mod 4) = 1 (mod 4).
ThenP + 2 ≅3 (mod 4), has to have a prime factor of the form 4k + 3. But pₓ≠P + 2 for all 1 ≤ i ≤ k as pₓ| P and pₓ≠2. This is a contradiction.
Suppose k is odd. Then P ≅ 3^k (mod 4) = 9^k/2 (mod 4) = 1 (mod 4).
Then P + 4 ≅3 (mod 4), has to have a prime factor of the form 4k + 3. But pₓ≠P + 4 for all 1 ≤ i ≤ k as pₓ| P and pₓ≠4. This is a contradiction.
So this indicates that there are infinite prime numbers of the form 4k+3.
Answer:
Your answer would be 3/5 or 3:5
Step-by-step explanation:
This is 1/2)2=1/4 hope this helps
Answer:
Sandy should have evaluated (-2)³ as -8
Step-by-step explanation:
(-2)³ (6 - 3) - 5(2 + 3) = -8 · 3 - 5 · 5 = - 24 - 25 = - 49
Because:
(-2) · (-2) · (-2) = 4 · (-2) = - 8
God with you!!!
Answer: 
Step-by-step explanation:
Get rid of parentheses: 
Add all the numbers together, and all the variables: 
Move all terms containing x to the left, all other terms to the right:


