Answer:
The answer is c i think
Step-by-step explanation:
they arent congruent pairs and they all have angles
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.
9514 1404 393
Answer:
(x, y) = (-11, 4)
Step-by-step explanation:
You can subtract 3 times the second equation from the first to eliminate y.
(7x +9y) -3(x +3y) = (-41) -3(1)
4x = -44
x = -11
Substituting into the second equation gives ...
-11 +3y = 1
3y = 12
y = 4
The solution is (x, y) = (-11, 4).
Answer:
total: 3x + 7
Step-by-step explanation:
Nancy = x
Bill = 2x + 7
together = 3x + 7
I think it's $11.79 but I'm not entirely sure
