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:
PQ = 16.04cm
Step-by-step explanation:
θ = 76°
QR = 4cm
PQ = ?
Since it's a right angle triangle,
QR = adjacent
PQ = opposite
Tanθ = opposite/adjacent
Tan76 = pq / 4
PQ = 4tan76
PQ = 4 * 4.010
PQ = 16.04cm
3x - y = 0 ⇒ 2(3x - y = 0 ) ⇒ 6x - 2y = 0
5x + 2y = 22 ⇒ 1(5x + 2y = 22) ⇒ <u>5x + 2y</u> = <u>22 </u>
11x = 22
x = 2
3x - y = 0 ⇒ 3(2) - y = 0 ⇒ 6 - y = 0 ⇒ 6 = y
Answer: x=2, y=6
Answer: 
Step-by-step explanation:
A perfect square trinomial is written as 
, where
first term 
= square of first term of binomial
second term=
=twice the product of both terms of binomial.
and third term 'c'=square of last term of binomial
Thus to create a perfect square trinomial put 'a' and 'c' a square number
Let a=4 and c=9
The required trinomial will be
![=(2x)^2+2(2x)(3)+3^2\\=(2x+3)^2.......\text{[using pattern}(a+b)^2=a^2+2ab+b^2]\\=(2x+3)(2x+3)](https://tex.z-dn.net/?f=%3D%282x%29%5E2%2B2%282x%29%283%29%2B3%5E2%5C%5C%3D%282x%2B3%29%5E2.......%5Ctext%7B%5Busing%20pattern%7D%28a%2Bb%29%5E2%3Da%5E2%2B2ab%2Bb%5E2%5D%5C%5C%3D%282x%2B3%29%282x%2B3%29)
Step-by-step explanation:
I think this might help
