The square of a prime number is not prime.
a) let x ∈ R, If x ∈ {prime numbers}, then ∉{prime numbers}
there says that if x is a real and x is in the set of the prime numbers, then the square of x isn't in the set of prime numbers.
b) Prove or disprove the statement.
ok, if x is a prime number, then x only can be divided by himself. Now is easy to see that = x*x can be divided by himself and x, then x*x is not a prime number, because can be divided by another number different than himself
Answer:
see explanation
Step-by-step explanation:
The y- intercept is at (0, 6) ⇒ c = 6
The x- coordinate of the vertex is at the midpoint of the zeros, that is
m = =
= - 2
Since the zeros are x = - 3 and x = - 1 then the factors are (x + 3), (x + 1) and
f(x) = a(x + 3)(x + 1) = a(x² + 4x + 3)
Comparing with f(x) = ax² + 8x + 6 ⇒ a = 2
Thus
f(x) = 2x² + 8x + 6
Substitute m = - 2 into the equation for value of n
n = 2(- 2)² + 8(- 2) + 6 = 8 - 16 + 6 = - 2
Hence
c = 6, m = - 2, a = 2, n = - 2