The rational root theorem states that the rational roots of a polynomial can only be in the form p/q, where p divides the constant term, and q divides the leading term.
In your case, both the leading term 5 and the constant term 11 are primes, so their only divisors are 1 and themselves.
So, the only feasible solutions are

For the record, in this case, none of the feasible solutions are actually a root of the polynomial.
Answer:
sin= opposite leg/hypotenuse
Sin(a)=12/37
Step-by-step explanation:
Answer:
not sure if they really want numbers here
x and 14x??
Step-by-step explanation:
I'll just name the lines as A, B, C, D, E.
A : Corresponding angles
B : Alternate Interior angles
C : Co - Interior angles
D : Vertical angles
E : Alternate Exterior angles
The answer to this question is 0