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:
6
Step-by-step explanation:
3x+6+5x+10=64
3x+5x+6+10=64
8x+16=64
8x=64-16
8x=48
divide both sides by 8
x=6
radians = degrees x PI/180
radians = 60 x pi/180 = 1.04719 radians
round answer as needed
if you need it in terms of pi it would be PI/3 radians
Answer:
Cos<42 sin<16 :)
Step-by-step explanation: