According to the real rational root theorem, what are all the potential rational roots of f(x)=5x3-7x+11

x = -11, -11/5, -1, -1/5, 1/5, 1, 11/5,11

Step-by-step explanation:

The general formula for a third-degree polynomial is

f(x) = ax³ + bx² + cx + d

Your polynomial is  

f(x) = 5x³ + 7x + 11

a = 5; d = 11

p/q = Factors of d/Factors of a

Factors of d = ±1, ±11

Factors of a = ±1, ±5

Potential roots are x = ±1/1, ±1/5, ±11/1, ±11/5

Putting them in order, we get the potential roots

x = -11, -11/5, -1, -1/5, 1/5, 1, 11/5, 11

(There are no rational roots. There is one irrational root and two imaginary roots.)

the answer would be 0.54

step-by-step explanation:

tangent z: 4/3

cosine y: 4/5

cotangent y: 4/3