Find a polynomial function of degree 3 such that f(10)=17 and the zeros of f are 0, 5 and 8 Please help giving brainliest answer

Step-by-step explanation:

Since f(0) = f(5) = f(8) = 0, we have f(x) = Ax(x - 5)(x - 8), where A is a real constant.

We know that f(10) = 17.

=> A(10)(10 - 5)(10 - 8) = 17

=> A(10)(5)(2) = 17

=> 100A = 17, A = 0.17.

Hence the answer is f(x) = 0.17x(x - 5)(x - 8).

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2