Find the fifth roots of 243(cos 260° + i sin 260°).

z1=3 cos (52 + i sin 52)

z2 =3 cos (124 + i sin 124)

z3 = 3 cos (196 + i sin 196)

z4 =3 cos (268 + i sin 268)

z5= 3 cos (340 + i sin 340)

Step-by-step explanation:

To find the fifth roots of 243 (cos 260° + i sin 260°).

z ^ 1/5 = r^1/5 ( cis ( theta + 360 *k)/5)  where k=0,1,2,3,4

So the first root of 243 (cos 260° + i sin 260°)  

is z1 =  243^1/5 ( cis ( 260 + 360 *0)/5)  

          3 cis ( 260/5)

        = 3 cis (52)

        = 3 cos (52 + i sin 52)

The second root of  243 (cos 260° + i sin 260°)  

is z2 =  243^1/5 ( cis ( 260 + 360 *1)/5)  

          3 cis ( 620/5)

        = 3 cis (124)

        = 3 cos (124 + i sin 124)

The third root of  243 (cos 260° + i sin 260°)  

is z3 =  243^1/5 ( cis ( 260 + 360 *2)/5)  

          3 cis ( 980/5)

        = 3 cis (196)

        = 3 cos (196 + i sin 196)

The fourth root of  243 (cos 260° + i sin 260°)  

is z4 =  243^1/5 ( cis ( 260 + 360 *3)/5)  

          3 cis ( 1340/5)

        = 3 cis (268)

        = 3 cos (268 + i sin 268)

The fifth root of  243 (cos 260° + i sin 260°)  

is z5 =  243^1/5 ( cis ( 260 + 360 *4)/5)  

          3 cis ( 1700/5)

        = 3 cis (340)

        = 3 cos (340 + i sin 340)

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50