quick math help

Question

1.Evaluate 6s^0 when S = -12

2.simplify (a^5b)^7 (a^8b^5)

3. simplify with no negatives 12y^-2 / 4y^-8

4. use radical notation to rewrite the expression then simplify 36 1/2

5.add and simplify (6-2i)+(8-8i)

6. preform the indicated operation and simplify 9i *5i

Answer 1

1.  s⁰ = 1 for all values except s=0 (which is undefined).  Since s=-12, 6*s⁰ = 6*(-12)⁰ = 6*1 = 6

 

2.  (a^5b)⁷(a^8b)⁵ = a^35b*a^40b = a^75b        [I'm guessing you meant (a^8b)⁵ vice a^{8b^5}]

 

3.  12y^-2 / 4y^-8 = 12y⁸/4y² = 3y⁶

 

4.  36^1/2 = √36 = ±6                                [Guessing again you meant 36^1/2]

 

5.  (6-2i)+(8-8i) = (6+8)+(-2i-8i) = 14 - 10i

 

6.  9i*5i = 45i² = 45*(-1) = -45