Write the additive inverse and multiplicative inverse of 3-7i?

Step-by-step explanation:

additive inverse means multiplying 3-7i by it's conjugate which would be

(3-7i)(3+7i)=3^2+7^2=9+49=58

the multiplicative inverse of a number z is 1/z

1/(3-7i)

multiply numerator and denominator by 3-7i's conjugate and it equals 58

(3-7i)/58

3/58-7/58i

Hope that helps :)

y > 1/3x

y < = -x + 4

hope this : )

a and c

step-by-step explanation: