Answer:
y-3
Problem:
What is the remainder when the dividend is xy-3, the divisor is y, and the quotient is x-1. ?
Step-by-step explanation:
Dividend=quotient×divisor+remainder
So we have
xy-3=(x-1)×(y)+remainder
xy-3=(xy-y)+remainder *distributive property
Now we just need to figure out what polynomial goes in for the remainder so this will be a true identity.
We need to get rid of minus y so we need plus y in the remainder.
We also need minus 3 in the remainder.
So the remainder is y-3.
Let's try it out:
xy-3=(xy-y)+remainder
xy-3=(xy-y)+(y-3)
xy-3=xy-3 is what we wanted so we are done here.
Answer:
2/3 per brother
Step-by-step explanation:
2/3 plus 2/3 plus 2/3 is 6
(7 + 5i) - ( -9 + i)
= 7 + 5i - (-9) - (i)
= 7 + 5i + 9 - i
= 16 + 4i
= 4i + 16
Answer:
it's D
Step-by-step explanation:
You use distribute property, then add all like terms
