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.
Y= |x+1|-2
———— is the answer hope that helps
x
Answer:
i can only say that you should use this forumla, the sins law
sin(a)/A = sin(b)/B = sin(c)/C
Step-by-step explanation:
Answer:
3 units up
Step-by-step explanation:
when the y intercept goes u, so does the line
I don’t see the answers on the screen
