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:
a, ax3 =b, bx3= c , cx3=d, dx3=e
Step-by-step explanation:
2, 2×3=6, 6x3=18,18 x3 =54, 54x3= 162
Multiplication of three with previous number will give you next number in series.
Answer:
I think the answer would be the 9 pack.
Step-by-step explanation:
Think, if you divide each of the packs by their price the 9 pack is cheaper for one roll.
