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:
x=7
Step-by-step explanation:
<u>If P is the midpoint of XY, then XP = PY:</u>
- 8x - 2 = 12x - 30
- 12x -8x = 30 -2
- 4x = 28
- x= 28/4
- x= 7
$139.23 you take the 239+34.70 then take 10% off
Volume is L×W×H so when you plug in your variables you get 4×4×4 or 60
