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:
-3x
Step-by-step explanation:
Your aanswer is 13.9 because the 13 stays the same and in decimals .1 is a tenth so there for .9 is your tenth 13.9
Answer:
50 Persent
Step-by-step explanation:
The unit rate is -1/3.
In direct variation, y varies directly to x by a constant ratio k.
Its equation looks like this: y = kx ; where k stands for the constant ratio.
In the above equation. x = -3y, it should be converted to the proper equation. both sides must be transferred and its accompanying sign must be changed.
3y = -x
3y/3 = -x/3
y = (-1/3)x
where: y is the total value; -1/3 is the unit rate; and x is the number of units
