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.
180.6 is the answer 240 4/5 is 240.8
240.8/4 is 60.2
60.2 x 3 is your answer
Considering the place value of the digit 9, in the first number, the decimal value of 9 is thousands. In the second number, the decimal value of 9 is hundredths. The conclusion we are drawing is: The second decimal is 10 times larger than the first one
(2x+3y)+(7x-3y)
First, group like terms together, so (2x+7x) + (3y-3y). Then simplify within the parentheses. 2x+7x = 9x and 3y-3y = 0, so you have 9x, which cannot be simplified any further. The answer is 9x.
