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.
21
E (225-25n)
n=1
So its D.
It would be 40•.25=10 which means 40-10=30 so it’d be .25 and you’d move the decimal over 2 spots so it’d be 25% off
Answer: m=6, m=-4
Step-by-step explanation:
To solve this proportion, we have to cross multiply.
Now that we have cross multiplied, we actually need to FOIL the left side to expand the equation.
Combine like terms.
We can move all terms to one side and then solve for m.
We can actually factor this to:
We set each factor equal to 0 to find m.
m-6=0
m=6
m+4=0
m=-4
Answer:
what is r90 and I will be able to answer
Step-by-step explanation:
