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:
The equation of the line passing through the points (-2,0) and (0,-4)
The equation of the line passing through the points (0,-2) and (4,0)
Option B
x minus 2 y = 4 and 2 x + y = negative 4
Answer:
Substitute y=x-2y=x−2 into y=-0.5x+4y=−0.5x+4.
x-2=-0.5x+4x−2=−0.5x+4
2 Solve for xx in x-2=-0.5x+4x−2=−0.5x+4.
x=4x=4
3 Substitute x=4x=4 into y=x-2y=x−2.
y=2y=2
4 Therefore,
\begin{aligned}&x=4\\&y=2\end{aligned}
x=4
y=2
Step-by-step explanation:
The answer is approx. 10.8 per hour.
If you want to round it up, it would be 11. I prefer you not to round though as this is Money. Money should have decimals included.
