Answer:
7 to the right
Step-by-step explanation:
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.
y = 0.6x - 3.2
The general form of the desired equation is
y = mx + b
where
m = slope of the line
b = y intercept of the line
If two lines are parallel, their slopes will be the same, Since the slope of the
given line "y = 0.6x +3" is 0.6, that will also be the slope of the desired line.
So our equation becomes:
y = 0.6x + b
Now we can substitute the x and y value of the desired point we want the new line to pass through and find b. So
y = 0.6x + b
-5 = 0.6(-3) + b
-5 = -1.8 + b
-3.2 = b
So the desired equation is now
y = 0.6x - 3.2
