Deena has included the discount on the wrong side of the equation.
simplified is 
I got this by dividing both the numerator and denominator by 4
Hope this helps
-AaronWiseIsBae
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:
5 + 3/4x = -19
Step-by-step explanation:
more than means adding
Answer:
The first five terms in Pattern B are 3, 7, 11, 15, 19.
Step-by-step explanation:
