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:
10 numbers
Step-by-step explanation:
a.
In order to find the common ratio, we just need to divide a term by the term that comes before it.
So using the terms 20 and -5, we have:
b.
The recursive rule can be found with the formula:
Where an is the nth term and q is the ratio. So we have:
c.
The explicit rule can be written as:
Where an is the nth term, a1 is the first term and q is the ratio. So:
The correct is the second choice
Answer is in the photo. I can't attach it here, but I uploaded it to a file hosting. link below! Good Luck!
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