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:
Step-by-step explanation:
As the mean is 20.2 and the standard deviation is 2.4
The range within the first standard deviation is
(17.8,22.6)
This means that we can safely use the range of (18,22) as we cannot confirm whether the other values fall within the range.
Within this range there are
We can substitute<span> y in the second </span>equation<span> with the first </span>equation<span> since y = y. The solution of the </span>linear<span> system is (1, 6). You can use the </span>substitution method<span> even if both </span>equations<span> of the </span>linear<span> system are in standard form. Just begin by solving one of the </span>equations<span> for one of its variables.</span>
