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:
k
2
+
k
+
1
Step By Step:
Cancel the common factor.
Then.
k
2
+
k
+1
by 1
Gets you k2+k+1
Hello,
Answer B.
Using Thales, 6/9=B/8==>B=48/9=5+1/3.
9x+14x+4x+90 = 360
27x = 270
x = 10
<F = 14x = 14(10) = 140
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