Question: "If y > 3, what is the value of n ?"

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.

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

susie has 7 apples