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For all real numbers x and y
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04 Nov 2019, 15:32
04 Nov 2019, 15:32
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Question Stats:
94% (00:53) correct
5% (01:48) wrong based on 19 sessions
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For all real numbers x and y, if \(x#y = x(x-y)\), then \(x#(x#y)\) =
A)\(x^2 - xy\)
B) \(x^2 - 2xy\)
C) \(x^3 - x^2 - xy\)
D) \(x^3 - (xy)^2\)
E) \(x^2 - x^3 +x^2y\)
My practice book indicates that E is the correct answer, but I can't quite grasp this one firmly. I see easily enough how to plug in arbitrary x and y values to produce a number to then check against the answer options, but I'm getting stuck after calculating (x#y) . . . I can't seem to fully understand what is truly meant by "x#(x#y)"
Any advice at all will be sincerely appreciated. Can anyone help a passionate social worker and prospective doctoral student (who is committed to social research but hasn't practiced this kind of functions arithmetic in a while) with this problem? You'll have my heartfelt thanks.
A)\(x^2 - xy\)
B) \(x^2 - 2xy\)
C) \(x^3 - x^2 - xy\)
D) \(x^3 - (xy)^2\)
E) \(x^2 - x^3 +x^2y\)
My practice book indicates that E is the correct answer, but I can't quite grasp this one firmly. I see easily enough how to plug in arbitrary x and y values to produce a number to then check against the answer options, but I'm getting stuck after calculating (x#y) . . . I can't seem to fully understand what is truly meant by "x#(x#y)"
Any advice at all will be sincerely appreciated. Can anyone help a passionate social worker and prospective doctoral student (who is committed to social research but hasn't practiced this kind of functions arithmetic in a while) with this problem? You'll have my heartfelt thanks.
Re: Tough function problem - request for assistance
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04 Nov 2019, 16:52
04 Nov 2019, 16:52
Expert Reply
\(x#y=x(x-y)\)
Basically the x on the LHS is the x on the RHS and the y is (x-y) alike.
so when you do have x#(x#y) means that the \((x#y)= x(x-y)=x^2-xy\)
Then \(x#(x#y) = x x^2-xy=x^3-x^2y\) this should be the answer
It is just a matter of substitution. The answer and the stem are transcribed correctly ??
Basically the x on the LHS is the x on the RHS and the y is (x-y) alike.
so when you do have x#(x#y) means that the \((x#y)= x(x-y)=x^2-xy\)
Then \(x#(x#y) = x x^2-xy=x^3-x^2y\) this should be the answer
It is just a matter of substitution. The answer and the stem are transcribed correctly ??
Re: Tough function problem - request for assistance
[#permalink]
27 Apr 2020, 19:24
27 Apr 2020, 19:24
1
Carcass wrote:
\(x#y=x(x-y)\)
Basically the x on the LHS is the x on the RHS and the y is (x-y) alike.
so when you do have x#(x#y) means that the \((x#y)= x(x-y)=x^2-xy\)
Then \(x#(x#y) = x x^2-xy=x^3-x^2y\) this should be the answer
It is just a matter of substitution. The answer and the stem are transcribed correctly ??
Basically the x on the LHS is the x on the RHS and the y is (x-y) alike.
so when you do have x#(x#y) means that the \((x#y)= x(x-y)=x^2-xy\)
Then \(x#(x#y) = x x^2-xy=x^3-x^2y\) this should be the answer
It is just a matter of substitution. The answer and the stem are transcribed correctly ??
We already solved x#y = x^2-xy
then x#(x#y) becomes x # (x^2-xy) here consider x^2-xy = y
then x#(x^2-xy) becomes x^2-x(x^2-xy) = x^2-x^3+x^2y
so the answer is E.
