Answered

Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

For all real numbers  x  and  y, if  x # y = x(x-y), then x # (x # y) =

My question is, how does x # (x # y) = 8 and how does it also equal to x(squared) - x(cubed) + x(squared) y?  

Is there a rule of the function that I have forgotten?  

Thanks

Sagot :

crisforp
if x # y = x(x-y) => x # (x # y) = x # [x(x-y)] =x[ x - x(x-y)] = x( x - x^2 +x*y ) = x^2 - x^3 + ( x^2 ) * y;
[tex]x\#y=x(x-y)\\\\x\#(x\#y)\\\\x\#y=x(x-y)=x^2-xy\\\\x\#(x\#y)=x\#(x^2-xy)=x[x-(x^2-xy)]=x(x-x^2+xy)\\\\=x^2-x^3+x^2y[/tex]


[tex]x^2-x^3+x^2y=8\\\\x^2(1-x+y)=8\\\\x^2(1-x+y)=2^2\cdot4\iff x^2=2^2\ and\ 1-x+y=4\\\\x=2\ and\ y=4-1+x\\\\x=2\ and\ y=3+2\\\\x=2\ and\ y=5[/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.