SOLUTION: please solve this for me: xy2(squared)+xyz-z2(square) x=36, y=1/6 and z=9

Question 32176: please solve this for me: xy2(squared)+xyz-z2(square)
x=36, y=1/6 and z=9
Answer by ikdeep(226) (Show Source): You can put this solution on YOUR website!
xy2(squared)+xyz-z2(square)
put the respective values of x ,y and z in the equation,,,, and we get
= (36*1/6)^2 + (36*1/6*9) - (9)^2....here ^ denotes power sign
on solving brackets we get,,,,,,,
= (6)^2 + 54 - 81
= 36 + 54 - 81
= 90 - 81
= 9 ...........
Hope this will help you
Please feel free to revert back for any further queries.

RELATED QUESTIONS
Solve xyz=x + y + z for... (answered by stanbon)

Please someone help me with these. I am not sure I am doing them right. Solve: 1. y... (answered by stanbon,curioustoknow)

Please help me solve this question If arg (z conjugate) = arg (z2), then A) z2=kz1 (answered by ikleyn)

Please help me to solve this problem : if x, y, z are unequal positive qualities, prove... (answered by Alan3354)

how do you solve the equation by completing the square. z2+5z-6=0 (z is... (answered by ewatrrr)

Please help me with this question, Z1=e^iπ/4 and Z2=e^-iπ/3. Find the arg/z/ of z1*z2... (answered by ikleyn)

Solve for z: z2-8z=9 (answered by Cromlix,stanbon)

I am confused on how to solve this problem as it is squaring and them cubing. thxs... (answered by rfer)

Hi, I have my homework already done but I really need to get a good grade [100%] on this... (answered by scott8148)