what is the maximum value of v^x-yz
if the value of v,x,y,z have to be choosen from the set A where A(-3,-2,-1,0,1,2,3)

• 15
  (-3)^2 - 3(-2)=15
• 9 years agoHelpfull: Yes(52) No(5)
• the options are a) 9 b) 12 c) 15 d) none of these
• 9 years agoHelpfull: Yes(7) No(0)
• the question posted is ambiguous.can b interpreted in multiple angles.so the ans is varying
• 9 years agoHelpfull: Yes(6) No(4)
• Actual question is :
  (yz is not in power of x)
  What is the maximum value of (v^x) - yz. If the value of v,x,y,z have to be chosen from the set A where A(-3,-2,-1,0,1,2,3)
  a) 9
  b) 12
  c) 15
  d) none of these
  Explanation is
  To maximize the value of (v^x) - yz, we make yz negative and v^x as maximum as possible using given value.
  (v^x)−yz=(−3)2−(−3×2) = 15
  
• 8 years agoHelpfull: Yes(5) No(2)
• (-3)^2-(-2)*3=15
• 9 years agoHelpfull: Yes(4) No(0)
• 36 (if values can be repeated)
• 9 years agoHelpfull: Yes(3) No(8)
• (-3)^2-(-3*3)=18
• 9 years agoHelpfull: Yes(2) No(4)
• if we take v=3,x=2,y=-3,z=1
  then we get 3^5=243 (MAX)
  
• 9 years agoHelpfull: Yes(2) No(7)
• for question v^(x-yz)
  
  -3^( 2 - 3 * -2 )
  => -3^8
  => 6561
• 9 years agoHelpfull: Yes(2) No(3)
• let take v=3...x=1,y=-3,z=2...now solving the above equatn result is 2187 which is max. den any value taken in otherway
• 9 years agoHelpfull: Yes(1) No(2)
• are is the right answer
  
• 9 years agoHelpfull: Yes(1) No(0)
• v^x-yz
  for max x-yz>0
  x>yz
  max(x)=6,min(y)=0
  max(x-yz)=6
  v^6=?
  if max(v)=3
  then max(3^6)=729
• 9 years agoHelpfull: Yes(0) No(2)
• v=3, x=2, y=-3,z=-1
• 9 years agoHelpfull: Yes(0) No(2)
• (-3)^3-2(-2)=27+4=31
• 9 years agoHelpfull: Yes(0) No(7)
• Put x = 2,v = 3 then
  v^x = 9
  Now put y = -3 , z = 3
  then yz= -9
  thus,we have
  v^x - yz = 9 - (-9) = 18.
  
• 8 years agoHelpfull: Yes(0) No(0)
• 3^3-0*1=27
  
• 8 years agoHelpfull: Yes(0) No(0)
• (-3)^2-3(-3)=9+9=18
• 8 years agoHelpfull: Yes(0) No(0)