z divided by 3
(1/x) - (1/y) = (1/z) Get the left-hand side over a common denominator:- (y-x)/xy = 1/z Take the reciprocal of both sides:- z = xy / (y-x)
z / (1/7) = 7z.
4x cubed y cubed z divided by x negative squared y negative 1 z sqaured = 4
The average is the sum of variables divided by their count. For an instance:x = 6y = 102z = 9The average will be( 6 + 102 + 9 ) / 3 = 117 / 3 = 39
z divided by 3
6/z
z = 3. If you mean (4*z + 6 )/3 then it is 6. (4*3 + 6) /3 = (12 + 6) /3 = 18/3 = 6 If you meant 4*z + (6/3) then it is 14. 4*3 + (6/3) = 12 + 2 = 14.
That depends entirely on what the value of z is.
Yes. If x is not divisible by 3 then x leaves a remainder of 1 or 2 when it is divided by 3. That is, x is of the form 3y+z where z = 1 or 2. Then x2 = (3y+z)2 = 9y2 + 6yz + z2 = 3(3y2 + 2yz) + z2 The first part of this expression is clearly a multiple of 3, but z2 is not. Whether z = 1 or 2, z2 leaves a remainder of 1 when divided by 3.
z = 77
x+y+z/3 (x plus y plus z divided by 3)
6+z = 3 z = 3-6 z = -3
84/z
Z = (x minus mu) divided by sigma.
Z=153.
(1/x) - (1/y) = (1/z) Get the left-hand side over a common denominator:- (y-x)/xy = 1/z Take the reciprocal of both sides:- z = xy / (y-x)