Z=153.
2nd z=17 1st Z=22 Easy as ABC: Just do the following 39-5=___ ___divide by 2=z z+5=Z
(5z - 3)(z - 3)
Without an equality sign and not knowing the plus or minus value of 81 it can't be solved. If you mean: z+81 = 9z-7 Then: z-9z = -7-81 => -8z = -88 By dividing both sides by -8: z = 11
Yes, if [x-(-y)=z] then it can be turned into [x+y=z] because the two negatives cancel each other out, causing you to be left with the opposite, a positive of plus sign.
It can't be factored because its discriminant is less than zero.
(z-9)(z+3
Using the quadratic formula you get z≅4.91547594742265 or z≅-0.91547594742265
6xyz(3x + 2y + z)
(c - z)(c^2 + cz + z^2)
The only common factor to all terms is yz. → xy³z² + y²z + xyz = yz(xy²z + y + x)
z^2
ax + ay - AZ = a(x + y - z)
4z squared (z+4) (z+4)
If that were z^2 - 16z + 64, it would factor very neatly to (z - 8)(z - 8) or (z - 8)^2 As it is, you could use z(z^2 - 16) + 64 or the much more cumbersome -1/27 (3 z+2 2^(2/3) (3 (9-sqrt(69)))^(1/3)+4 3^(2/3) (2/(9-sqrt(69)))^(1/3)) (-9 z^2+(12 3^(2/3) (2/(9-sqrt(69)))^(1/3)+6 2^(2/3) (3 (9-sqrt(69)))^(1/3)) z-8 2^(1/3) (3 (9-sqrt(69)))^(2/3)-48 3^(1/3) (2/(9-sqrt(69)))^(2/3)+48)
x squared + y squared = z squared.
(z=-3) and (z=7)