Raymond B. answered 09/04/21
Math, microeconomics or criminal justice
1 h^3 = 8
h=2 is one solution, but with a 3rd degree equation, there are 3 solutions. the other two are imaginary.
factor h^3-8= (h-2)(h^2 +2h +4) =0 and set each factor =0
h-2=0, h=2
h^2 +2h+4 = 0
complete the square
h^2 +2h +1 =-4+1 =-3
(h+1)^2 = -3
h+1 = sqr(-3)
h= -1 + or - isqr3
2 x- 1/5 = 6
x= 6 + 1/5 = 31/5
x= 31/5 = 6.2
3 5-z = 1/3
-z = 1/3 -5
z = 5-1/3 = 4 2/3 or 4.666...
4 1/2 k + 4 = 12
k/2 = 12-4 = 8
k = 16
5 5/t = 7
t= 5/7
6 b/3a = 5
b=15a
or 3a =5/b
a =5/3b
7 c^4 = 16 -2b
c= 4th root of (16-2b) = (16-2b)^(1/4)
or 2b = 16-c^4
b = 8 -c^4/2
8 4+2x = xy
y = 4/x +2
or
xy-2x =-4
x(y-2) =-4
x= -4/(y-2) = 4/(2-y)
9 5 = x(x+3)
5=x^2 +3x
x^2 +3x -5 = 0
x=-3/2 + or - (1/2)sqr(9-4(-5))
= -3/2 + or - (1/2)sqr29
10 x-1 = 2y
y= (x-1)/2 = x/2 -1/2
or x = 2y+1