(2/3)x^2 -6x + 15 = 0
Using the quadratic formula:
x = [-b +-sq root(b^2 - 4 *a*c)] / 2a
x= [--6 +-sq root(36 -4*(2/3)*15] / 2*(2/3)
x= [6 +-sq root 36 -40] / (4/3)
x1 = 4.5 + (2i / (4/3))
x1 = 4.5 + 1.5i
x2 = 4.5 - (2i / (4/3))
x2 = 4.5 - 1.5i
Answer:
So simple question ask some complicated na yrr
Answer:
x = 10
MB = 16
CM = 16
CB = 32
Step-by-step explanation:
Since it's the mispoint
2x - 4 = x + 6
Subtract x from both sides
x - 4 = 6
Add 4 to both sides
x = 10
MB = 2x - 4
MB = 20 - 4
MB = 16
CM = x + 6
CM = 16
CB = 16 + 16
CB = 32
(X-2)(x^2-6x+2) so yes, (x-2) is in there
