Answer:
SUM OF ALL ANGLES OF A TRIANGLE =180
THEREFORE,
X+8+2X-3+6X-5=180
9X=180
X=20.
THEREFORE,<J=X+8=20+8=28
<K=6X-5=6*20-5=120-5=115
<L=2X-3=2*20-3=40-3=37.
Strictly speaking, x^2 + 2x + 4 doesn't have solutions; if you want solutions, you must equate <span>x^2 + 2x + 4 to zero:
</span>x^2 + 2x + 4= 0. "Completing the square" seems to be the easiest way to go here:
rewrite x^2 + 2x + 4 as x^2 + 2x + 1^2 - 1^2 = -4, or
(x+1)^2 = -3
or x+1 =i*(plus or minus sqrt(3))
or x = -1 plus or minus i*sqrt(3)
This problem, like any other quadratic equation, has two roots. Note that the fourth possible answer constitutes one part of the two part solution found above.
(2k² + 5k - 6)(3k - 1)
(6k³ - 2k² + 15k² - 5k - 18k + 6)
6k² + 13k² - 23k + 6
Use the FOIL method to simplify the problem.
(FOIL stands for first outer, inner, last)
