Answer:
Step-by-step explanation:
f(x) = 6x³-x²+4x+5
g(x) = 9x³-1
(f+g)(x) = {6x³-x²+4x+5} + {9x³-1}
open brackets
6x³-x²+4x+5 + 9x³-1
choose like terms
6x³+ 9x³-x²+4x+5 -1
15x³-x²+4x+4= x²(15x-1)+4(x+1)= (x²+4)(x+1)(15x-1)
(f+g)(x)= 15x³-x²+4x+4 = (x²+4)(x+1)(15x-1)
(f+g)(2) = 15(2)³-(2)²+4(2)+4 = 15(8)-4+8+4 = 120+8 = 128
(f-g)(x) = {6x³-x²+4x+5} - {9x³-1}
open brackets
6x³-x²+4x+5 - 9x³+1
choose like terms
6x³- 9x³-x²+4x+5 +1
-3x³-x²+4x+6= -x²(3x+1)+2(2x+3)= (-x²+2)(3x+1)(2x+3) = (2-x²)(3x+1)(2x+3)
(f-g)(x) = -3x³-x²+4x+6 = (2-x²)(3x+1)(2x+3)
(f-g)(-3) = -3(-3)³-(-3)²+4(-3)+6 = -3(-3x-3x-3)-(-3x-3)-12+6=-3(-27)-3(9)-12+6= 81-27-12+6 = 54-6= 48
(f-g)(-3) = 48
The answer is D) 11/21
Your welcome
Correct. In the case where you are given enough information to use the law of cosines you could in fact then use the law of sines afterwards to find your remaining angle. That being said beware of solutions that don't make a feasible triangle (if you were using the law of cosines you only have one angle, so that means whatever your second angle is that you found using the law of sines can't make your sum go over 180, because you still need some angle left for the last angle).
Answer:3rd
Step-by-step explanation:
