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
Answer:
1.5y - 5
Step-by-step explanation:
I used the distributive approach and distributed -1/2 to the -3y and 10.
(If you try this and it's incorrect, you may just need to distribute -1/2 to the 10, creating -3y - 5.)
Based on the given conditions, formulate:
If the fraction is defined, the denominator cannot equal 0.:
Append the inequality as the domain:
Reduce the fraction:
Multiply both sides of the equation by the common denominator:
Reduce the fractions:
Swap the sides:
Divide both sides of the equation by the coefficient of variable:
Cross out the common factor:
Find the intersection:
get the result:
Answer: x=560
E’(5,-2)
F’(1,-2)
G’(5,1)
H’(5,5)
Please check if I understand the question correct
