Answer:
Part A: 6x + 20
Part B: Expression G and H are not equivalent for any value of x.
Step-by-step explanation:
Part A:
G = 2(3x + 10)
G = 6x + 20
Part B:
6x + 20 = 6x + 12
-20 -20
6x = 6x - 8
-6x -6x
0 = -8
No solution. This means that expressions G and H are not equivalent for any value of x.
Answer:
Step-by-step explanation:
Given the functions
F(x) = 6-x²
G(x) = x²+4x-12
A) we are to find F(x)+g(x)
F(x)+G(x) = 6-x²+x²+4x-12
F(x)+G(x) = 6+0+4x-12
F(x)+G(x) = 4x+6-12
F(x)+G(x) = 4x-6
The domain of the function is the values of x for which the expression exists. The expression exists on all real interval i.e xER
F(x)-g(x) = 6-x²-(x²+4x-12)
F(x)-g(x) = 6-x²-x²-4x+12
F(x)-g(x) = 6-2x²+4x+12
F(x)-g(x) = -2x²+4x+6
The domain of the function is the values of x for which the expression exists. The expression exists on all real interval i.e xER
3) F(x)/G(x)
= 6-x²/x²+4x-12
The domain of the function is the values of x for which the expression exists. The expression exists on all real interval i.e xER
She could arrange them into 6 tens and 3 ones
Answer:
?=58
h=48
g=84
Step-by-step explanation:
?=58) 180-122=58
done the same thing
