-2y>12x-42
y<21-6x
so a is true
Iiiiiiiiiiiiiiiiiii[ooioj09uij09uikj
Answer:
A.) gf(x) = 3x^2 + 12x + 9
B.) g'(x) = 2
Step-by-step explanation:
A.) The two given functions are:
f(x) = (x + 2)^2 and g(x) = 3(x - 1)
Open the bracket of the two functions
f(x) = (x + 2)^2
f(x) = x^2 + 2x + 2x + 4
f(x) = x^2 + 4x + 4
and
g(x) = 3(x - 1)
g(x) = 3x - 3
To find gf(x), substitute f(x) for x in g(x)
gf(x) = 3( x^2 + 4x + 4 ) - 3
gf(x) = 3x^2 + 12x + 12 - 3
gf(x) = 3x^2 + 12x + 9
Where
a = 3, b = 12, c = 9
B.) To find g '(12), you must first find the inverse function of g(x) that is g'(x)
To find g'(x), let g(x) be equal to y. Then, interchange y and x for each other and make y the subject of formula
Y = 3x + 3
X = 3y + 3
Make y the subject of formula
3y = x - 3
Y = x/3 - 3/3
Y = x/3 - 1
Therefore, g'(x) = x/3 - 1
For g'(12), substitute 12 for x in g' (x)
g'(x) = 12/4 - 1
g'(x) = 3 - 1
g'(x) = 2.
M can be any positive real number.
Explanation:
From f(x) = √(mx) ; if x is posive m has to be positive; if x is negative m has to be negative; if x is cero m can have any value, and the range will always be 0 or positve
From g(x) = m √x; x can only be 0 or positive and the range will have the sign of m.
Given we concluded that the range of f(x) can only be 0 or positive, then me can only be 0 or positive.
Adjacent angles :
IGJ and IGH
KGL and LGM
vertical angles :
JGK and HGM
IGJ and MGL
hope this helps
