This is a strange question, and f(x) may not even exist. Why do I say that? Well..
[1] We know that f(a+b) = f(a) + f(b). Therefore, f(0+0) = f(0) + f(0). In other words, f(0) = f(0) + f(0). Subtracting, we see, f(0) - f(0) = f(0) or 0 = f(0).
[2] So, what's the problem? We found the answer, f(0) = 0, right? Maybe, but the second rule says that f(x) is always positive. However, f(0) = 0 is not positive!
Since there is a contradiction, we must either conclude that the single value f(0) does not exist, or that the entire function f(x) does not exist.
To fix this, we could instead say that "f(x) is always nonnegative" and then we would be safe.
T-6=-4
Just simply add 6 to both sides, it will cancel out the -6 and leave T alone, and it will become T=2, There's your answer.
equation of f(x)
point of intersection of f(x) and x axis - (-4,0)
point of intersection of f(x) and y axis - (0,4)
Equation y = x +4
g(x) = f( kx)
= kx +4
g(-2) = -2
-2k +4= -2
-2k = -2 -4
-2k = -6
k= 3
Its hard to see... try again
Answer:
16
Step-by-step explanation:
v=lwh
v=4x2x2
