Question 1
Because the period is 2π, and the amplitude is 1obtain
f(x) = sin(x)
Because the horizontal shift is π, obtain
f(x) = sin(x - π)
Because the vertical shift is -4, obtain
f(x) = sin(x - π) - 4
Answer: 1. f(x) = sin(x - π) - 4
Question 2
The radius is 36/2 = 18 in.
1 revolution (360°) is the circumference, which is
2π(18) = 36π in
When the revolution is 62°, the distance traveled is
(62/360)*(36π) = (31/5)π in
Answer: 3. (31π)/5
Question 3.
Consider f(x) = 3cos(2x-π) - 1
f(0) = 3cos(-π) - 1 = -4
f(π/2) = 3cos(0) - 1 = 2
Rate of change = (2+4)/(π/2) = 12/π
From the graph, the rate of change of g(x) is
3/(π/2) = 6/π
Consider h(x) = sin(x) - 4
h(0) = 0 - 4 = -4
h(π/2) = 1 - 4 = -3
Rate of change = (-3+4)/(π/2) = 2/π
Therefore h(x) has the smallest rate of change
Answer: h(x)
Hi,
first you should divide 130 (speed) by 2 (time), then divide 390 by 65 (result of 130 / 2)
No need to fear, thehotdogman93 is here!
The first step is to get rid of those very large numbers. It's going to be very difficult to factor unless we can bring those high numbers down. So lets see if we can factor each term.
So after dividing 49 with every single digit. The only number that divides evenly is 7 and one, and 16 isnt divisible evenly by 7 so that didn't work. Looks like we're gonna have to work with these big numbers.
There is something interesting though about these numbers. 16 and 49 are both perfect squares. 16 is the same as 4^2 and 49 is the same as 7^2. So we can factor the whole trinomial as:

If we were to expand this out as:

and multiply it back into the original form. It would match with the expression we started with. The 4's would multiply back into 16x^2 and the 7's would multiply back into 49.
Additionally 4 * -7 is -28, so you can combine two -28x's into the -56x term in the original trinomial.
Thus, the answer is yes you can, and the answer is:

Answer:
Remember,
and the range of g must be in the domain of f.
a)


The domain of f(g(x)) and g(f(x)) is the set of reals.
b)


The domain of f(g(x)) is the set of nonnegative reals and the domain of g(f(x)) is the set of number such that 
c)


The domain of f(g(x)) is the set of reals except the 1 and the domain of g(f(x)) is the set of reals except the 1 and -1
d)


The domain of f(g(x)) is the set of reals except 2, and the domain of g(f(x)) is the set of reals except -1.
e)


The domain of f(g(x)) is the set of nonnegative reals except -3. The domain of g(f(x)) is the set of nonnegative reals except -2.