Answer:
we can draw the graph of cosine with amplitude 2.
Step-by-step explanation:
In the given function g(x) = 2cosx we know the following parameters which are helpful to draw the graph.
Its a cosine graph given in the form of y = acos(Bx + c) + d
Here a is amplitude, period = , c = phase shift, d = vertical shift
We know a = 2, B = , c = 0, d = 0.
Answer:
g(-4) = -1
g(-1) = -1
g(1) = 3
Explanation:
If you are given a function that is defined by a system of equations associated with certain intervals of x, just find which interval makes x true, and then substitute x into the equation of that interval.
For example, given g(-4), this is an expression which is asking for the value of the equation when x = -4. So -4 is not ≥ 2, so ¼x - 1 will not be used. -4 is also not ≤ -1 and ≤ 2, so -(x - 1)² + 3 will not be used either. So in turn, we will just use -1 which is always -1 so g(-4) will just be -1, right because there is no x variable in -1 so it will always be the same.
Using the same idea as before g(-1) is g(x) when x = -1 so -1 will not be a solution because -1 is not less than -1 (< -1). -1 is not ≥ 2 either so we will be using the second equation because -1 is part of the interval -1≤x≤2 (it is a solution to this inequality), therefore -(x - 1)² + 3 will be used.
As x = -1, -(x - 1)² + 3 = -(-1 - 1)² + 3 = -(-2)² + 3 = -4 + 3 = -1.
It is a coincidence that g(-1) = -1.
Now for g(1), where g(x) has an input of 1 or the value of the function where x = 1, we will not use the first equation because x = 1 → x < -1 → 1 < -1 [this is false because 1 is never less than -1], so we will not use -1.
We will use -(x - 1)² + 3 again because 1 is not ≥ 2, 1≥2 [this is also false]. And -1 ≤ 1 < 2 [This is a true statement]. Therefore g(1) = -(1 - 1)² + 3 = -(0)² + 3 = 3
Answer:
$2 per ride
Step-by-step explanation:
$32-$20=$12
McKenna rode 6 rides so $12 divided by 6 rides is $2 for every ride
Vcyl = Vcone
pi×x^2×y = 1/3×pi×(3x)^2 (h)
pi's cancel--> x^2•y = 3x^2 (h)
h = y/3
Answer:
7 and 1/3
Step-by-step explanation:
