we have the number
![6-\sqrt[]{-40}](https://tex.z-dn.net/?f=6-%5Csqrt%5B%5D%7B-40%7D)
Remember that
i^2=-1
so
substitute
![6-\sqrt[]{(i^2)40}](https://tex.z-dn.net/?f=6-%5Csqrt%5B%5D%7B%28i%5E2%2940%7D)
![6-2i\sqrt[]{10}](https://tex.z-dn.net/?f=6-2i%5Csqrt%5B%5D%7B10%7D)
therefore
the real part is 6
the imaginary part is -2√10
The parabola's vertex would not be on the x-axis or y-axis and there would be no x-intercepts.
We are given the function:
g(x) = 6 (4)^x
Part A.
To get the average rate of change, we use the formula:
average rate of change = [g(x2) – g(x1)] / (x2 – x1)
Section A:
average rate of change = [6 (4)^1 – 6 (4)^0] / (1 – 0) =
18
Section B:
average rate of change = [6 (4)^3 – 6 (4)^2] / (3 – 2) =
288
Part B.
288 / 18 = 16
Therefore the average rate of change of Section B is 16 times
greater than in Section A.
<span>The average rate of change is greater between x = 2 to x = 3 than between
x = 1 and x = 0 because an exponential function's rate of change increases
with increasing x (not constant).</span>
The answer is C
4(3-x)+6x=3x+12-x All real numbers are solutions.
