Answer:
John ski down the mountain is 1285.37 feet.
Step-by-step explanation:
Given : John is skiing on a mountain with an altitude of 1200 feet. The angle of depression is 21.
To find : About how far does John ski down the mountain ?
Solution :
We draw a rough image of the question for easier understanding.
Refer the attached figure below.
According to question,
Let AB be the height of mountain i.e. AB=1200 feet
The angle of depression is 21 i.e. 
We have to find how far does John ski down the mountain i.e. AC = ?
Using trigonometric,
Therefore, John ski down the mountain is 1285.37 feet.
Part A. You have the correct first and second derivative.
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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.
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Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out
To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0.
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well.
The opposite of left is right, so the number opposite k is k units to the right.
This is a linear function.
A linear function is a straight line that has a slope, and follows the formula: <em>
y = mx + b</em>
In which:
y = y
m = slope
x = x
b = y -intercept
hope this helps<em />
