A graph's correlation between two variables reveals the relationship between them. The correlation is given a name based on the characteristics of the variables represented on the graphs.
What does the term correlation mean?
The data points of a person with two different factors are related in a scatterplot. The term "correlation" refers to this connection.
A correlation is the relationship between any two variables represented on any graph.
The correlation is divided into three main categories based on the sort of relationship between the variables. Those are
i) Positively correlation
ii) Negative correlation
iii) No correlation
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Answer 79,000
explanation 1 ton is 32,000 ounces and you have 2 so you would have 64,000 and you add the extra 15,000 to get 79,000
Divide 1/16 by 3/5 and you get .104 we can check by multiplying our answer by 3/5 and we get 1/16
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.
