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.
In this question, both tickets cost 2$ per ticket.
The answer to this question would be: $0
In WinOne scenario, you need to match a ticket that has to pick from A-J(10 possibilities) and 0-9 (10 possibilities). The chance to win would be: 1/10* 1/10= 1/100
The expected value must be:
E= chance to win * win amount - ticket price
E= 1//100*$200 - $2= $2-$2= 0
Answer:
( intransitive) To tend steadily upward or downward. ...
( transitive) To form with a slope; to give an oblique or slanting direction to; to incline or slant. ...
( colloquial, usually followed by a preposition)
Step-by-step explanation:
the answer to ur problem will be -7/5
