I had this before and I think is was A.
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.
Answer:
Quadratic Equation:


From the standard form of a Quadratic Function, we get:

Discriminant:



From the discriminant, we conclude that the equation will have two real solutions.
State that:



By the way, solving the equation given:





Answer:
D
Step-by-step explanation:
first of all, what is mode: it is the highest occuring number in a set of numbers.
now since 12 is supposed to be the mean and it occurs twice but 9 also occurs twice so for 12 to be the mode, we must add one more 12.
therefore the answer is 12
So basically when you are adding or subtracting fractions the denominators the number on the bottom of the fraction 12 in this case and 8 as well the two denominators sharing the least common multiple so what is the lowest multiple of 12 and 8 so count off 12: 12, 24, 36... 8: 8, 16 , 24 does that help?