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nalin [4]
2 years ago
5

B^2 – 4ac if a = 2, b = 9, and c = 4

Mathematics
1 answer:
babymother [125]2 years ago
4 0

Answer:

49

Step-by-step explanation:

(b)^{2} - 4(a)(c)  \\(9)^{2} - 4(2)(4)   \\81 - 4(2)(4)  \\81 - 32 \\49

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I think I know this much so far for A (but I could be wrong):
Eduardwww [97]
Part A. You have the correct first and second derivative.

---------------------------------------------------------------------

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
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so either
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3 years ago
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erastovalidia [21]
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6 0
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topjm [15]
0.

It is impossible for 151 to be selected, as it is outside the range of numbers (1 to 100) that can be selected. Therefore the probability is 0.
4 0
4 years ago
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