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ArbitrLikvidat [17]
3 years ago
10

HELP ME PLZZZ ASAP PLZZZ

Mathematics
1 answer:
AnnyKZ [126]3 years ago
6 0
I believe that B is the answer bc -2-1 is -3 but with absolute value it it's 3, whish is the correct distance
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The other question got deleted. Pls due tmm November 26. Pls will mark BRAINLIEST. Pls show work and have answer rt
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Finaly Finished!!!!

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Find the area of the rhombus.<br><br> Find the area. The figure is not drawn to scale.
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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.

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

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. 
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3 years ago
What is 2 +4s=16 on a graph
Annette [7]
That is the answer !! hope this helped

4 0
2 years ago
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