Answer:
K = 50
L = 90
KML = 40
LMN = 140
Step-by-step explanation:
The sum of the angles in a triangle will always be 180 degrees, and in the triangle pictured, we know the value of each angle. We can say that
5x + 4x + 9x = 180
18x = 180
x = 10
Now that we know the value of x, we can figure out the value of the angles.
K = 5x = 50
L = 9x = 90
KML = 4x = 40
A straight line is 180 degrees, so to find angle KMN, you have to subtract KML (4x) from 180
180 - 4x = 180 - 40 = 140
The answer is c2 have a good day !!!! Bc you add all the numbers up and you will get it
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.
It would be c, y = 2/3x + 6
