The slope-intercept form of (-5,2) (-5,-1) is undefined
<u>Solution:</u>
Need to find the slope intercept form of line passing through two point (-5,2) (-5,-1).
Slope intercept form of line passing through 
is given by
On Substituting the values we get the slope intercept form of given points,
= undefined
Hence the slope intercept form of given points is undefined
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.
Using pythagorean identities,tan(sin^-1(-5/13))= tan(arcsin(-5/13))
Let A = arcsin(-5/13) = -arcsin(5/13)
Thus, sin A = -5/13 and cos A = sqrt(1 - (5/13)^2) = 12/13.= tan(A)= sin(A) / cos(A)= -5/13 / (12/13)= -5/12
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Step-by-step explanation: hhhhhhhhh
Answer:
im pretty sure all you have to do is switch the signs (if its negative it will become positive) but only for the points on the x axis so basically (-12,13) would become (12,13)
