1000 kg/m3 for fresh water
hope this helps
<h3>
Answer: 55</h3>
==============================================
The 125 degree angle and angle 6 are supplementary. This is because of the same side interior angles theorem.
Let x be the measure of angle 6. Add this to 125, set the sum equal to 180, and solve for x.
x+125 = 180
x = 180-125
x = 55
------------
Or you could approach it this way:
y = measure of angle 2
y+125 = 180
y = 55
angle 6 = angle 2 (corresponding angles)
angle 6 = 55 degrees
-------------
Yet another way you could solve:
z = measure of angle 3
z+125 = 180
z = 55
angle 6 = angle 3 (alternate interior angles)
angle 6 = 55 degrees
A similar approach using alternate interior angles would involve angle 5 = 125, and then noticing that x+125 = 180 solves to x = 55
5.6*1 2/5 = 5 3/5 * 1 2/5 =28/5 * 7/5 = 196/25 = 784/100 = 7.84 or 7 21/25
The more plot point the better but you must have at least three points, a labeled X-axis and Y-axis, and a table for the data to be organized into.
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.
