Answer:
Workdone = 3200 lb.ft
Step-by-step explanation:
We are told that the bucket is filled with 40 lb of water but water leaks out of a hole in the bucket at a rate of 0.2lb/s
Thus,
Weight of water at any given time (t) would be;
w(t) = 40 - 0.2t - - - - (1)
We are told the bucket is pulled up at a rate of 2ft/s.
Thus, height at time (t); y = 0 + 2t = 2t
Since y = 2t,
Then,t = y/2
Put y/2 for t in eq 1
Thus; w(y) = 40 - 0.2(y/2)
w(y) = 40 - 0.1y
Now, at y = 80 ft, we have;
w(80) = 40 - 0.1(80)
w(80) = 40 - 8 = 32 lb
Since 32 lbs are left, it means there is always water in the bucket.
Thus, work done is;
W = 80,0[∫(Total weight).dy]
W = 80,0[∫[(weight of rope) + (weight of bucket) + (weight of water)]dy]
W = 80,0[∫[0 + 4 + 40 - 0.1y]dy]
Integrating, we have;
W = [44y - y²/20] at boundary of 80 and 0
So,
W = [44(80) - 80²/20] - [0 - 0²/20]
W = 3200 lb.ft
The answer is B, the x-axis crosses over to -8 the way the line goes also means it is a negative. I don't know for sure because i haven't done this in a while but I hope it helps.
Answer:
k = (6/15)
Step-by-step explanation:
The equation is:
6*(x + 1) + 2 = 3*(k*5*x + 1) + 3
To have no solutions, we need to have something like:
x + 7 = x + 4
where we can remove x in both sides and end with
7 = 4
So this equation is false, meaning that there is no value of x such that this equation is true, then the equation has no solutions.
First, let's try to simplify our equation:
6*(x + 1) + 2 = 3*(k*5*x + 1) + 3
6*x + 6 + 2 = 3*k*5*x + 3*1 + 3
6*x + 8 = 15*k*x + 6
if 15*k = 6, then the system clerly has no solution.
then:
k = 6/15
then we get:
6*x + 8 = (6/15)*15*x + 6
6*x + 8 = 6*x + 6
8 = 6
The system has no solutions.
Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.
A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'
B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)
C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)
D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n
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* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.
The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
