Answer:
V' = -0.11552 *V\\= -0.11552(1.8)\\ \\=-0.20794 million per year
Step-by-step explanation:
Given that oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 3 million barrels of oil in the well; six years later 1,500,000 barrels remain.
i.e. if V stands for volume of oil, then

To find A and k
V(0) = A = 3 million
Hence V = 
V(6) = 1.5
i.e. 

a) Using the above value of k , we have
million per year.
Distributive Property: a ( b + c) = (a*b) + (a*c)
b) x(x + 9) = x*x + 9*x
= x² + 9x
c) x² - 18x
x² = x * x
18x = 18 *x
G C F = x
Greatest common factor in both the terms is 'x'.Take the common variable from both the terms.
x² - 18x = x*x - 18*x
= x (x - 18)
Standard form. Factored form
x² x*x
x² + 9x x(x +9)
x² - 18x x(x - 18)
-x² + 10x x(-x + 10) = x(10 -x)
-x² - 2.75x -x(x + 2.75)
The probability of it happening once is 1/5, since 10/50 simplifies to 1/5. Because the ticket goes back in the bag, the probability stays constant at 1/5.
For it to happen twice, you multiply the probability of it happening once, twice.
1/5 • 1/5 = 1/25 probability of it being Gary's name twice in a row.
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2
abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.
d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.