PartialMathematician

If you want to check, b = 2, P = 0, and Q = 0. 

\((0\cdot 2^2) = 0\cdot 0\cdot2\) simplifies to \(0=0\) because anything multiplied by 0 is 0.

- PM

Let's try to express \((Pb^2) = QPb\) in terms of b. Since Q = P/2 and P = b - 2, \(((b - 2) \cdot b^2) = \dfrac{b-2}{2} (b - 2) b \Rightarrow b^3 - 2b^2 = \dfrac{1}{2}b^3 -2b^2 + 2b\). Combining like terms, we have \(\dfrac{1}{2}b^3 = 2b \Rightarrow b^3 = 4b\). Dividing both sides by b, we have \(b^2 = 4 \Rightarrow b = \pm 2\), which in this case, changes to b = 2. We know that P is two less than b (P = b - 2), so P = 2-2 = 0. 

\(\boxed{P = 0}\).

 - PM

So there you have it... the height of the water decreases around \(t = 2.25\).

Hope this helps,

 - PM

That is where the graph caught you off guard! The graph is actually not a straight line. Zoom out, and you will see that the graph curves like an "N" at around y = 58! 

- PM

Please provide more information.

I am assuming that you want the distance of the hour and minute hands at 1 o'clock because there are different distances at a different time. Also, when the time goes from 6:00 to 7:00, the distance actually decreases. 

If the hour hand is 20mm, then the minute hand is 40mm. At 1:00, the angle formed from the hour and minute hands is 30 degrees. I think the triangle formed might be an isosceles triangle, but I am not 100% sure. If it is an isosceles triangle, then the degrees are 30, 30, and 120. 

For future reference to the triangle, the tip of the minute hand is point M, the tip of the hour hand is point H, and the center of the clock is point O.

Please tell me if my assumptions about the problem are correct or incorrect. I do not want to go too far if my explanation might not be suitable for this problem. 

- PM