answer:Let's denote the number of lollipops Alison has as A and the number of lollipops Diane has as D. According to the problem, Henry has 30 more lollipops than Alison, so Henry has A + 30 lollipops. We are given that Alison has 60 lollipops, so A = 60. We are also told that if they combine their lollipops and eat 45 lollipops each day, it will take them 6 days to finish the lollipops. This means the total number of lollipops they have together is 45 lollipops/day * 6 days = 270 lollipops. Now, we can write the total number of lollipops as the sum of Alison's, Henry's, and Diane's lollipops: A + (A + 30) + D = 270 Since we know A = 60, we can substitute that in: 60 + (60 + 30) + D = 270 60 + 90 + D = 270 150 + D = 270 D = 270 - 150 D = 120 Now we have the number of lollipops Diane has, which is 120. The ratio of the number of lollipops Alison has to the number of lollipops Diane has is A:D, which is 60:120. This ratio can be simplified by dividing both numbers by the greatest common divisor, which is 60 in this case: 60/60 : 120/60 1 : 2 So the ratio of the number of lollipops Alison has to the number of lollipops Diane has is boxed{1:2} .

answer:Let's calculate the net amount of water filled in the tank after each cycle of opening and closing the pipes A, B, and C. In the first minute, pipe A fills the tank with 40 liters. In the second minute, pipe B fills the tank with 30 liters. In the third minute, pipe C drains the tank by 20 liters. So, the net amount of water filled in the tank after one complete cycle (3 minutes) is: 40 liters (from A) + 30 liters (from B) - 20 liters (from C) = 50 liters. Now, we need to find out how many such cycles are required to fill the 950-liter tank. Let's divide the total capacity of the tank by the net amount of water filled in one cycle: 950 liters / 50 liters per cycle = 19 cycles. Since each cycle takes 3 minutes, the total time to complete 19 cycles is: 19 cycles * 3 minutes per cycle = 57 minutes. However, after 19 cycles, the tank will be full, and we don't need to drain it. So we need to subtract the last minute (which would have been for draining) from the total time. Therefore, the tank will be full in 57 minutes - 1 minute = boxed{56} minutes.

answer:1. Let ( O_1 ) be the center of the inscribed circle in (triangle ABC) with radius ( R ), and ( O_2 ) be the center of the inscribed circle in (triangle ADE) with radius ( r ). Denote ( p_1 ) as the semiperimeter of (triangle ABC) and ( p_2 ) as the semiperimeter of (triangle ADE). 2. We are given that ( AK = p_2 ) and ( AK = p_1 - BC ). Since the quadrilateral ( BDEC ) can have a circle inscribed in it, the sum of opposite angles must be 180 degrees, i.e., (angle ABC + angle DEC = 180^circ). 3. We use the property of tangent segments from a common external point to the points of tangency with an inscribed circle: [ 2AK = AK + AH = AD + DK + EH + AE = AD + (DM + ME) + AE = 2p_2 ] [ 2AK = (AB - BK) + (AC - CH) = AB + AC + (BF + CF) - 2(BF + CF) = 2(p_1 - BC) ] Given ( AK = 3 ) and ( BC = 15 ): [ p_1 - 15 = 3 Rightarrow p_1 = 18 ] 4. Determine the radius ( r ) of the inscribed circle in (triangle ADE): [ r = frac{S_{text{ADE}}}{p_2} = frac{S_{text{ADE}}}{AK} = frac{0.5}{3} = frac{1}{6} ] 5. Since (angle ADE = angle ABC) and triangles ( ADE ) and ( ABC ) are similar, we have: [ frac{r}{R} = frac{p_2}{p_1} Rightarrow R = frac{r cdot (AK + BC)}{AK} = frac{frac{1}{6} cdot (3 + 15)}{3} = 1 ] 6. Let (angle BAC = alpha). Using the tangent function: [ tanleft(frac{alpha}{2}right) = frac{R}{AK} = frac{1}{3} ] Therefore, [ tan alpha = frac{2 tanleft(frac{alpha}{2}right)}{1 - tan^2left(frac{alpha}{2}right)} = frac{2 cdot frac{1}{3}}{1 - left(frac{1}{3}right)^2} = frac{frac{2}{3}}{frac{8}{9}} = frac{2}{3} cdot frac{9}{8} = frac{3}{4} ] Conclusion: [ boxed{frac{3}{4}} ]

answer:To solve the given expression step-by-step, we start with the original expression and simplify it as follows: [ begin{align*} frac{sqrt{18}}{3}+|sqrt{2}-2|+2023^{0}-left(-1right)^{1} &= frac{3sqrt{2}}{3}+|sqrt{2}-2|+1-1 &= frac{3sqrt{2}}{3}+|-sqrt{2}+2|+1-1 &= sqrt{2}+2-sqrt{2}+1-1 &= sqrt{2}+2-sqrt{2}+1-1 &= 2+1-1 &= 2. end{align*} ] Therefore, the final answer is boxed{2}.