answer：First, I need to determine the points for a single line and a tetris. A single line is worth 1000 points, and a tetris is worth 8 times that amount, which is 8000 points.Next, I'll calculate the total points from the singles. Tim scored 6 singles, so multiplying 6 by 1000 gives 6000 points.Then, I'll calculate the total points from the tetrises. Tim scored 4 tetrises, so multiplying 4 by 8000 gives 32000 points.Finally, I'll add the points from the singles and the tetrises to find the total score. Adding 6000 and 32000 results in a total of 38000 points.

answer：First, the little monkey picked 6 peaches. The monkey's mother gave the little monkey 3 additional peaches, making the total number of peaches the little monkey has 6 + 3 = 9 peaches.At this point, the monkey's mother has 3 times as many peaches as the little monkey. Therefore, the number of peaches the monkey's mother has is 3 × 9 = 27 peaches.To find out how many peaches the monkey's mother originally had, we need to add the 3 peaches she gave to the little monkey to the number of peaches she has after giving them away. So, the original number of peaches the monkey's mother had is 27 + 3 = 30 peaches.

answer：Okay, so I've got this problem about a marching band with 110 members. It says there are 4 times as many percussion players as woodwind players, and twice as many woodwinds as brass. I need to find out how many brass players there are. Hmm, let's break this down step by step.First, I think I need to define some variables to represent the number of players in each section. Let's say:- Let B be the number of brass players.- Let W be the number of woodwind players.- Let P be the number of percussion players.Now, according to the problem, there are twice as many woodwinds as brass. So, that means W = 2B. Got that down.Next, it says there are 4 times as many percussion players as woodwind players. So, P = 4W. Since we already have W in terms of B, we can substitute that in. So, P = 4*(2B) = 8B. Okay, so now everything is in terms of B, which is what we're trying to find.The total number of band members is 110, so B + W + P = 110. Substituting the expressions we have in terms of B, that becomes B + 2B + 8B = 110. Let's add those up: 1B + 2B + 8B = 11B. So, 11B = 110.To find B, we divide both sides by 11: B = 110 / 11 = 10. So, there are 10 brass players.Wait, let me double-check that. If there are 10 brass players, then woodwinds would be 2*10 = 20, and percussion would be 4*20 = 80. Adding those up: 10 + 20 + 80 = 110. Yep, that matches the total number of band members. So, it looks like 10 is the correct number of brass players.

answer：First, calculate the total number of toes for the Hoopit students. Each Hoopit has 4 hands with 3 toes on each hand, so:4 hands * 3 toes/hand = 12 toes per HoopitWith 7 Hoopit students:7 Hoopit students * 12 toes/Hoopit = 84 toesNext, subtract the total number of Hoopit toes from the overall total to find the number of Neglart toes:164 total toes - 84 Hoopit toes = 80 Neglart toesEach Neglart has 5 hands, and there are 8 Neglart students:8 Neglart students * 5 hands/Neglart = 40 handsFinally, divide the total number of Neglart toes by the total number of Neglart hands to find the number of toes per hand:80 Neglart toes / 40 Neglart hands = 2 toes/hand

answer：Okay, so I have this geometry problem here. It says that triangle ABC is isosceles with AB equal to BC. Hmm, wait, if AB equals BC, then that means triangle ABC has two equal sides: AB and BC. So, point B is the vertex, and AC is the base. Therefore, angles at A and C should be equal because in an isosceles triangle, the angles opposite the equal sides are equal. So, angle BAC equals angle BCA. Got that.Now, the problem says that on the arc AC of its circumscribed circle, which does not contain point B, an arbitrary point M is taken. I need to prove that MA plus MC equals 2 times MB times cosine of angle A. So, MA + MC = 2MB cos A.Alright, let me visualize this. Triangle ABC is isosceles with AB = BC. So, drawing triangle ABC with AB and BC equal. Then, the circumcircle of triangle ABC would pass through all three points A, B, and C. The arc AC that does not contain point B would be the arc from A to C that goes the other way around the circle, avoiding point B. So, point M is somewhere on that arc.I think I need to use some properties of cyclic quadrilaterals or maybe some trigonometric identities here. Since M is on the circumcircle, quadrilateral ABMC is cyclic. That might be useful.Let me recall that in a cyclic quadrilateral, the sum of opposite angles is 180 degrees. So, angle AMB plus angle ACB equals 180 degrees. Similarly, angle AMC plus angle ABC equals 180 degrees. Hmm, not sure if that's directly helpful yet.Wait, since triangle ABC is isosceles with AB = BC, angle ABC is equal to angle BAC? No, wait, AB = BC, so the sides opposite angles at C and A are equal. So, angle at A equals angle at C. So, angle BAC = angle BCA.Let me denote angle BAC as angle A. So, angle BAC = angle BCA = angle A. Then, angle ABC would be 180 - 2A degrees.Since ABMC is cyclic, angle AMB is equal to angle ACB because they subtend the same arc AB. Wait, is that right? Let me think. In a cyclic quadrilateral, angles subtended by the same chord are equal. So, angle AMB and angle ACB both subtend chord AB. Therefore, angle AMB = angle ACB = angle A.Similarly, angle AMC subtends chord AC, and angle ABC subtends chord AC as well. So, angle AMC = angle ABC = 180 - 2A.Hmm, maybe I can use the Law of Sines here. In triangle AMB and triangle CMB, perhaps.Wait, let me consider triangles AMB and CMB. Since M is on the circumcircle, maybe I can relate the sides using the Law of Sines.In triangle AMB, by the Law of Sines, we have:MA / sin(angle AMB) = MB / sin(angle BAM)Similarly, in triangle CMB:MC / sin(angle CMB) = MB / sin(angle BCM)But angle AMB is equal to angle ACB, which is angle A, as we established earlier. Similarly, angle CMB is equal to angle CAB, which is also angle A. So, both angles AMB and CMB are equal to angle A.Wait, is that correct? Let me double-check. In cyclic quadrilateral ABMC, angle AMB and angle ACB are subtended by the same chord AB. So, angle AMB = angle ACB = angle A. Similarly, angle CMB and angle CAB are subtended by chord CB, but wait, chord CB is not the same as chord AB. Hmm, maybe I made a mistake here.Wait, chord CB is the same as chord BC, which is equal to AB. So, maybe angle CMB is equal to angle CAB as well? Hmm, not sure. Let me think again.Alternatively, maybe I should consider the arcs. Since M is on arc AC not containing B, the arcs AM and MC are parts of the circumcircle. Maybe I can express the lengths MA and MC in terms of the arcs.Alternatively, perhaps using coordinates or vectors might help, but that might be more complicated.Wait, another approach: since triangle ABC is isosceles with AB = BC, maybe we can place it in a coordinate system to simplify calculations.Let me place point B at the origin (0,0). Since AB = BC, let me place point A at (a, b) and point C at (-a, b) so that AB = BC. Then, the circumcircle can be determined, and point M can be parameterized on arc AC.But this might get messy. Maybe there's a better way.Wait, going back to the Law of Sines idea. In triangle AMB, angle at M is angle AMB = angle A. So, angle at A is angle BAM, which is part of angle BAC. Since angle BAC is angle A, angle BAM would be some portion of it, say angle φ. Then, angle ABM would be 180 - angle A - angle φ.Similarly, in triangle CMB, angle at M is angle CMB = angle A. Angle at C is angle BCM, which is part of angle BCA, which is also angle A. Let me denote angle BCM as angle ψ. Then, angle CBM would be 180 - angle A - angle ψ.But since AB = BC, maybe angles at B in triangles AMB and CMB are related.Wait, in triangle ABC, AB = BC, so it's an isosceles triangle with AB = BC, so angle at B is 180 - 2A.In triangle AMB, angles are angle at A: φ, angle at M: A, angle at B: 180 - φ - A.Similarly, in triangle CMB, angles are angle at C: ψ, angle at M: A, angle at B: 180 - ψ - A.But since AB = BC, maybe the lengths from B to M are the same? Wait, no, because M is arbitrary on arc AC.Hmm, maybe I need to relate the sides MA, MC, and MB using the Law of Sines.In triangle AMB:MA / sin(angle ABM) = MB / sin(angle BAM) = AB / sin(angle AMB)Similarly, in triangle CMB:MC / sin(angle CBM) = MB / sin(angle BCM) = BC / sin(angle CMB)But AB = BC, and angle AMB = angle CMB = angle A.So, from triangle AMB:MA = (MB * sin(angle ABM)) / sin(angle BAM)From triangle CMB:MC = (MB * sin(angle CBM)) / sin(angle BCM)But angle ABM = 180 - angle BAM - angle AMB = 180 - φ - ASimilarly, angle CBM = 180 - angle BCM - angle CMB = 180 - ψ - ABut I'm not sure if this is leading me anywhere.Wait, maybe I can use the fact that in triangle ABC, since it's isosceles with AB = BC, the circumradius can be expressed in terms of the sides.The formula for the circumradius R of triangle ABC is R = (a) / (2 sin A), where a is the length of side BC (or AB, since they are equal). Wait, actually, in triangle ABC, sides AB = BC = c, and AC = b.Then, the circumradius R = (c) / (2 sin A). Wait, no, more accurately, R = (a) / (2 sin A), where a is the side opposite angle A. In triangle ABC, side AC is opposite angle B, which is 180 - 2A. So, R = (AC) / (2 sin B) = (AC) / (2 sin(180 - 2A)) = (AC) / (2 sin 2A).But AB = BC = c, so using the Law of Sines on triangle ABC:AB / sin C = BC / sin A = AC / sin BSince AB = BC = c, and angles at A and C are equal (both angle A), then:c / sin A = AC / sin(180 - 2A) = AC / sin 2ASo, AC = c * sin 2A / sin A = c * 2 cos ATherefore, AC = 2c cos A.So, the circumradius R = AC / (2 sin 2A) = (2c cos A) / (2 * 2 sin A cos A) ) = (2c cos A) / (4 sin A cos A) ) = c / (2 sin A)So, R = c / (2 sin A)Okay, so now, in triangle AMB, using the Law of Sines:MA / sin(angle ABM) = MB / sin(angle BAM) = AB / sin(angle AMB)We have AB = c, angle AMB = angle A, angle BAM = φ, angle ABM = 180 - φ - A.So, MA = (c * sin(angle ABM)) / sin(angle AMB) = (c * sin(180 - φ - A)) / sin A = (c * sin(φ + A)) / sin ASimilarly, in triangle CMB:MC = (c * sin(angle CBM)) / sin ABut angle CBM = 180 - angle BCM - angle CMB = 180 - ψ - ABut since AB = BC, and angles at A and C are equal, maybe angle BAM = angle BCM? Let me think.Wait, since M is on arc AC not containing B, the arcs AM and MC are such that angles subtended by them at B are equal? Hmm, not necessarily.Alternatively, maybe the sum of angles BAM and BCM is equal to angle BAC + angle BCA = 2A.Wait, angle BAM + angle BCM = φ + ψ = 2A?Wait, no, because in triangle ABC, angles at A and C are each A, so angle BAC = angle BCA = A.But when we take point M on arc AC, the angles BAM and BCM are parts of angles BAC and BCA respectively.So, angle BAM = φ, angle BCM = ψ, and φ + ψ = 2A? Hmm, not necessarily. Because depending on where M is, φ and ψ can vary, but their sum might not necessarily be 2A.Wait, maybe I need to consider the arcs.In the circumcircle, the measure of an arc is twice the measure of the inscribed angle subtended by it.So, arc AM corresponds to angle ABM, and arc MC corresponds to angle CBM.But I'm getting confused.Wait, let me think differently. Since M is on arc AC not containing B, the arcs AM and MC are such that the angles subtended by them at B are equal to half the measure of the arcs.Wait, the measure of arc AM is equal to 2 * angle ABM, and measure of arc MC is equal to 2 * angle CBM.But since M is on arc AC, the total measure of arc AC is equal to 2 * angle ABC, because in the circumcircle, the measure of an arc is twice the measure of the inscribed angle opposite it.In triangle ABC, angle ABC is 180 - 2A, so arc AC is 2*(180 - 2A) = 360 - 4A degrees.Wait, that can't be right because the total circumference is 360 degrees, so arc AC should be less than that.Wait, no, actually, in the circumcircle, the measure of arc AC is equal to 2 * angle ABC. Since angle ABC is 180 - 2A, arc AC is 2*(180 - 2A) = 360 - 4A degrees.But since M is on arc AC not containing B, the arc AC is actually the major arc AC, which is 360 - 4A degrees.Wait, but in a triangle, the sum of angles is 180 degrees, so angle ABC is 180 - 2A, which is less than 180 degrees, so arc AC is 2*(180 - 2A) = 360 - 4A degrees, which is greater than 180 degrees, so it is indeed the major arc.Therefore, the measure of arc AM plus arc MC equals 360 - 4A degrees.But since M is arbitrary on arc AC, the measures of arcs AM and MC can vary, but their sum is fixed.Wait, but how does that help me?Alternatively, maybe I can use Ptolemy's theorem on cyclic quadrilateral ABMC.Ptolemy's theorem states that in a cyclic quadrilateral, the product of the diagonals equals the sum of the products of opposite sides.So, in quadrilateral ABMC, we have:AB * MC + BC * MA = AC * MBBut AB = BC = c, and AC = 2c cos A, as we found earlier.So, substituting:c * MC + c * MA = 2c cos A * MBDivide both sides by c:MC + MA = 2 cos A * MBWhich is exactly what we needed to prove: MA + MC = 2MB cos A.Wait, that was much simpler! I think I overcomplicated it earlier with the Law of Sines.So, using Ptolemy's theorem directly gives the result.But let me make sure I didn't skip any steps. Ptolemy's theorem applies to cyclic quadrilaterals, which ABMC is because all points lie on the circumcircle of triangle ABC.Therefore, AB * MC + BC * MA = AC * MBSince AB = BC = c, and AC = 2c cos A,c * MC + c * MA = 2c cos A * MBDivide both sides by c:MA + MC = 2 cos A * MBWhich is the desired result.So, the key was recognizing that quadrilateral ABMC is cyclic and applying Ptolemy's theorem.I think that's the solution. It was a bit tricky at first, but using Ptolemy's theorem simplified everything.

answer：First, I need to determine the total water delivery rate of all four hoses combined. There are two hoses delivering water at a rate of 2 gallons per minute each. So, the combined rate for these hoses is:2 hoses * 2 gallons/minute/hose = 4 gallons/minuteThe other two hoses deliver water at a rate of 3 gallons per minute each. Therefore, their combined rate is:2 hoses * 3 gallons/minute/hose = 6 gallons/minuteAdding both rates together gives the total delivery rate:4 gallons/minute + 6 gallons/minute = 10 gallons/minuteNext, I need to calculate the time required to fill the pool with a volume of 15,000 gallons. Using the total delivery rate:15,000 gallons / 10 gallons/minute = 1,500 minutesFinally, to convert the time from minutes to hours:1,500 minutes / 60 minutes/hour = 25 hours