💡First, I need to determine the total volume of water that needs to be drained from the pool. The pool is at 80% capacity, so I'll calculate 80% of the total volume.The volume of a rectangular pool is given by the formula:[ text{Volume} = text{width} times text{length} times text{depth} ]Let's denote the length of the pool as ( L ). The total volume of the pool when it's full is:[ text{Total Volume} = 50 text{ ft (width)} times L text{ ft (length)} times 10 text{ ft (depth)} = 500L text{ cubic feet} ]Since the pool is at 80% capacity, the volume of water in the pool is:[ text{Volume of Water} = 0.80 times 500L = 400L text{ cubic feet} ]Next, I know that the hose can remove 60 cubic feet of water per minute and it takes 1000 minutes to drain the pool. Therefore, the total volume of water removed is:[ text{Total Water Removed} = 60 text{ cubic feet/minute} times 1000 text{ minutes} = 60000 text{ cubic feet} ]Since the total volume of water removed should be equal to the volume of water in the pool, I can set them equal to each other and solve for ( L ):[ 400L = 60000 ][ L = frac{60000}{400} = 150 text{ feet} ]Therefore, the length of the pool is 150 feet.

💡The greatest integer function, denoted as [x], returns the largest integer that is less than or equal to x. For the given value x = 2.7, I need to find the greatest integer that does not exceed 2.7.First, I identify the integers that are less than or equal to 2.7. These integers are 0, 1, 2, and so on. Among these, the largest integer that is less than or equal to 2.7 is 2.Therefore, applying the greatest integer function to 2.7 results in 2.

💡Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem gives me an equation: (3a - b)/cos B = c / cos C. I need to find sin C first, and then if c is sqrt(3), find the maximum area of the triangle.Alright, let's start with part (1). I remember that in triangles, the Law of Sines is often useful. It says that a/sin A = b/sin B = c/sin C. Maybe I can use that here.Given the equation: (3a - b)/cos B = c / cos C. Hmm, so maybe I can express a, b, and c in terms of sines. Let me denote the common ratio as 2R, where R is the circumradius. So, a = 2R sin A, b = 2R sin B, c = 2R sin C.Let me substitute these into the equation:(3*(2R sin A) - 2R sin B) / cos B = (2R sin C) / cos CSimplify numerator on the left:(6R sin A - 2R sin B) / cos B = 2R sin C / cos CFactor out 2R:2R*(3 sin A - sin B) / cos B = 2R sin C / cos CSince 2R is non-zero, we can divide both sides by 2R:(3 sin A - sin B) / cos B = sin C / cos CSo, (3 sin A - sin B)/cos B = tan CHmm, okay. Let's write that as:3 sin A / cos B - sin B / cos B = tan CWhich is:3 sin A / cos B - tan B = tan CBut I'm not sure if that helps. Maybe I can express sin A in terms of other angles. Since in a triangle, A + B + C = π, so A = π - B - C.So, sin A = sin(π - B - C) = sin(B + C). And sin(B + C) = sin B cos C + cos B sin C.So, sin A = sin B cos C + cos B sin C.Let me substitute that back into the equation:3*(sin B cos C + cos B sin C)/cos B - sin B / cos B = tan CSimplify term by term:First term: 3*(sin B cos C + cos B sin C)/cos B = 3 sin B cos C / cos B + 3 cos B sin C / cos B = 3 sin B tan C + 3 sin CSecond term: - sin B / cos B = - tan BSo putting it all together:3 sin B tan C + 3 sin C - tan B = tan CLet me write that:3 sin B tan C + 3 sin C - tan B = tan CBring all terms to one side:3 sin B tan C + 3 sin C - tan B - tan C = 0Factor tan C:tan C (3 sin B - 1) + 3 sin C - tan B = 0Hmm, this seems a bit complicated. Maybe I can express tan B in terms of sin and cos:tan B = sin B / cos BSo, substituting back:tan C (3 sin B - 1) + 3 sin C - sin B / cos B = 0I'm not sure if this is the right path. Maybe I should try another approach.Wait, going back to the equation:(3 sin A - sin B)/cos B = sin C / cos CWhich is:(3 sin A - sin B)/cos B = tan CLet me rearrange this:3 sin A - sin B = tan C * cos BBut tan C = sin C / cos C, so:3 sin A - sin B = (sin C / cos C) * cos B = sin C cos B / cos CSo,3 sin A - sin B = sin C cos B / cos CHmm, maybe I can use the Law of Sines again. Since a/sin A = b/sin B = c/sin C = 2R.So, sin A = a/(2R), sin B = b/(2R), sin C = c/(2R)But I'm not sure if that helps here. Alternatively, maybe express sin A in terms of sin(B + C).Wait, I already did that earlier: sin A = sin(B + C) = sin B cos C + cos B sin C.So, substituting back into 3 sin A - sin B:3*(sin B cos C + cos B sin C) - sin B = 3 sin B cos C + 3 cos B sin C - sin BFactor sin B:sin B (3 cos C - 1) + 3 cos B sin CSo, we have:sin B (3 cos C - 1) + 3 cos B sin C = sin C cos B / cos CWait, that seems complicated. Maybe I can collect like terms.Let me write the equation again:3 sin A - sin B = sin C cos B / cos CSubstituting sin A:3*(sin B cos C + cos B sin C) - sin B = sin C cos B / cos CExpanding:3 sin B cos C + 3 cos B sin C - sin B = sin C cos B / cos CBring all terms to the left:3 sin B cos C + 3 cos B sin C - sin B - sin C cos B / cos C = 0Hmm, maybe factor sin B and sin C:sin B (3 cos C - 1) + sin C (3 cos B - cos B / cos C) = 0Wait, that might not be helpful. Let me see:Alternatively, let's multiply both sides by cos C to eliminate the denominator:(3 sin A - sin B) cos C = sin C cos BSo,3 sin A cos C - sin B cos C = sin C cos BBring all terms to one side:3 sin A cos C - sin B cos C - sin C cos B = 0Hmm, maybe factor terms:3 sin A cos C - (sin B cos C + sin C cos B) = 0But sin B cos C + sin C cos B = sin(B + C) = sin(π - A) = sin ASo, we have:3 sin A cos C - sin A = 0Factor sin A:sin A (3 cos C - 1) = 0Since sin A ≠ 0 (because A is an angle in a triangle, so 0 < A < π), we have:3 cos C - 1 = 0 ⇒ cos C = 1/3Therefore, sin C = sqrt(1 - cos² C) = sqrt(1 - 1/9) = sqrt(8/9) = 2√2 / 3So, sin C = 2√2 / 3.Okay, that seems to be the answer for part (1).Now, part (2): If c = sqrt(3), find the maximum value of the area S of triangle ABC.The area S of a triangle can be given by (1/2)ab sin C, where a and b are sides, and C is the included angle.Given that c = sqrt(3), and from part (1), we know sin C = 2√2 / 3 and cos C = 1/3.We can use the Law of Cosines to relate a, b, and c:c² = a² + b² - 2ab cos CSubstituting c = sqrt(3), cos C = 1/3:(√3)² = a² + b² - 2ab*(1/3)So,3 = a² + b² - (2/3)abWe need to maximize the area S = (1/2)ab sin C = (1/2)ab*(2√2/3) = (√2/3)abSo, to maximize S, we need to maximize ab, given the constraint 3 = a² + b² - (2/3)ab.Let me denote ab as a variable, say, let’s set x = ab. We need to express the constraint in terms of x.But it's a bit tricky because the constraint is quadratic in a and b. Maybe we can use the method of Lagrange multipliers or complete the square.Alternatively, we can express a² + b² in terms of x.From the constraint:a² + b² = 3 + (2/3)ab = 3 + (2/3)xWe know that for any real numbers a and b, a² + b² ≥ 2ab (by AM ≥ GM). So,3 + (2/3)x ≥ 2xSimplify:3 ≥ 2x - (2/3)x = (6x - 2x)/3 = (4x)/3Multiply both sides by 3:9 ≥ 4x ⇒ x ≤ 9/4So, ab ≤ 9/4.Therefore, the maximum value of ab is 9/4.Thus, the maximum area S is (√2/3)*(9/4) = (9√2)/12 = (3√2)/4.So, the maximum area is 3√2 / 4.Wait, let me double-check that.We have S = (√2/3)ab, and ab ≤ 9/4, so S ≤ (√2/3)*(9/4) = (9√2)/12 = (3√2)/4. Yes, that seems correct.Alternatively, we can use the method of expressing a and b in terms of variables and use calculus to find the maximum.Let me consider a and b as variables with the constraint a² + b² - (2/3)ab = 3.We can set up the function to maximize: f(a,b) = abSubject to g(a,b) = a² + b² - (2/3)ab - 3 = 0Using Lagrange multipliers:∇f = λ∇gSo,df/da = b = λ*(2a - (2/3)b)df/db = a = λ*(2b - (2/3)a)So, we have:b = λ*(2a - (2/3)b) ...(1)a = λ*(2b - (2/3)a) ...(2)Let me solve these equations.From equation (1):b = λ*(2a - (2/3)b)Multiply both sides by 3 to eliminate fractions:3b = λ*(6a - 2b)Similarly, from equation (2):a = λ*(2b - (2/3)a)Multiply both sides by 3:3a = λ*(6b - 2a)So now, we have:3b = λ*(6a - 2b) ...(1a)3a = λ*(6b - 2a) ...(2a)Let me write these as:6a - 2b = 3b / λ ...(1b)6b - 2a = 3a / λ ...(2b)Let me denote 1/λ as k for simplicity.Then,6a - 2b = 3b k ...(1c)6b - 2a = 3a k ...(2c)Now, let's solve for a and b.From (1c):6a = 3b k + 2b ⇒ 6a = b(3k + 2)From (2c):6b = 3a k + 2a ⇒ 6b = a(3k + 2)So, from (1c): a = [b(3k + 2)] / 6From (2c): b = [a(3k + 2)] / 6Substitute a from (1c) into (2c):b = [ ( [b(3k + 2)] / 6 ) * (3k + 2) ] / 6Simplify:b = [ b(3k + 2)^2 ] / 36Assuming b ≠ 0 (since it's a side length), we can divide both sides by b:1 = (3k + 2)^2 / 36Multiply both sides by 36:36 = (3k + 2)^2Take square roots:±6 = 3k + 2So,Case 1: 3k + 2 = 6 ⇒ 3k = 4 ⇒ k = 4/3Case 2: 3k + 2 = -6 ⇒ 3k = -8 ⇒ k = -8/3But since k = 1/λ, and λ is a multiplier, we can consider both cases.Let's take Case 1: k = 4/3Then, from (1c): 6a - 2b = 3b*(4/3) ⇒ 6a - 2b = 4b ⇒ 6a = 6b ⇒ a = bSo, a = b.From the constraint equation:a² + a² - (2/3)a² = 3 ⇒ 2a² - (2/3)a² = 3 ⇒ (6a² - 2a²)/3 = 3 ⇒ (4a²)/3 = 3 ⇒ 4a² = 9 ⇒ a² = 9/4 ⇒ a = 3/2So, a = b = 3/2Thus, ab = (3/2)^2 = 9/4, which matches our earlier result.Case 2: k = -8/3From (1c): 6a - 2b = 3b*(-8/3) ⇒ 6a - 2b = -8b ⇒ 6a = -6b ⇒ a = -bBut since a and b are lengths, they must be positive, so a = -b is impossible. Thus, Case 2 is invalid.Therefore, the maximum occurs when a = b = 3/2, giving ab = 9/4, and thus the maximum area is (√2/3)*(9/4) = 3√2/4.So, the maximum area is 3√2/4.**Final Answer**(1) boxed{dfrac{2sqrt{2}}{3}} (2) boxed{dfrac{3sqrt{2}}{4}}

💡Okay, so I have this geometry problem about a trapezoid inscribed in a circle. Let me try to visualize it first. The trapezoid is named ABCD with bases AB and CD. Since it's inscribed in a circle, it must be an isosceles trapezoid because only isosceles trapezoids can be cyclic. That means the non-parallel sides AD and BC are equal in length, and the base angles are equal. Now, there's another circle ω that passes through points C and D. This circle intersects segments CA and CB at points A₁ and B₁, respectively. So, ω intersects CA at A₁ and CB at B₁. Then, points A₂ and B₂ are defined as the symmetric points of A₁ and B₁ with respect to the midpoints of segments CA and CB. That means if I take the midpoint of CA, A₂ is the reflection of A₁ over this midpoint. Similarly, B₂ is the reflection of B₁ over the midpoint of CB.The goal is to prove that points A, B, A₂, and B₂ lie on the same circle. So, I need to show that these four points are concyclic.Let me start by recalling some properties of cyclic quadrilaterals and reflections. Since ABCD is cyclic, opposite angles sum to 180 degrees. Also, the reflection over the midpoint of a segment essentially means that the midpoint is the center of symmetry for the two points. So, if M is the midpoint of CA, then MA₁ = MA₂, and similarly for B₁ and B₂ with the midpoint of CB.Maybe I can use power of a point or some properties of similar triangles here. Let me think about the power of point C with respect to circle ω. The power of C with respect to ω should be equal for both intersections CA₁ and CB₁. So, CA₁ * CA = CB₁ * CB. Wait, but since A₂ is the reflection of A₁ over the midpoint of CA, then CA₂ = CA - 2*CA₁ or something like that? Hmm, maybe not exactly. Let me think more carefully.If M is the midpoint of CA, then MA = MC. So, if A₂ is the reflection of A₁ over M, then MA₂ = MA₁, but in the opposite direction. So, CA₂ = 2*CM - CA₁? Wait, maybe I should express it in terms of vectors or coordinates.Alternatively, maybe I can use coordinate geometry. Let me assign coordinates to the points to make it more concrete. Let me place point C at the origin (0,0), and point D somewhere on the x-axis, say (d,0). Since ABCD is an isosceles trapezoid, points A and B will be symmetric with respect to the y-axis. Let me assign point A as (a,b) and point B as (-a,b). Then, since AB and CD are the bases, AB is parallel to CD, which lies on the x-axis.Now, the circle ω passes through C(0,0) and D(d,0), and intersects CA and CB at A₁ and B₁. Let me parametrize CA and CB. The line CA goes from C(0,0) to A(a,b), so any point on CA can be written as (ta, tb) for t between 0 and 1. Similarly, CB goes from C(0,0) to B(-a,b), so any point on CB can be written as (-ta, tb) for t between 0 and 1.Since ω passes through C and D, its equation can be written as x² + y² + px + qy + r = 0. But since it passes through C(0,0), substituting gives r = 0. So, the equation simplifies to x² + y² + px + qy = 0.It also passes through D(d,0), so substituting D into the equation gives d² + pd = 0, which implies p = -d. So, the equation becomes x² + y² - dx + qy = 0.Now, let's find the points A₁ and B₁ where ω intersects CA and CB. For point A₁ on CA, which is (ta, tb), substitute into the circle equation:(ta)² + (tb)² - d(ta) + q(tb) = 0t²(a² + b²) - t(da - qb) = 0t(t(a² + b²) - (da - qb)) = 0So, t = 0 corresponds to point C, and the other solution is t = (da - qb)/(a² + b²). Therefore, point A₁ is at ((da - qb)/(a² + b²) * a, (da - qb)/(a² + b²) * b).Similarly, for point B₁ on CB, which is (-ta, tb), substitute into the circle equation:(-ta)² + (tb)² - d(-ta) + q(tb) = 0t²(a² + b²) + t(da + qb) = 0t(t(a² + b²) + (da + qb)) = 0Again, t = 0 is point C, and the other solution is t = -(da + qb)/(a² + b²). Therefore, point B₁ is at ( (da + qb)/(a² + b²) * a, -(da + qb)/(a² + b²) * b ). Wait, actually, since it's (-ta, tb), the x-coordinate would be -ta, so substituting t gives:x = - [ (da + qb)/(a² + b²) ] * ay = [ (da + qb)/(a² + b²) ] * bSo, point B₁ is ( -a(da + qb)/(a² + b²), b(da + qb)/(a² + b²) )Now, points A₂ and B₂ are symmetric to A₁ and B₁ with respect to the midpoints of CA and CB, respectively.Let me find the midpoints first. The midpoint of CA is M1 = (a/2, b/2). Similarly, the midpoint of CB is M2 = (-a/2, b/2).To find A₂, which is the reflection of A₁ over M1, we can use the midpoint formula. If M1 is the midpoint between A₁ and A₂, then:M1 = ( (x₁ + x₂)/2, (y₁ + y₂)/2 )So, solving for A₂:x₂ = 2*M1_x - x₁y₂ = 2*M1_y - y₁Similarly for B₂:x₂ = 2*M2_x - x₁y₂ = 2*M2_y - y₁So, let's compute A₂.Coordinates of A₁: ( (da - qb)a/(a² + b²), (da - qb)b/(a² + b²) )Midpoint M1: (a/2, b/2)So,x₂ = 2*(a/2) - ( (da - qb)a/(a² + b²) ) = a - (a(da - qb))/(a² + b²)= [a(a² + b²) - a(da - qb)] / (a² + b²)= [a³ + ab² - a²d + aqb] / (a² + b²)Similarly,y₂ = 2*(b/2) - ( (da - qb)b/(a² + b²) ) = b - (b(da - qb))/(a² + b²)= [b(a² + b²) - b(da - qb)] / (a² + b²)= [a²b + b³ - b²d + qb²] / (a² + b²)Similarly, for B₂:Coordinates of B₁: ( -a(da + qb)/(a² + b²), b(da + qb)/(a² + b²) )Midpoint M2: (-a/2, b/2)So,x₂ = 2*(-a/2) - ( -a(da + qb)/(a² + b²) ) = -a + a(da + qb)/(a² + b²)= [ -a(a² + b²) + a(da + qb) ] / (a² + b²)= [ -a³ - ab² + a²d + aqb ] / (a² + b²)y₂ = 2*(b/2) - ( b(da + qb)/(a² + b²) ) = b - b(da + qb)/(a² + b²)= [ b(a² + b²) - b(da + qb) ] / (a² + b²)= [ a²b + b³ - b²d - qb² ] / (a² + b²)Now, we have coordinates for A₂ and B₂. The next step is to check if points A, B, A₂, and B₂ lie on the same circle. To do this, we can use the determinant condition for concyclic points.The determinant for four points (x₁,y₁), (x₂,y₂), (x₃,y₃), (x₄,y₄) to be concyclic is:|x₁ y₁ x₁² + y₁² 1||x₂ y₂ x₂² + y₂² 1||x₃ y₃ x₃² + y₃² 1||x₄ y₄ x₄² + y₄² 1| = 0But this determinant might be too cumbersome to compute directly with these coordinates. Maybe there's a better approach.Alternatively, since ABCD is cyclic, perhaps we can use some properties of cyclic quadrilaterals or harmonic division.Wait, another idea: since A₂ and B₂ are reflections of A₁ and B₁ over midpoints, maybe there's a homothety or inversion that maps A₁ to A₂ and B₁ to B₂, and preserves the circle through A and B.Alternatively, since A₂ is the reflection of A₁ over the midpoint of CA, then CA₂ = CA - 2*CA₁ or something like that? Wait, actually, since M1 is the midpoint, then CA₂ = 2*CM1 - CA₁. But CM1 is half of CA, so CA₂ = CA - CA₁. Similarly, CB₂ = CB - CB₁.But from the power of point C with respect to circle ω, we have CA₁ * CA = CB₁ * CB. So, CA₁ * CA = CB₁ * CB.Let me denote CA = c and CB = b. Then, CA₁ * c = CB₁ * b.But from the reflection, CA₂ = c - 2*CA₁ and CB₂ = b - 2*CB₁.Wait, is that correct? If M1 is the midpoint of CA, then CA₂ = 2*CM1 - CA₁. Since CM1 = c/2, then CA₂ = c - CA₁. Similarly, CB₂ = b - CB₁.So, CA₂ = c - CA₁ and CB₂ = b - CB₁.From the power of point C, CA₁ * CA = CB₁ * CB, so CA₁ * c = CB₁ * b.Let me denote CA₁ = k, then CA₂ = c - k, and CB₁ = (c*k)/b, so CB₂ = b - (c*k)/b = (b² - c*k)/b.Now, I need to relate these to show that A, B, A₂, B₂ lie on a circle.Maybe using power of a point from A and B with respect to the circle passing through A₂ and B₂.Alternatively, perhaps using angles. If I can show that angles at A and B subtended by A₂B₂ are equal or supplementary, then the points lie on a circle.Alternatively, maybe using spiral similarity or something.Wait, another approach: since ABCD is cyclic, angles at A and B are supplementary with angles at D and C. Maybe we can relate angles involving A₂ and B₂.Alternatively, maybe consider triangle CA₁A₂ and CB₁B₂. Since A₂ is reflection of A₁ over midpoint of CA, triangle CA₁A₂ is such that CA₁ = CA₂ and it's symmetric. Similarly for CB₁B₂.Wait, actually, since A₂ is reflection of A₁ over midpoint of CA, then CA₁ = CA₂ and the midpoint is M1. Similarly, CB₁ = CB₂.Wait, no, reflection over midpoint doesn't necessarily mean CA₁ = CA₂. It means that M1 is the midpoint between A₁ and A₂. So, CA₂ = 2*CM1 - CA₁. Since CM1 is half of CA, CA₂ = CA - CA₁.Similarly, CB₂ = CB - CB₁.So, CA₂ = CA - CA₁ and CB₂ = CB - CB₁.From power of point C with respect to ω: CA₁ * CA = CB₁ * CB.So, CA₁ * CA = CB₁ * CB ⇒ (CA - CA₂) * CA = (CB - CB₂) * CB.Let me write this as:(CA - CA₂) * CA = (CB - CB₂) * CBWhich simplifies to:CA² - CA * CA₂ = CB² - CB * CB₂Rearranged:CA² - CB² = CA * CA₂ - CB * CB₂Hmm, not sure if that helps directly.Alternatively, maybe consider the circle through A, B, A₂, B₂. Let me compute the power of point C with respect to this circle.If points A, B, A₂, B₂ lie on a circle, then the power of point C with respect to this circle should satisfy:CA * CA₂ = CB * CB₂From earlier, we have CA * CA₂ = CA * (CA - CA₁) = CA² - CA * CA₁Similarly, CB * CB₂ = CB * (CB - CB₁) = CB² - CB * CB₁But from power of point C with respect to ω, CA * CA₁ = CB * CB₁.So, CA² - CA * CA₁ = CB² - CB * CB₁ ⇒ CA² - (CA * CA₁) = CB² - (CB * CB₁)But since CA * CA₁ = CB * CB₁, let's denote this common value as k.Then, CA² - k = CB² - k ⇒ CA² = CB² ⇒ CA = CBBut in an isosceles trapezoid, CA = CB because it's symmetric. So, CA = CB.Therefore, CA² - k = CB² - k ⇒ same value.Thus, CA * CA₂ = CB * CB₂Therefore, the power of point C with respect to the circle through A, B, A₂, B₂ is equal for both CA and CB, which suggests that C lies on the radical axis of the two circles, but since C is fixed, it implies that the power condition is satisfied, meaning that A, B, A₂, B₂ lie on a circle.Wait, maybe I need to think differently. Since CA * CA₂ = CB * CB₂, then point C has equal power with respect to the circle passing through A, B, A₂, B₂. But since C is outside this circle, it means that CA * CA₂ = CB * CB₂ is the power of C with respect to the circle. Therefore, the points A, B, A₂, B₂ lie on a circle.Alternatively, using the power of point C, since CA * CA₂ = CB * CB₂, then by the converse of the power of a point theorem, points A, B, A₂, B₂ lie on a circle.Yes, that seems to be the key. Since the power of C with respect to the circle through A, B, A₂, B₂ is equal for both CA and CB, which are the lengths from C to A and B, it implies that A, B, A₂, B₂ are concyclic.So, putting it all together, by using the power of point C and the reflection properties, we can conclude that points A, B, A₂, and B₂ lie on the same circle.

💡First, I need to determine the total distance the second train must cover to completely pass the first train. This includes the lengths of both trains and the distance between them.Next, I'll calculate the relative speed at which the second train is approaching the first train by dividing the total distance by the time it takes to pass.Finally, since the trains are moving in the same direction, I'll find the speed of the first train by subtracting the relative speed from the speed of the second train.

💡First, I need to determine the time Javier spends on each activity. He starts by outlining his speech for 30 minutes. After that, he takes a 10-minute break.Next, he spends 28 minutes more on writing than he did on outlining. Since he outlined for 30 minutes, the writing time is 30 + 28 = 58 minutes.After writing, Javier realizes he needs to rewrite the opening of his speech, which takes an additional 15 minutes.Following the rewriting, he takes another 5-minute break to clear his mind.Finally, Javier practices his speech for half the time he spent writing, including the additional rewriting time. The total writing time is 58 + 15 = 73 minutes, so the practice time is 73 / 2 = 36.5 minutes.To find the total time Javier spends on his speech, including breaks, I add up all the individual times:30 minutes (outlining) + 10 minutes (first break) + 58 minutes (writing) + 15 minutes (rewriting) + 5 minutes (second break) + 36.5 minutes (practicing) = 154.5 minutes.