💡Okay, so I have this problem where the lateral surface of a cone is unfolded into a semicircle with a radius of 2. I need to find the volume of the cone. Hmm, let's see. I remember that when you unfold the lateral surface of a cone, you get a sector of a circle. In this case, it's a semicircle, which is half of a full circle. First, I should recall some formulas related to cones. The volume of a cone is given by ( V = frac{1}{3}pi r^2 h ), where ( r ) is the radius of the base and ( h ) is the height. So, I need to find both ( r ) and ( h ) to calculate the volume.Now, the lateral surface area of a cone is ( pi r l ), where ( l ) is the slant height. When this lateral surface is unfolded into a semicircle, the radius of that semicircle is actually the slant height ( l ) of the cone. The problem says the radius of the semicircle is 2, so that means ( l = 2 ).Wait, let me make sure. The semicircle has a radius of 2, which corresponds to the slant height of the cone. So, yes, ( l = 2 ). Next, the circumference of the base of the cone is related to the length of the arc of the semicircle. The circumference of the base is ( 2pi r ), and the length of the arc of the semicircle is half the circumference of a full circle with radius 2. So, the arc length is ( pi times 2 = 2pi ). Therefore, the circumference of the base of the cone is equal to the arc length of the semicircle. So, ( 2pi r = 2pi ). Dividing both sides by ( 2pi ), I get ( r = 1 ). Alright, so the radius of the base of the cone is 1. Now, I need to find the height ( h ). I remember that in a right circular cone, the slant height ( l ), radius ( r ), and height ( h ) form a right triangle, so by the Pythagorean theorem, ( l^2 = r^2 + h^2 ). Plugging in the known values, ( 2^2 = 1^2 + h^2 ), which simplifies to ( 4 = 1 + h^2 ). Subtracting 1 from both sides, I get ( h^2 = 3 ), so ( h = sqrt{3} ). Now that I have both ( r = 1 ) and ( h = sqrt{3} ), I can plug these into the volume formula. Calculating the volume: ( V = frac{1}{3}pi (1)^2 (sqrt{3}) = frac{1}{3}pi sqrt{3} ). Hmm, that seems right. Let me double-check my steps. The slant height is 2, the arc length is ( 2pi ), which equals the base circumference, giving ( r = 1 ). Then, using Pythagoras, I found ( h = sqrt{3} ). Plugging into the volume formula, I get ( frac{sqrt{3}}{3}pi ). Wait, is there another way to express ( frac{sqrt{3}}{3} )? It can also be written as ( frac{pi sqrt{3}}{3} ). I think both are correct, but maybe the problem expects it in a specific form. Alternatively, I can rationalize the denominator if needed, but in this case, ( frac{sqrt{3}}{3} ) is already simplified. So, I think that's the final answer.**Final Answer**The volume of the cone is boxed{dfrac{sqrt{3}}{3} pi}.

💡First, determine the amount Jason needs to borrow by subtracting the down payment from the total cost of the car.Amount to borrow = Total cost of the car - Down paymentAmount to borrow = 32,000 - 8,000Amount to borrow = 24,000Next, calculate the monthly interest rate by dividing the annual interest rate by 12.Monthly interest rate = Annual interest rate / 12Monthly interest rate = 5% / 12Monthly interest rate = 0.05 / 12Monthly interest rate = 0.00416667Then, calculate the monthly payment without interest by dividing the amount to borrow by the number of payments.Monthly payment without interest = Amount to borrow / Number of paymentsMonthly payment without interest = 24,000 / 48Monthly payment without interest = 500Finally, calculate the interest per month by multiplying the monthly payment without interest by the monthly interest rate, and add it to the monthly payment without interest to find the total monthly payment.Interest per month = Monthly payment without interest * Monthly interest rateInterest per month = 500 * 0.00416667Interest per month = 2.083335Total monthly payment = Monthly payment without interest + Interest per monthTotal monthly payment = 500 + 2.083335Total monthly payment = 502.083335Rounded to the nearest cent, Jason's monthly payments will be approximately boxed{502.08}.

💡First, determine the filling rates of each pipe. Pipe 1 fills the tank in 20 hours, so its rate is 1/20 per hour. Pipe 2 fills the tank in 30 hours, so its rate is 1/30 per hour.Next, calculate the combined filling rate of both pipes:1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12 per hour.Since there is a leak that allows one-third of the water to escape, only two-thirds of the supplied water remains in the tank. Therefore, the effective filling rate is:(2/3) * (1/12) = 2/36 = 1/18 per hour.Finally, to find the total time required to fill the tank at this effective rate:Time = 1 / (1/18) = 18 hours.

💡Alright, so I have this problem here where I need to find the equation of a line ( l_2 ) that passes through the point ( A(1, -1) ) and intersects another line ( l_1: 2x + y - 6 = 0 ) at point ( B ). The condition is that the distance between points ( A ) and ( B ) should be exactly 5 units. Hmm, okay, let me try to break this down step by step.First, I know that line ( l_2 ) passes through ( A(1, -1) ), so if I can find the slope of ( l_2 ), I can write its equation. But since I don't know the slope yet, maybe I can represent it in a general form. Let me recall that the equation of a line passing through a point ( (x_1, y_1) ) with slope ( m ) is ( y - y_1 = m(x - x_1) ). So, applying that here, the equation of ( l_2 ) would be ( y + 1 = m(x - 1) ).Now, this line ( l_2 ) intersects ( l_1 ) at point ( B ). To find the coordinates of ( B ), I need to solve the system of equations given by ( l_1 ) and ( l_2 ). Let me write them down:1. ( 2x + y - 6 = 0 ) (equation of ( l_1 ))2. ( y + 1 = m(x - 1) ) (equation of ( l_2 ))I can substitute equation 2 into equation 1 to find the x-coordinate of ( B ). Let me rearrange equation 2 to express ( y ) in terms of ( x ):( y = m(x - 1) - 1 )Now, substitute this into equation 1:( 2x + [m(x - 1) - 1] - 6 = 0 )Simplify this:( 2x + m(x - 1) - 1 - 6 = 0 )( 2x + mx - m - 7 = 0 )Combine like terms:( (2 + m)x - (m + 7) = 0 )Solving for ( x ):( (2 + m)x = m + 7 )( x = frac{m + 7}{2 + m} )Okay, so that's the x-coordinate of ( B ). Now, let's find the y-coordinate using equation 2:( y = m(x - 1) - 1 )Substitute ( x = frac{m + 7}{2 + m} ):( y = mleft( frac{m + 7}{2 + m} - 1 right) - 1 )Simplify inside the parentheses:( frac{m + 7}{2 + m} - 1 = frac{m + 7 - (2 + m)}{2 + m} = frac{m + 7 - 2 - m}{2 + m} = frac{5}{2 + m} )So, ( y = m cdot frac{5}{2 + m} - 1 = frac{5m}{2 + m} - 1 )Therefore, the coordinates of ( B ) are:( left( frac{m + 7}{2 + m}, frac{5m}{2 + m} - 1 right) )Hmm, let me write that more neatly:( B left( frac{m + 7}{m + 2}, frac{5m - (m + 2)}{m + 2} right) ) because ( frac{5m}{m + 2} - 1 = frac{5m - (m + 2)}{m + 2} = frac{4m - 2}{m + 2} )So, ( B left( frac{m + 7}{m + 2}, frac{4m - 2}{m + 2} right) )Now, I need the distance between ( A(1, -1) ) and ( B ) to be 5. The distance formula is:( |AB| = sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = 5 )Plugging in the coordinates:( sqrt{left( frac{m + 7}{m + 2} - 1 right)^2 + left( frac{4m - 2}{m + 2} + 1 right)^2} = 5 )Let me simplify each term inside the square root.First term: ( frac{m + 7}{m + 2} - 1 = frac{m + 7 - (m + 2)}{m + 2} = frac{5}{m + 2} )Second term: ( frac{4m - 2}{m + 2} + 1 = frac{4m - 2 + (m + 2)}{m + 2} = frac{5m}{m + 2} )So, plugging these back into the distance formula:( sqrt{left( frac{5}{m + 2} right)^2 + left( frac{5m}{m + 2} right)^2} = 5 )Factor out ( frac{5}{m + 2} ) from both terms inside the square root:( sqrt{left( frac{5}{m + 2} right)^2 (1 + m^2)} = 5 )Simplify the square root:( frac{5}{|m + 2|} sqrt{1 + m^2} = 5 )Divide both sides by 5:( frac{sqrt{1 + m^2}}{|m + 2|} = 1 )Multiply both sides by ( |m + 2| ):( sqrt{1 + m^2} = |m + 2| )Now, square both sides to eliminate the square root:( 1 + m^2 = (m + 2)^2 )Expand the right side:( 1 + m^2 = m^2 + 4m + 4 )Subtract ( m^2 ) from both sides:( 1 = 4m + 4 )Subtract 4 from both sides:( -3 = 4m )Divide both sides by 4:( m = -frac{3}{4} )Okay, so the slope ( m ) of line ( l_2 ) is ( -frac{3}{4} ). Therefore, the equation of ( l_2 ) is:( y + 1 = -frac{3}{4}(x - 1) )Let me write this in standard form. Multiply both sides by 4 to eliminate the fraction:( 4(y + 1) = -3(x - 1) )Expand both sides:( 4y + 4 = -3x + 3 )Bring all terms to one side:( 3x + 4y + 1 = 0 )So, the equation of line ( l_2 ) is ( 3x + 4y + 1 = 0 ).Wait a second, but earlier I considered the case where ( l_2 ) is vertical. Let me check if there's another possible line. If ( l_2 ) is vertical, its equation would be ( x = 1 ). Let me see where this intersects ( l_1 ):Substitute ( x = 1 ) into ( l_1 ):( 2(1) + y - 6 = 0 )( 2 + y - 6 = 0 )( y = 4 )So, point ( B ) would be ( (1, 4) ). The distance between ( A(1, -1) ) and ( B(1, 4) ) is:( |AB| = sqrt{(1 - 1)^2 + (4 - (-1))^2} = sqrt{0 + 25} = 5 )That's exactly the distance we need! So, ( x = 1 ) is another possible equation for ( l_2 ).Therefore, there are two possible lines: one with slope ( -frac{3}{4} ) and the vertical line ( x = 1 ).Let me just double-check my calculations to make sure I didn't make any mistakes. When I solved for ( m ), I got ( m = -frac{3}{4} ), which seems correct. And when I considered the vertical line, it also satisfied the distance condition. So, I think both solutions are valid.**Final Answer**The equations of line ( l_2 ) are boxed{x = 1} and boxed{3x + 4y + 1 = 0}.

💡Alright, so I've got this trapezoid problem here. Let me try to visualize it first. The trapezoid is named ABCD, with AB parallel to CD. Also, AC is perpendicular to CD. Hmm, okay, so AC is one of the diagonals, and it's perpendicular to the base CD. That must mean that triangle ACD is a right triangle, right? Because if AC is perpendicular to CD, then angle C is a right angle in triangle ACD.Given that CD is 10 units long, and they also gave me the tangents of angles D and B. Specifically, tan D is 2, and tan B is 1.25. I need to find the length of BC. Hmm, okay, let's break this down step by step.First, since AB is parallel to CD, and AC is perpendicular to CD, that should also make AC perpendicular to AB, right? Because if two lines are parallel, a line perpendicular to one will be perpendicular to the other. So, that means triangle ABC is also a right triangle, with AC perpendicular to AB. So, both triangles ACD and ABC are right triangles, which is helpful.Starting with triangle ACD, which is a right triangle at C. They gave me CD = 10 and tan D = 2. Tan of an angle in a right triangle is opposite over adjacent. So, in triangle ACD, angle D is at point D, so the sides opposite and adjacent to angle D would be AC and CD, respectively. So, tan D = AC / CD. Plugging in the numbers, that would be 2 = AC / 10. Solving for AC, I get AC = 2 * 10 = 20. So, AC is 20 units long.Now, moving on to triangle ABC, which is also a right triangle at C. They gave me tan B = 1.25. Again, tan of an angle is opposite over adjacent. In triangle ABC, angle B is at point B, so the sides opposite and adjacent to angle B would be AC and AB, respectively. So, tan B = AC / AB. Plugging in the values, 1.25 = 20 / AB. Solving for AB, I get AB = 20 / 1.25. Let me calculate that: 20 divided by 1.25 is the same as 20 * (4/5) because 1.25 is 5/4, so dividing by 5/4 is multiplying by 4/5. So, 20 * 4/5 is 16. Therefore, AB is 16 units long.Now, I need to find the length of BC. Since triangle ABC is a right triangle at C, I can use the Pythagorean theorem. The theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Here, BC is the hypotenuse because it's opposite the right angle at C. So, BC squared equals AB squared plus AC squared.Plugging in the numbers, BC² = AB² + AC² = 16² + 20². Calculating each term: 16 squared is 256, and 20 squared is 400. Adding those together, 256 + 400 equals 656. So, BC squared is 656. To find BC, I take the square root of 656. Let me see if I can simplify that. 656 divided by 16 is 41, so 656 is 16 times 41. Therefore, the square root of 656 is the square root of (16 * 41), which is 4 times the square root of 41. So, BC is 4√41 units long.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting with triangle ACD, tan D = 2, CD = 10, so AC = 20. That seems right. Then, in triangle ABC, tan B = 1.25, which is 5/4, so AC / AB = 5/4, meaning AB = (4/5)*AC = (4/5)*20 = 16. That also checks out.Then, using Pythagoras: BC² = 16² + 20² = 256 + 400 = 656. Square root of 656 is indeed 4√41 because 16*41 is 656. So, yeah, that seems correct.I think that's all. I don't see any mistakes in my reasoning. So, the length of BC is 4√41.**Final Answer**The length of ( BC ) is (boxed{4sqrt{41}}).

💡Okay, so I'm trying to understand this problem about the expected lifetime of a flashlight that uses two lightbulbs, one blue and one red. The flashlight fails when the last bulb burns out, which means the lifetime of the flashlight is the maximum of the lifetimes of the two bulbs. The problem states that the expected lifetime of the red bulb, η, is 4 years, and it wants to show that the expected lifetime of the flashlight, which is the maximum of ξ (blue bulb) and η (red bulb), is at least 4 years.Alright, let's break this down step by step. First, I need to recall some basic probability concepts. The maximum of two random variables, in this case, ξ and η, represents the later time when either of the two bulbs burns out. So, if I have two bulbs, one might last longer than the other, and the flashlight will stop working when the last one dies. That makes sense.Now, the problem mentions that max(ξ, η) is greater than or equal to η. That seems obvious because the maximum of two numbers is always at least as big as each of the numbers individually. So, if I have two lifetimes, the maximum one can't be shorter than either of the individual lifetimes. That makes sense intuitively.Next, the problem takes expectations on both sides. So, if max(ξ, η) ≥ η, then taking the expectation, we get E[max(ξ, η)] ≥ E[η]. That also seems straightforward because expectation is a linear operator, and if one random variable is always greater than or equal to another, their expectations should follow the same inequality.Given that E[η] = 4, it follows that E[max(ξ, η)] ≥ 4. Therefore, the expected lifetime of the flashlight is at least four years. That seems like a solid conclusion based on the given information.But wait, let me think if there's anything missing here. The problem doesn't specify anything about the distribution of ξ or η, just that E[η] = 4. So, does this conclusion hold regardless of the distribution of ξ? Hmm, I think it does because we're only using the fact that max(ξ, η) is always at least η, and taking expectations preserves that inequality.Is there a scenario where E[max(ξ, η)] could be exactly 4? Well, if ξ is always less than or equal to η, then max(ξ, η) would just be η, and thus E[max(ξ, η)] would be exactly 4. But if ξ sometimes is greater than η, then E[max(ξ, η)] would be greater than 4. So, the expected lifetime of the flashlight is at least 4 years, but it could be more depending on the distribution of ξ.I wonder if there's a way to calculate the exact expectation if we had more information about ξ and η. For example, if both ξ and η were independent and identically distributed, we could use properties of order statistics to find E[max(ξ, η)]. But since we don't have that information, we can only say that it's at least 4.Another thought: this seems related to the concept of the expectation of the maximum of two random variables. In general, E[max(X, Y)] can be calculated if we know the joint distribution of X and Y. But again, without more information, we can't compute it exactly here.So, to summarize my understanding: since the flashlight's lifetime is determined by the last bulb to burn out, which is the maximum of ξ and η, and since this maximum is always at least as large as η, the expectation of this maximum must be at least as large as the expectation of η, which is 4 years. Therefore, the expected lifetime of the flashlight is at least 4 years.I think that covers the reasoning. I don't see any flaws in this logic, and it seems to align with the principles of probability and expectation. Maybe I should also consider if there are any edge cases or special distributions where this might not hold, but I can't think of any off the top of my head. If ξ and η are both non-negative random variables, which they are since they represent lifetimes, then the inequality should hold.In conclusion, based on the given information and the properties of expectation, it's valid to state that the expected lifetime of the flashlight is at least four years.