answer：First, I need to determine the number of days the nomad actually travels in a year. A typical year has 365 days, and the nomad takes 45 days off for rest and reading. So, the number of travel days is 365 minus 45, which equals 320 days.Next, I know that the nomad travels 15 miles each day. To find the total distance traveled in a year, I multiply the number of travel days by the daily distance. That would be 320 days multiplied by 15 miles per day.Calculating that gives me 4,800 miles. Therefore, the nomad travels a total of 4,800 miles in a year.

answer：First, I need to calculate the total number of each type of medical supply that Dr. Lee is donating.For the stethoscopes, there are 5 boxes with 12 stethoscopes each. Multiplying 5 by 12 gives 60 stethoscopes.Next, for the blood pressure monitors, there are 7 boxes with 10 monitors each. Multiplying 7 by 10 results in 70 blood pressure monitors.Then, for the thermometers, there are 9 boxes with 15 thermometers each. Multiplying 9 by 15 equals 135 thermometers.Finally, to find the total number of medical supplies donated, I add up the totals of each category: 60 stethoscopes + 70 blood pressure monitors + 135 thermometers, which equals 265 medical supplies.

answer：Okay, so I have this problem about John and the decline in Pears Soap usage. It's split into two parts. Let me try to tackle them one by one.First, the function given is ( U(t) = 50 e^{-0.05t} ), where ( t ) is the number of years since 1950. So, in 1950, ( t = 0 ), and the usage is 50 million bars per year. As time goes on, the usage decreases exponentially because of the negative exponent.**Problem 1: Determine the year when the usage rate dropped below 1 million bars per year.**Alright, so we need to find the value of ( t ) when ( U(t) = 1 ). Let's set up the equation:( 1 = 50 e^{-0.05t} )Hmm, okay. I need to solve for ( t ). Let's divide both sides by 50 to isolate the exponential part:( frac{1}{50} = e^{-0.05t} )Which simplifies to:( 0.02 = e^{-0.05t} )Now, to solve for ( t ), I can take the natural logarithm of both sides. Remember that ( ln(e^{x}) = x ), so that should help.Taking ln on both sides:( ln(0.02) = ln(e^{-0.05t}) )Simplify the right side:( ln(0.02) = -0.05t )Now, solve for ( t ):( t = frac{ln(0.02)}{-0.05} )Let me compute ( ln(0.02) ). I know that ( ln(1) = 0 ), ( ln(e) = 1 ), and ( ln(0.02) ) is a negative number because 0.02 is less than 1.Calculating ( ln(0.02) ). Let me recall that ( ln(1/50) = ln(1) - ln(50) = 0 - ln(50) = -ln(50) ). So, ( ln(0.02) = -ln(50) ).I remember that ( ln(50) ) is approximately... let's see, ( ln(50) ) is between ( ln(25) = 3.218 ) and ( ln(100) = 4.605 ). Since 50 is halfway between 25 and 100 on a logarithmic scale, but actually, it's not linear. Let me compute it more accurately.Alternatively, I can use the calculator approximation. ( ln(50) approx 3.9120 ). So, ( ln(0.02) = -3.9120 ).So, plugging back into the equation:( t = frac{-3.9120}{-0.05} )The negatives cancel out, so:( t = frac{3.9120}{0.05} )Dividing 3.9120 by 0.05. Let me compute that. 0.05 goes into 3.9120 how many times?Well, 0.05 * 78 = 3.90, and 0.05 * 78.24 = 3.912. So, t ≈ 78.24 years.So, approximately 78.24 years after 1950. Since 0.24 of a year is roughly 0.24 * 365 ≈ 87.6 days, which is about 2 months and 27 days. So, roughly 78 years and 2 months.Therefore, adding 78 years to 1950 brings us to 2028. Adding 2 months would be around February or March 2028. But since the question asks for the year, we can say 2028.Wait, but let me double-check my calculation for ( ln(0.02) ). Maybe I was too quick. Let me compute it more accurately.Using a calculator, ( ln(0.02) ) is approximately -3.91202. So, that's correct.Then, ( t = (-3.91202)/(-0.05) = 78.2404 ). So, yes, 78.24 years.So, 1950 + 78 years is 2028. But wait, 1950 + 78 is 2028? Wait, 1950 + 70 is 2020, plus 8 is 2028. Yes, correct.But wait, hold on. Let me think again. The usage drops below 1 million bars per year in 2028. But let me verify with the function.Compute ( U(78) = 50 e^{-0.05*78} ).Compute exponent: -0.05*78 = -3.9.So, ( e^{-3.9} ). Let me compute that. ( e^{-3} ) is about 0.0498, ( e^{-4} ) is about 0.0183. So, ( e^{-3.9} ) is somewhere between 0.0183 and 0.0498.Let me compute it more accurately. 3.9 is 3 + 0.9.We know that ( e^{-3} ≈ 0.0498 ), and ( e^{-0.9} ≈ 0.4066 ). So, ( e^{-3.9} = e^{-3} * e^{-0.9} ≈ 0.0498 * 0.4066 ≈ 0.02025 ).So, ( U(78) = 50 * 0.02025 ≈ 1.0125 ) million bars per year.So, at t=78, the usage is approximately 1.0125 million, which is just above 1 million.So, we need to find when it drops below 1 million. So, t needs to be a bit more than 78.Wait, so maybe 78.24 years is the exact point where it crosses 1 million.But since we can't have a fraction of a year in the year, we need to see whether in 2028, the usage is still above 1 million, and in 2029, it's below.Wait, but 78.24 years is 78 years and about 3 months. So, 1950 + 78 years is 2028, and adding 3 months would be around April 2028.But the question is asking for the year when the usage rate dropped below 1 million. So, if in 2028, the usage is still above 1 million in the beginning, but drops below 1 million later in the year, so the year when it first drops below is 2028.Alternatively, if we consider the usage rate as a continuous function, the exact time is 78.24 years after 1950, which is mid-2028. So, the year would still be 2028.But let me confirm this with another method.Alternatively, let's compute U(78) and U(79).Compute U(78):( U(78) = 50 e^{-0.05*78} = 50 e^{-3.9} ≈ 50 * 0.02025 ≈ 1.0125 ) million.Compute U(79):( U(79) = 50 e^{-0.05*79} = 50 e^{-3.95} ).Compute ( e^{-3.95} ). Let's see, 3.95 is 3.9 + 0.05.We know ( e^{-3.9} ≈ 0.02025 ), and ( e^{-0.05} ≈ 0.9512 ). So, ( e^{-3.95} = e^{-3.9} * e^{-0.05} ≈ 0.02025 * 0.9512 ≈ 0.01926 ).Thus, ( U(79) = 50 * 0.01926 ≈ 0.963 ) million.So, in 2028 (t=78), the usage is ~1.0125 million, which is above 1 million.In 2029 (t=79), the usage is ~0.963 million, which is below 1 million.Therefore, the usage drops below 1 million in 2029.Wait, so which year is it? The question is a bit ambiguous. It says "the year when the usage rate dropped below 1 million bars per year."If we consider the continuous model, the exact time is around t=78.24, which is mid-2028. However, if we are talking about annual usage, perhaps we need to consider the usage at the end of each year.But the function U(t) is given as the usage rate in millions of bars per year. So, it's a continuous function, not necessarily discretized per year.But the question is about when it "dropped below 1 million bars per year." So, if it's a continuous decline, it crosses 1 million in the middle of 2028, so the year when it first drops below is 2028.But in reality, if we think in terms of annual usage, perhaps each year's usage is U(t) averaged over the year. But since it's a continuous function, it's more precise to say that it drops below 1 million partway through 2028, so the year is 2028.But let me see, the problem says "the year when the usage rate dropped below 1 million bars per year." So, if the rate is below 1 million at any point during the year, then the year is 2028.Alternatively, if we consider the usage at the end of each year, then in 2028, it's still above 1 million, and in 2029, it's below. So, depending on interpretation, it could be 2028 or 2029.But since the function is continuous, I think the correct answer is 2028, because the rate drops below 1 million during that year.But to be thorough, let me compute U(78.24):( t = 78.24 )( U(78.24) = 50 e^{-0.05*78.24} )Compute exponent: 0.05*78.24 = 3.912So, ( e^{-3.912} ). Let's compute that.We know that ( e^{-3.912} = e^{-3} * e^{-0.912} )( e^{-3} ≈ 0.0498 )( e^{-0.912} ). Let's compute that.We can write 0.912 as 0.9 + 0.012.( e^{-0.9} ≈ 0.4066 )( e^{-0.012} ≈ 1 - 0.012 + (0.012)^2/2 - ... ≈ 0.9881 ) (using Taylor series approximation)So, ( e^{-0.912} ≈ e^{-0.9} * e^{-0.012} ≈ 0.4066 * 0.9881 ≈ 0.4025 )Thus, ( e^{-3.912} ≈ 0.0498 * 0.4025 ≈ 0.02005 )So, ( U(78.24) = 50 * 0.02005 ≈ 1.0025 ) million.Wait, that's still just above 1 million. Hmm, did I do that correctly?Wait, 0.0498 * 0.4025.Let me compute 0.0498 * 0.4 = 0.019920.0498 * 0.0025 = 0.0001245So, total is approximately 0.01992 + 0.0001245 ≈ 0.0200445So, 50 * 0.0200445 ≈ 1.002225 million.So, still just above 1 million.Wait, so maybe I need a slightly higher t.Wait, let's compute t such that ( 50 e^{-0.05t} = 1 )So, ( e^{-0.05t} = 0.02 )Take natural log:( -0.05t = ln(0.02) ≈ -3.91202 )Thus, ( t = 3.91202 / 0.05 ≈ 78.2404 )So, t ≈ 78.2404 years.So, 78 years is 1950 + 78 = 2028.0.2404 years is approximately 0.2404 * 365 ≈ 87.7 days, so about March 2028.So, the usage rate drops below 1 million in March 2028. Therefore, the year is 2028.But let me check U(78.2404):( U(78.2404) = 50 e^{-0.05*78.2404} = 50 e^{-3.91202} )But ( e^{-3.91202} = 0.02 ), so 50 * 0.02 = 1. So, exactly 1 million.Wait, that's confusing. Earlier, when I computed with approximate values, I got 1.002225, but actually, it's exactly 1 million at t=78.2404.Wait, that must be because I used approximate values earlier. So, actually, at t=78.2404, it's exactly 1 million.So, the usage rate is 1 million at t≈78.24, which is mid-2028. So, before that time in 2028, it's above 1 million, and after that, it's below.Therefore, the year when the usage rate dropped below 1 million is 2028.But wait, in reality, the usage rate is a continuous function, so it's 1 million in mid-2028, so the year is 2028.But let me confirm with the exact calculation.If t = 78.2404, then 1950 + 78.2404 = 2028.2404, which is approximately March 2028.So, the usage rate is 1 million in March 2028, so the year is 2028.Therefore, the answer is 2028.**Problem 2: Calculate the total number of Pears Soap bars used from 1950 to 2000 using the given model.**Alright, so we need to find the total usage from t=0 to t=50 (since 2000 - 1950 = 50 years).The usage rate is given by ( U(t) = 50 e^{-0.05t} ) million bars per year.To find the total usage over a period, we need to integrate the usage rate over that time.So, total usage ( T ) is:( T = int_{0}^{50} U(t) dt = int_{0}^{50} 50 e^{-0.05t} dt )Let me compute this integral.First, factor out the constant 50:( T = 50 int_{0}^{50} e^{-0.05t} dt )The integral of ( e^{kt} ) is ( frac{1}{k} e^{kt} ). So, here, k = -0.05.Thus,( int e^{-0.05t} dt = frac{1}{-0.05} e^{-0.05t} + C = -20 e^{-0.05t} + C )Therefore, evaluating from 0 to 50:( T = 50 [ -20 e^{-0.05t} ]_{0}^{50} )Compute at t=50:( -20 e^{-0.05*50} = -20 e^{-2.5} )Compute at t=0:( -20 e^{0} = -20 * 1 = -20 )Thus,( T = 50 [ (-20 e^{-2.5}) - (-20) ] = 50 [ -20 e^{-2.5} + 20 ] )Factor out 20:( T = 50 * 20 [ -e^{-2.5} + 1 ] = 1000 (1 - e^{-2.5}) )Now, compute ( e^{-2.5} ). Let me recall that ( e^{-2} ≈ 0.1353 ), and ( e^{-0.5} ≈ 0.6065 ). So, ( e^{-2.5} = e^{-2} * e^{-0.5} ≈ 0.1353 * 0.6065 ≈ 0.0821 ).Thus, ( 1 - e^{-2.5} ≈ 1 - 0.0821 = 0.9179 ).Therefore, ( T ≈ 1000 * 0.9179 = 917.9 ) million bars.But let me compute ( e^{-2.5} ) more accurately.Using a calculator, ( e^{-2.5} ≈ 0.082085 ).So, ( 1 - 0.082085 = 0.917915 ).Thus, ( T = 1000 * 0.917915 ≈ 917.915 ) million bars.So, approximately 917.915 million bars used from 1950 to 2000.But let me write it as 917.915 million, which is 917,915,000 bars.But the question says "the total number of Pears Soap bars used", so we can write it as approximately 917.92 million bars.Alternatively, if we want to be precise, we can write it as 917.915 million, but usually, two decimal places are sufficient.So, the total usage is approximately 917.92 million bars.But let me verify the integral calculation again.We had:( T = 50 int_{0}^{50} e^{-0.05t} dt )Integral of ( e^{-0.05t} ) is ( -20 e^{-0.05t} ), correct.So,( T = 50 [ -20 e^{-0.05*50} + 20 e^{0} ] = 50 [ -20 e^{-2.5} + 20 ] )Which is 50 * 20 (1 - e^{-2.5}) = 1000 (1 - e^{-2.5})Yes, that's correct.So, 1000*(1 - e^{-2.5}) ≈ 1000*(1 - 0.082085) ≈ 1000*0.917915 ≈ 917.915 million.So, that's correct.Therefore, the total number of Pears Soap bars used from 1950 to 2000 is approximately 917.92 million bars.But wait, let me think about units. The function U(t) is in millions of bars per year. So, integrating over 50 years gives us millions of bars.So, yes, 917.92 million bars is correct.Alternatively, if we want to express it in numerical terms without the unit, it's 917,920,000 bars.But since the question says "the total number of Pears Soap bars", and the model is in millions, it's probably acceptable to leave it in millions.So, the answer is approximately 917.92 million bars.But let me compute it more accurately.Compute ( e^{-2.5} ):Using a calculator, ( e^{-2.5} ≈ 0.082085 )Thus, 1 - 0.082085 = 0.917915Multiply by 1000: 917.915So, 917.915 million bars.Rounded to three decimal places, it's 917.915, which is 917.915 million.Alternatively, if we want to round to the nearest whole number, it's 918 million bars.But the question doesn't specify, so probably 917.92 million is fine.But let me check if I did everything correctly.Wait, the integral was from 0 to 50, correct?Yes, because 1950 to 2000 is 50 years.And the integral of 50 e^{-0.05t} from 0 to 50 is 1000*(1 - e^{-2.5}), which is approximately 917.915 million.Yes, that seems correct.Alternatively, if I compute it step by step:Compute ( e^{-2.5} ):We can use the Taylor series for ( e^{-x} ) around x=0:( e^{-x} = 1 - x + x^2/2! - x^3/3! + x^4/4! - ... )But for x=2.5, it's better to compute it numerically.Alternatively, use the fact that ( e^{-2.5} = 1 / e^{2.5} ).Compute ( e^{2.5} ):We know that ( e^{2} ≈ 7.3891 ), and ( e^{0.5} ≈ 1.6487 ).So, ( e^{2.5} = e^{2} * e^{0.5} ≈ 7.3891 * 1.6487 ≈ 12.1825 ).Thus, ( e^{-2.5} ≈ 1 / 12.1825 ≈ 0.082085 ), which matches our previous calculation.Therefore, 1 - e^{-2.5} ≈ 0.917915.Multiply by 1000: 917.915 million.Yes, that's correct.So, the total usage is approximately 917.92 million bars.Therefore, the answers are:1. The year is 2028.2. The total usage is approximately 917.92 million bars.But let me write them in the required format.For problem 1, the year is 2028.For problem 2, the total usage is approximately 917.92 million bars.But let me check if the integral was correctly set up.Yes, because the usage rate is in millions per year, integrating over 50 years gives millions of bars.Yes, correct.So, I think that's it.

answer：First, I need to determine the annual carbon dioxide emissions for both gasoline cars and electric cars.A traditional gasoline car emits 4.6 metric tons of CO₂ per year, while an electric car emits 1.2 metric tons annually.Next, I'll calculate the total emissions for 15 gasoline cars by multiplying 4.6 metric tons by 15, which equals 69 metric tons.Then, I'll calculate the total emissions for 15 electric cars by multiplying 1.2 metric tons by 15, resulting in 18 metric tons.Finally, to find the reduction in CO₂ emissions, I'll subtract the total emissions of electric cars from the total emissions of gasoline cars: 69 metric tons minus 18 metric tons equals 51 metric tons.Therefore, selling 15 electric cars reduces annual carbon dioxide emissions by 51 metric tons compared to selling gasoline cars.

answer：First, determine the time each person spends on one page. The designer takes 3 hours per page, while the developer takes 1.5 times longer, which is 4.5 hours per page.Next, calculate the total time each person spends on all 12 pages. The designer will spend 3 hours multiplied by 12 pages, totaling 36 hours. The developer will spend 4.5 hours multiplied by 12 pages, totaling 54 hours.Since both are working simultaneously, the total time spent on the project is determined by the longer duration, which is 54 hours.

answer：First, I need to determine the total improvement in mood and cognitive abilities after three music therapy sessions.Each session improves mood by 20%, so for three sessions, the total mood improvement is 20% multiplied by 3, which equals 60%.Similarly, each session improves cognitive abilities by 15%, so for three sessions, the total cognitive improvement is 15% multiplied by 3, resulting in 45%.Therefore, after one week of three music therapy sessions, Emma's parent experiences a 60% improvement in mood and a 45% improvement in cognitive abilities.