answer：First, determine the work rates of A and B individually. A can complete the job in 15 days, so A's work rate is 1/15 of the job per day. Similarly, B's work rate is 1/20 of the job per day.Next, calculate their combined work rate by adding their individual rates:[frac{1}{15} + frac{1}{20} = frac{4}{60} + frac{3}{60} = frac{7}{60} text{ of the job per day}]Let ( d ) be the number of days they worked together. The amount of work completed together in ( d ) days is:[frac{7}{60} times d]Given that the fraction of the work left is 0.06666666666666665, which is approximately ( frac{1}{15} ), the fraction of the work completed is:[1 - frac{1}{15} = frac{14}{15}]Set up the equation to solve for ( d ):[frac{7}{60} times d = frac{14}{15}]Solve for ( d ):[d = frac{frac{14}{15}}{frac{7}{60}} = frac{14}{15} times frac{60}{7} = frac{14 times 60}{15 times 7} = frac{840}{105} = 8]Therefore, A and B worked together for 8 days.

answer：Okay, so I have this problem where a company increases the price of a product by 20%, and they want to increase their total income by 10% even though the demand might decrease. I need to figure out by what proportion the demand should decrease to meet this new income goal. The options are A. 1/10, B. 1/11, C. 1/12, D. 1/13.Alright, let me break this down step by step. First, let's recall that total income (or revenue) is calculated by multiplying the price of a product by the quantity sold (demand). So, if the price increases, but the quantity sold decreases, the total income will depend on the balance between these two changes.Let me denote the original price as P and the original quantity demanded as Q. Therefore, the original total income (I) would be:I = P * QNow, the company increases the price by 20%. So, the new price (P_new) is:P_new = P + 20% of P = P * (1 + 0.20) = 1.20 * PThey want the total income to increase by 10%. So, the new total income (I_new) should be:I_new = I + 10% of I = I * (1 + 0.10) = 1.10 * IBut since I = P * Q, we can write:I_new = 1.10 * P * QNow, the new income is also equal to the new price multiplied by the new quantity demanded (Q_new):I_new = P_new * Q_newSubstituting the values we have:1.10 * P * Q = 1.20 * P * Q_newHmm, okay, let me write that out:1.10 * P * Q = 1.20 * P * Q_newI notice that P appears on both sides of the equation, so I can divide both sides by P to simplify:1.10 * Q = 1.20 * Q_newNow, I can solve for Q_new:Q_new = (1.10 / 1.20) * QLet me compute 1.10 divided by 1.20. Well, 1.10 divided by 1.20 is the same as 11/12, because 1.10 is 11/10 and 1.20 is 12/10, so dividing them gives (11/10) / (12/10) = 11/12.So, Q_new = (11/12) * QThis means that the new quantity demanded is 11/12 of the original quantity. To find out the decrease in demand, I subtract Q_new from Q:Decrease in demand = Q - Q_new = Q - (11/12)Q = (1 - 11/12)Q = (1/12)QSo, the demand decreases by 1/12 of the original demand. Therefore, the proportion by which the demand must decrease is 1/12.Let me double-check my steps to make sure I didn't make a mistake. I started with the original income, increased the price by 20%, set up the equation for the new income, solved for the new quantity demanded, and then found the decrease in demand. Everything seems to add up. The calculations look correct, and I didn't skip any steps that could potentially lead to an error.So, the answer should be 1/12, which is option C.

answer：Okay, so I have this problem where I need to find the value of the expression:[frac{p}{pq + 2} + frac{q}{pr + 2} + frac{r}{qp + 2}]where ( p ), ( q ), and ( r ) are the roots of the polynomial equation ( x^3 - 6x^2 + 11x - 6 = 0 ). Hmm, let me think about how to approach this.First, I remember that for a cubic equation ( x^3 + ax^2 + bx + c = 0 ), the sums and products of the roots can be found using Vieta's formulas. In this case, the equation is ( x^3 - 6x^2 + 11x - 6 = 0 ), so comparing it to the standard form, I can see that:- The sum of the roots ( p + q + r = 6 )- The sum of the products of the roots two at a time ( pq + pr + qr = 11 )- The product of the roots ( pqr = 6 )Okay, so I have these three pieces of information. Now, looking at the expression I need to evaluate:[frac{p}{pq + 2} + frac{q}{pr + 2} + frac{r}{qp + 2}]This looks a bit complicated, but maybe I can manipulate it somehow using the information from Vieta's formulas.Let me try to rewrite each term in the expression. For the first term, ( frac{p}{pq + 2} ), I can factor out a ( p ) from the denominator if I multiply numerator and denominator by something. Wait, actually, let me think differently.If I consider each term, say ( frac{p}{pq + 2} ), I can write it as:[frac{p}{pq + 2} = frac{p}{pq + 2} times frac{r}{r} = frac{pr}{pqr + 2r}]Wait, that might not be helpful. Alternatively, maybe I can write the denominator in terms of ( pqr ), since I know ( pqr = 6 ).Let me try that. For the first term:[frac{p}{pq + 2} = frac{p}{pq + 2} = frac{p}{pq + 2} times frac{r}{r} = frac{pr}{pqr + 2r}]But ( pqr = 6 ), so substituting that in:[frac{pr}{6 + 2r}]Hmm, that seems a bit better. Similarly, for the second term:[frac{q}{pr + 2} = frac{q}{pr + 2} = frac{q}{pr + 2} times frac{q}{q} = frac{q^2}{pqr + 2q} = frac{q^2}{6 + 2q}]And the third term:[frac{r}{qp + 2} = frac{r}{qp + 2} = frac{r}{qp + 2} times frac{r}{r} = frac{r^2}{pqr + 2r} = frac{r^2}{6 + 2r}]Wait, so now the entire expression becomes:[frac{pr}{6 + 2r} + frac{q^2}{6 + 2q} + frac{r^2}{6 + 2r}]Hmm, that doesn't seem immediately helpful. Maybe I made a mistake in my manipulation. Let me try a different approach.Looking back at the original expression:[frac{p}{pq + 2} + frac{q}{pr + 2} + frac{r}{qp + 2}]I notice that each term has a similar structure. Maybe I can express each term as ( frac{p}{pq + 2} = frac{p}{pq + 2} times frac{r}{r} = frac{pr}{pqr + 2r} ), which again gives me ( frac{pr}{6 + 2r} ) since ( pqr = 6 ).Similarly, the other terms become ( frac{pq}{6 + 2q} ) and ( frac{qr}{6 + 2p} ). Wait, no, let me check that again.Wait, for the second term, ( frac{q}{pr + 2} ), if I multiply numerator and denominator by ( q ), I get ( frac{q^2}{pqr + 2q} = frac{q^2}{6 + 2q} ). Similarly, the third term ( frac{r}{qp + 2} ) becomes ( frac{r^2}{6 + 2p} ).So, the expression becomes:[frac{pr}{6 + 2r} + frac{q^2}{6 + 2q} + frac{r^2}{6 + 2p}]Hmm, this still looks complicated. Maybe I can factor out a 2 from the denominators:[frac{pr}{2(3 + r)} + frac{q^2}{2(3 + q)} + frac{r^2}{2(3 + p)}]So, factoring out 1/2:[frac{1}{2} left( frac{pr}{3 + r} + frac{q^2}{3 + q} + frac{r^2}{3 + p} right)]I'm not sure if this helps. Maybe I need to think of another way to manipulate the original expression.Let me consider the expression ( frac{p}{pq + 2} ). Since ( pq ) is a product of two roots, and I know ( pq + pr + qr = 11 ), maybe I can express ( pq ) as ( 11 - pr - qr ). But that might complicate things further.Alternatively, perhaps I can write each term as ( frac{p}{pq + 2} = frac{p}{pq + 2} ). Let me consider that ( pq = frac{pqr}{r} = frac{6}{r} ). So, substituting that in:[frac{p}{frac{6}{r} + 2} = frac{p}{frac{6 + 2r}{r}} = frac{pr}{6 + 2r}]Ah, that's the same as before. So, each term can be rewritten as ( frac{pr}{6 + 2r} ), ( frac{pq}{6 + 2q} ), and ( frac{qr}{6 + 2p} ).Wait, no, actually, for the first term, it's ( frac{pr}{6 + 2r} ), the second term is ( frac{pq}{6 + 2q} ), and the third term is ( frac{qr}{6 + 2p} ). So, the entire expression is:[frac{pr}{6 + 2r} + frac{pq}{6 + 2q} + frac{qr}{6 + 2p}]Hmm, maybe I can factor out a 2 from the denominators:[frac{pr}{2(3 + r)} + frac{pq}{2(3 + q)} + frac{qr}{2(3 + p)}]So, factoring out 1/2:[frac{1}{2} left( frac{pr}{3 + r} + frac{pq}{3 + q} + frac{qr}{3 + p} right)]Still not sure. Maybe I can think of each term as ( frac{pr}{3 + r} ). Let me consider that ( pr = frac{pqr}{q} = frac{6}{q} ). So, substituting that in:[frac{6}{q(3 + r)}]Similarly, the other terms become ( frac{6}{p(3 + q)} ) and ( frac{6}{r(3 + p)} ).So, the expression inside the parentheses becomes:[frac{6}{q(3 + r)} + frac{6}{p(3 + q)} + frac{6}{r(3 + p)}]Hmm, that seems a bit more manageable. Let me factor out the 6:[6 left( frac{1}{q(3 + r)} + frac{1}{p(3 + q)} + frac{1}{r(3 + p)} right)]Now, I need to evaluate this sum. Maybe I can find a common denominator or find a way to express this in terms of the known sums from Vieta's formulas.Alternatively, perhaps I can consider the expression ( frac{1}{q(3 + r)} ). Let me write it as ( frac{1}{q(3 + r)} = frac{1}{3q + qr} ). Similarly, the other terms are ( frac{1}{3p + pq} ) and ( frac{1}{3r + pr} ).So, the expression becomes:[6 left( frac{1}{3q + qr} + frac{1}{3p + pq} + frac{1}{3r + pr} right)]Hmm, maybe I can factor out a q from the first denominator:[frac{1}{q(3 + r)} = frac{1}{q(3 + r)}]But I don't see an immediate way to simplify this. Maybe I need to think differently.Let me go back to the original expression:[frac{p}{pq + 2} + frac{q}{pr + 2} + frac{r}{qp + 2}]I wonder if there's a symmetry or a substitution that can help. Maybe I can let ( x = p ), ( y = q ), ( z = r ) and see if I can find a pattern or a way to express the sum.Alternatively, perhaps I can consider the expression as a sum of fractions and try to combine them. Let me try to find a common denominator.The denominators are ( pq + 2 ), ( pr + 2 ), and ( qp + 2 ). Wait, actually, ( pq + 2 ) and ( qp + 2 ) are the same, so the denominators are ( pq + 2 ), ( pr + 2 ), and ( pq + 2 ). Wait, no, actually, the denominators are ( pq + 2 ), ( pr + 2 ), and ( qp + 2 ), which are all different unless ( p = r ), which I don't think is necessarily the case.Wait, actually, let me check the original polynomial. The polynomial is ( x^3 - 6x^2 + 11x - 6 = 0 ). Let me try to factor this polynomial to find the roots.Trying rational roots, possible roots are factors of 6, so 1, 2, 3, 6.Testing x=1: ( 1 - 6 + 11 - 6 = 0 ). So, x=1 is a root.Then, we can factor out (x - 1):Using polynomial division or synthetic division:Dividing ( x^3 - 6x^2 + 11x - 6 ) by (x - 1):Coefficients: 1 | -6 | 11 | -6Bring down 1.Multiply by 1: 1Add to next coefficient: -6 + 1 = -5Multiply by 1: -5Add to next coefficient: 11 + (-5) = 6Multiply by 1: 6Add to last coefficient: -6 + 6 = 0So, the quotient is ( x^2 - 5x + 6 ), which factors further into (x - 2)(x - 3).So, the roots are p=1, q=2, r=3.Wait, that's great! So, the roots are 1, 2, and 3. So, p, q, r are 1, 2, 3 in some order.Therefore, I can substitute p=1, q=2, r=3 into the expression.So, let's compute each term:First term: ( frac{p}{pq + 2} = frac{1}{1*2 + 2} = frac{1}{2 + 2} = frac{1}{4} )Second term: ( frac{q}{pr + 2} = frac{2}{1*3 + 2} = frac{2}{3 + 2} = frac{2}{5} )Third term: ( frac{r}{qp + 2} = frac{3}{2*1 + 2} = frac{3}{2 + 2} = frac{3}{4} )Wait, but hold on, the denominators are:- For the first term: pq + 2 = 1*2 + 2 = 4- For the second term: pr + 2 = 1*3 + 2 = 5- For the third term: qp + 2 = 2*1 + 2 = 4So, the expression becomes:[frac{1}{4} + frac{2}{5} + frac{3}{4}]Let me compute this:First, add ( frac{1}{4} + frac{3}{4} = 1 )Then, add ( 1 + frac{2}{5} = frac{5}{5} + frac{2}{5} = frac{7}{5} )Wait, that can't be right because earlier I thought the answer was 3/4. Did I make a mistake?Wait, no, because I substituted p=1, q=2, r=3, but the expression is:[frac{p}{pq + 2} + frac{q}{pr + 2} + frac{r}{qp + 2}]So, substituting p=1, q=2, r=3:First term: ( frac{1}{1*2 + 2} = frac{1}{4} )Second term: ( frac{2}{1*3 + 2} = frac{2}{5} )Third term: ( frac{3}{2*1 + 2} = frac{3}{4} )So, total is ( frac{1}{4} + frac{2}{5} + frac{3}{4} )Compute ( frac{1}{4} + frac{3}{4} = 1 ), then ( 1 + frac{2}{5} = frac{7}{5} ). Hmm, but earlier, using Vieta's formulas, I thought the answer was 3/4. There's a discrepancy here.Wait, maybe I made a mistake in the initial approach. Let me check.In the initial approach, I tried to manipulate the expression and ended up with ( frac{p + q + r}{8} = frac{6}{8} = frac{3}{4} ). But when I substituted the actual roots, I got ( frac{7}{5} ). That means one of the approaches is wrong.Wait, perhaps I made a mistake in the initial manipulation. Let me go back.Original expression:[frac{p}{pq + 2} + frac{q}{pr + 2} + frac{r}{qp + 2}]I tried to rewrite each term as ( frac{p^2}{pqr + 2p} ), which would be ( frac{p^2}{6 + 2p} ). Similarly for the others. Then, I thought that ( frac{p^2}{6 + 2p} = frac{p}{8} ), but that's not correct.Wait, let's see:If I have ( frac{p^2}{6 + 2p} ), that's not equal to ( frac{p}{8} ). That was a mistake.So, my initial approach was flawed. I thought that ( frac{p^2}{6p + 2p} = frac{p^2}{8p} = frac{p}{8} ), but that's incorrect because ( 6 + 2p ) is not ( 8p ). It's ( 6 + 2p ), which is different.So, that approach was wrong. Therefore, the correct way is to substitute the actual roots, which gives ( frac{7}{5} ).Wait, but let me double-check the substitution.Given p=1, q=2, r=3.First term: ( frac{1}{1*2 + 2} = frac{1}{4} )Second term: ( frac{2}{1*3 + 2} = frac{2}{5} )Third term: ( frac{3}{2*1 + 2} = frac{3}{4} )Adding them up: ( frac{1}{4} + frac{2}{5} + frac{3}{4} )Compute ( frac{1}{4} + frac{3}{4} = 1 ), then ( 1 + frac{2}{5} = frac{7}{5} ). So, the value is ( frac{7}{5} ).But wait, the initial polynomial has roots 1, 2, 3, so substituting them should give the correct result. Therefore, the initial approach was wrong, and the correct answer is ( frac{7}{5} ).But wait, let me check if I substituted correctly. The expression is:[frac{p}{pq + 2} + frac{q}{pr + 2} + frac{r}{qp + 2}]So, for p=1, q=2, r=3:First term: ( frac{1}{1*2 + 2} = frac{1}{4} )Second term: ( frac{2}{1*3 + 2} = frac{2}{5} )Third term: ( frac{3}{2*1 + 2} = frac{3}{4} )Yes, that's correct. So, the sum is ( frac{1}{4} + frac{2}{5} + frac{3}{4} = frac{1}{4} + frac{3}{4} + frac{2}{5} = 1 + frac{2}{5} = frac{7}{5} ).Therefore, the correct answer is ( frac{7}{5} ), not ( frac{3}{4} ) as I initially thought.Wait, but why did the initial approach give a different answer? Because I incorrectly manipulated the expression. I thought that ( frac{p}{pq + 2} = frac{p^2}{pqr + 2p} ), which is correct, but then I thought that ( pqr + 2p = 8p ), which is not correct. Instead, ( pqr = 6 ), so ( pqr + 2p = 6 + 2p ), not ( 8p ). Therefore, my initial approach was wrong.So, the correct way is to substitute the actual roots, which gives ( frac{7}{5} ).Wait, but let me think again. Maybe I can find another way to compute the expression without substituting the roots, just using Vieta's formulas.Let me denote the expression as S:[S = frac{p}{pq + 2} + frac{q}{pr + 2} + frac{r}{qp + 2}]I can write each term as:[frac{p}{pq + 2} = frac{p}{pq + 2} = frac{p}{pq + 2} times frac{r}{r} = frac{pr}{pqr + 2r} = frac{pr}{6 + 2r}]Similarly,[frac{q}{pr + 2} = frac{q}{pr + 2} times frac{q}{q} = frac{q^2}{pqr + 2q} = frac{q^2}{6 + 2q}]And,[frac{r}{qp + 2} = frac{r}{qp + 2} times frac{r}{r} = frac{r^2}{pqr + 2r} = frac{r^2}{6 + 2r}]So, S becomes:[S = frac{pr}{6 + 2r} + frac{q^2}{6 + 2q} + frac{r^2}{6 + 2r}]Wait, but this seems similar to what I did before. Let me try to write this as:[S = frac{pr}{6 + 2r} + frac{q^2}{6 + 2q} + frac{r^2}{6 + 2r}]Hmm, maybe I can combine the terms with the same denominator. Notice that the first and third terms have the same denominator ( 6 + 2r ). So,[frac{pr + r^2}{6 + 2r} = frac{r(p + r)}{6 + 2r}]Similarly, the second term is ( frac{q^2}{6 + 2q} ).So, S becomes:[frac{r(p + r)}{6 + 2r} + frac{q^2}{6 + 2q}]Hmm, not sure if this helps. Maybe I can express ( p + r ) in terms of the known sums. Since ( p + q + r = 6 ), then ( p + r = 6 - q ). So,[frac{r(6 - q)}{6 + 2r} + frac{q^2}{6 + 2q}]Hmm, still complicated. Maybe I can write ( 6 + 2r = 2(3 + r) ), so:[frac{r(6 - q)}{2(3 + r)} + frac{q^2}{2(3 + q)}]So, factoring out 1/2:[frac{1}{2} left( frac{r(6 - q)}{3 + r} + frac{q^2}{3 + q} right)]Hmm, maybe I can split the first fraction:[frac{r(6 - q)}{3 + r} = frac{6r - rq}{3 + r} = frac{6r}{3 + r} - frac{rq}{3 + r}]So, S becomes:[frac{1}{2} left( frac{6r}{3 + r} - frac{rq}{3 + r} + frac{q^2}{3 + q} right)]Hmm, not sure. Maybe I can consider that ( rq = frac{pqr}{p} = frac{6}{p} ), but that might not help.Alternatively, perhaps I can think of ( frac{6r}{3 + r} ) as ( 6 times frac{r}{3 + r} ). Let me denote ( frac{r}{3 + r} = frac{r}{3 + r} ). Similarly, ( frac{q}{3 + q} ).Wait, maybe I can write ( frac{r}{3 + r} = 1 - frac{3}{3 + r} ). Let me check:[frac{r}{3 + r} = frac{(3 + r) - 3}{3 + r} = 1 - frac{3}{3 + r}]Yes, that's correct. So,[frac{6r}{3 + r} = 6 left( 1 - frac{3}{3 + r} right) = 6 - frac{18}{3 + r}]Similarly, ( frac{q^2}{3 + q} ) can be written as ( q - frac{3q}{3 + q} ), because:[frac{q^2}{3 + q} = frac{q(3 + q) - 3q}{3 + q} = q - frac{3q}{3 + q}]So, substituting back into S:[frac{1}{2} left( 6 - frac{18}{3 + r} - frac{rq}{3 + r} + q - frac{3q}{3 + q} right)]Hmm, this is getting more complicated. Maybe I need to find another approach.Wait, perhaps I can consider the expression ( frac{p}{pq + 2} ). Let me write this as:[frac{p}{pq + 2} = frac{p}{pq + 2} = frac{p}{pq + 2} times frac{1}{p} times p = frac{1}{q + frac{2}{p}} times p]Wait, that might not help. Alternatively, maybe I can write ( pq + 2 = pq + 2 ), and since ( pq = frac{pqr}{r} = frac{6}{r} ), so:[pq + 2 = frac{6}{r} + 2 = frac{6 + 2r}{r}]Therefore,[frac{p}{pq + 2} = frac{p}{frac{6 + 2r}{r}} = frac{pr}{6 + 2r}]Which is what I had before. So, S becomes:[frac{pr}{6 + 2r} + frac{q^2}{6 + 2q} + frac{r^2}{6 + 2r}]Wait, maybe I can combine the terms with the same denominator:[frac{pr + r^2}{6 + 2r} + frac{q^2}{6 + 2q} = frac{r(p + r)}{6 + 2r} + frac{q^2}{6 + 2q}]Since ( p + r = 6 - q ), as before, this becomes:[frac{r(6 - q)}{6 + 2r} + frac{q^2}{6 + 2q}]Hmm, maybe I can write ( 6 + 2r = 2(3 + r) ) and ( 6 + 2q = 2(3 + q) ), so:[frac{r(6 - q)}{2(3 + r)} + frac{q^2}{2(3 + q)} = frac{1}{2} left( frac{r(6 - q)}{3 + r} + frac{q^2}{3 + q} right)]This seems similar to what I did earlier. Maybe I can consider that ( frac{r(6 - q)}{3 + r} = frac{r(6 - q)}{3 + r} ). Let me see if I can express this in terms of known sums.Alternatively, perhaps I can consider that ( 6 - q = p + r ), so:[frac{r(p + r)}{3 + r} = frac{rp + r^2}{3 + r}]Which is the same as before. Hmm.Wait, maybe I can write ( frac{rp + r^2}{3 + r} = r times frac{p + r}{3 + r} ). Since ( p + r = 6 - q ), this becomes:[r times frac{6 - q}{3 + r}]But I don't see an immediate way to simplify this further.Alternatively, perhaps I can consider that ( frac{6 - q}{3 + r} = frac{6 - q}{3 + r} ). Since ( p + q + r = 6 ), ( 6 - q = p + r ), so:[frac{p + r}{3 + r} = frac{p + r}{3 + r}]Hmm, not helpful.Wait, maybe I can think of ( frac{p + r}{3 + r} = frac{p}{3 + r} + frac{r}{3 + r} ). So,[r times left( frac{p}{3 + r} + frac{r}{3 + r} right) = frac{rp}{3 + r} + frac{r^2}{3 + r}]Which brings me back to where I was.Alternatively, perhaps I can consider that ( frac{rp}{3 + r} = frac{rp}{3 + r} ). Since ( rp = frac{pqr}{q} = frac{6}{q} ), so:[frac{6}{q(3 + r)}]Similarly, ( frac{r^2}{3 + r} = frac{r^2}{3 + r} ). Hmm, not sure.Wait, maybe I can consider that ( frac{r^2}{3 + r} = r - frac{3r}{3 + r} ). Let me check:[frac{r^2}{3 + r} = frac{r(3 + r) - 3r}{3 + r} = r - frac{3r}{3 + r}]Yes, that's correct. So,[frac{r^2}{3 + r} = r - frac{3r}{3 + r}]Similarly, ( frac{rp}{3 + r} = frac{6}{q(3 + r)} ), as before.So, putting it all together:[frac{rp}{3 + r} + frac{r^2}{3 + r} = frac{6}{q(3 + r)} + r - frac{3r}{3 + r}]Therefore, the expression inside the parentheses becomes:[frac{6}{q(3 + r)} + r - frac{3r}{3 + r} + frac{q^2}{3 + q}]Hmm, this is getting too convoluted. Maybe I need to abandon this approach and stick with substituting the actual roots, which gave me ( frac{7}{5} ).Alternatively, perhaps I can consider that since the roots are 1, 2, 3, the expression is symmetric, and I can compute it directly.Wait, but in the initial substitution, I got ( frac{7}{5} ), but when I tried to use Vieta's formulas, I got ( frac{3}{4} ). Clearly, one of them is wrong. Since substituting the actual roots is straightforward and gives a clear answer, I think that ( frac{7}{5} ) is the correct answer.Wait, but let me double-check the substitution again to make sure I didn't make a mistake.Given p=1, q=2, r=3.First term: ( frac{1}{1*2 + 2} = frac{1}{4} )Second term: ( frac{2}{1*3 + 2} = frac{2}{5} )Third term: ( frac{3}{2*1 + 2} = frac{3}{4} )Adding them up:( frac{1}{4} + frac{2}{5} + frac{3}{4} )Convert to common denominator, which is 20:( frac{5}{20} + frac{8}{20} + frac{15}{20} = frac{5 + 8 + 15}{20} = frac{28}{20} = frac{7}{5} )Yes, that's correct. So, the value is ( frac{7}{5} ).Therefore, the initial approach using Vieta's formulas was incorrect, and the correct answer is ( frac{7}{5} ).I think the mistake in the initial approach was incorrectly simplifying ( frac{p^2}{6 + 2p} ) to ( frac{p}{8} ), which is not valid. Instead, each term should be handled separately, and substituting the actual roots gives the correct result.

answer：Okay, so I have this math problem here: 4.3 plus 3.88. Hmm, I need to add these two decimal numbers together. Let me think about how to approach this. First, I remember that when adding decimals, it's important to line up the decimal points so that the digits in each place value are properly aligned. That way, I can add the tenths, hundredths, and so on correctly. So, I'll write down the numbers one above the other, making sure the decimal points are in line. 4.3+3.88------Wait, but 4.3 only has one digit after the decimal point, while 3.88 has two. I think I need to add a zero to 4.3 to make it 4.30 so that both numbers have the same number of decimal places. That should make the addition easier. Okay, so now it looks like this:4.30+3.88------Now, I'll start adding from the rightmost digit, which is the hundredths place. 0 (from 4.30) plus 8 (from 3.88) equals 8. So, I'll write down 8 in the hundredths place.Next, I'll move to the tenths place. 3 (from 4.30) plus 8 (from 3.88) equals 11. Hmm, that's more than one digit. I think I need to carry over the extra digit to the next place. So, I'll write down 1 in the tenths place and carry over 1 to the ones place.Now, I'll add the ones place. 4 (from 4.30) plus 3 (from 3.88) equals 7, and then I add the carried-over 1, making it 8. So, I'll write down 8 in the ones place.Putting it all together, the sum should be 8.18. Let me double-check to make sure I didn't make any mistakes. Starting from the right:- Hundredths place: 0 + 8 = 8- Tenths place: 3 + 8 = 11, write down 1, carry over 1- Ones place: 4 + 3 + 1 (carried over) = 8Yes, that seems correct. So, 4.3 plus 3.88 equals 8.18.

answer：Okay, so I have this problem where Steven subtracts the units digit from the tens digit for each two-digit number and then finds the sum of all his answers. I need to figure out what that total sum is. The options given are A) 30, B) 45, C) 55, D) 90, and E) 100. Hmm, let me think through this step by step.First, let's clarify what a two-digit number is. A two-digit number ranges from 10 to 99. Each of these numbers has a tens digit and a units digit. For example, in the number 47, 4 is the tens digit and 7 is the units digit. Steven is subtracting the units digit from the tens digit for each number, so for 47, he would calculate 4 - 7, which equals -3.Now, I need to find the sum of all these differences for every two-digit number from 10 to 99. That sounds like a lot of numbers to handle individually, so there must be a pattern or a formula to simplify this.Let me consider the range of two-digit numbers. The tens digit can be any number from 1 to 9, and the units digit can be any number from 0 to 9. For each tens digit, there are 10 possible units digits. So, for each tens digit, we'll have 10 numbers.Let's break it down:1. **Sum of Tens Digits:** For each tens digit from 1 to 9, it appears 10 times (once for each units digit). So, the total sum of all tens digits is: [ 10 times (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) ] Calculating the sum inside the parentheses: [ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = frac{9 times 10}{2} = 45 ] Therefore: [ 10 times 45 = 450 ]2. **Sum of Units Digits:** For each units digit from 0 to 9, it appears 9 times (once for each tens digit from 1 to 9). So, the total sum of all units digits is: [ 9 times (0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) ] Calculating the sum inside the parentheses: [ 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = frac{10 times 9}{2} = 45 ] Therefore: [ 9 times 45 = 405 ]3. **Difference Calculation:** Steven's answer for each two-digit number is the difference between the tens digit and the units digit. So, the total sum of all these differences is: [ text{Total Sum} = text{Sum of Tens Digits} - text{Sum of Units Digits} = 450 - 405 = 45 ]Wait a minute, so according to this calculation, the total sum is 45. That matches option B. But let me double-check to make sure I haven't missed anything.Let me consider a smaller range to test this method. Suppose we only consider two-digit numbers from 10 to 19. The tens digit is 1 for all these numbers, and the units digits range from 0 to 9.Calculating the differences:- 10: 1 - 0 = 1- 11: 1 - 1 = 0- 12: 1 - 2 = -1- 13: 1 - 3 = -2- 14: 1 - 4 = -3- 15: 1 - 5 = -4- 16: 1 - 6 = -5- 17: 1 - 7 = -6- 18: 1 - 8 = -7- 19: 1 - 9 = -8Adding these up:1 + 0 + (-1) + (-2) + (-3) + (-4) + (-5) + (-6) + (-7) + (-8) = 1 - 36 = -35Hmm, that's negative. But according to my earlier method, the sum of tens digits for 10-19 would be 10*1 = 10, and the sum of units digits would be 0+1+2+3+4+5+6+7+8+9 = 45. So, the difference would be 10 - 45 = -35, which matches the manual calculation. So, my method seems correct.Extending this to all two-digit numbers, the total sum is 45. Therefore, the answer should be B) 45.

answer：Okay, so I have this problem about triangle ABC where the three interior angles form an arithmetic sequence, and the reciprocals of the corresponding sides also form an arithmetic sequence. I need to find the angles A, B, and C.First, let me recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference. So, if the angles A, B, C are in an arithmetic sequence, I can denote them as A = α, B = α + d, and C = α + 2d, where α is the first term and d is the common difference.Since the sum of the interior angles of a triangle is always π (or 180 degrees), I can write:A + B + C = πSubstituting the expressions for A, B, and C:α + (α + d) + (α + 2d) = πSimplifying this:3α + 3d = πDivide both sides by 3:α + d = π/3So, that tells me that the middle angle B is π/3, which is 60 degrees. Therefore, the angles can be rewritten as:A = α, B = π/3, C = π/3 - αOkay, so now I have the angles in terms of α. Now, moving on to the sides. The reciprocals of the sides a, b, c form an arithmetic sequence. So, 1/a, 1/b, 1/c are in arithmetic sequence. That means:2/b = 1/a + 1/cHmm, so the reciprocal of the middle side is the average of the reciprocals of the other two sides.I remember from the Law of Sines that in any triangle, the sides are proportional to the sines of the opposite angles. So, a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle.Therefore, I can express the sides in terms of the sines of the angles:a = 2R sin Ab = 2R sin Bc = 2R sin CSo, substituting these into the reciprocals:1/a = 1/(2R sin A)1/b = 1/(2R sin B)1/c = 1/(2R sin C)Since 2/b = 1/a + 1/c, substituting the expressions above:2/(2R sin B) = 1/(2R sin A) + 1/(2R sin C)Simplify both sides by multiplying by 2R:2/(sin B) = 1/sin A + 1/sin CSo, 2/sin B = 1/sin A + 1/sin CWe already know that B = π/3, so sin B = sin(π/3) = √3/2. Therefore, 2/(√3/2) = 4/√3.So, 4/√3 = 1/sin A + 1/sin CBut we also have expressions for A and C in terms of α:A = αC = π/3 - αSo, sin A = sin αsin C = sin(π/3 - α)I can use the sine subtraction formula for sin(π/3 - α):sin(π/3 - α) = sin π/3 cos α - cos π/3 sin αWhich is:= (√3/2) cos α - (1/2) sin αSo, sin C = (√3/2) cos α - (1/2) sin αTherefore, 1/sin A + 1/sin C becomes:1/sin α + 1/[ (√3/2) cos α - (1/2) sin α ]So, let's compute this:1/sin α + 1/[ (√3/2 cos α - 1/2 sin α ) ]Let me write this as:1/sin α + 2/(√3 cos α - sin α )So, the equation becomes:4/√3 = 1/sin α + 2/(√3 cos α - sin α )Hmm, this looks a bit complicated. Maybe I can combine these fractions or find a common denominator.Let me denote x = sin α and y = cos α for simplicity.So, the equation becomes:4/√3 = 1/x + 2/(√3 y - x )But I know that x^2 + y^2 = 1, since sin^2 α + cos^2 α = 1.So, maybe I can express y in terms of x or vice versa.Alternatively, perhaps I can manipulate the equation:4/√3 = 1/x + 2/(√3 y - x )Let me combine the terms on the right-hand side:[ (√3 y - x ) + 2x ] / [x (√3 y - x ) ] = 4/√3Simplify numerator:√3 y - x + 2x = √3 y + xSo, numerator is √3 y + xDenominator is x (√3 y - x )Therefore:(√3 y + x ) / [x (√3 y - x ) ] = 4/√3Cross-multiplying:√3 (√3 y + x ) = 4 x (√3 y - x )Simplify left side:√3 * √3 y + √3 x = 3 y + √3 xRight side:4x √3 y - 4x^2So, equation becomes:3 y + √3 x = 4√3 x y - 4x^2Hmm, this is getting more complicated. Maybe I need another approach.Wait, perhaps instead of using the Law of Sines, I can use the Law of Cosines? Or maybe there's a trigonometric identity that can help here.Alternatively, maybe I can consider specific cases. Since the angles are in arithmetic progression, and one of them is π/3, maybe the triangle is equilateral? If all angles are π/3, then the sides are equal, so reciprocals would also be equal, which is a trivial arithmetic sequence.But the problem says the reciprocals form an arithmetic sequence, which could be non-trivial. So, maybe the triangle is equilateral? Let me check.If the triangle is equilateral, then all sides are equal, so 1/a, 1/b, 1/c are equal, which is an arithmetic sequence with common difference zero. So, that satisfies the condition.But the problem doesn't specify that the triangle is non-equilateral, so maybe the only solution is an equilateral triangle.But wait, let me think again. If the angles are in arithmetic progression and the reciprocals of the sides are in arithmetic progression, does that necessarily mean the triangle is equilateral?Suppose the triangle is not equilateral. Then, the sides are different, so their reciprocals are different. So, they form a non-trivial arithmetic sequence.But in that case, would the angles still be in arithmetic progression?Hmm, perhaps not necessarily. So, maybe the only solution is when all angles are equal, making the triangle equilateral.Alternatively, maybe there's another solution where the triangle is not equilateral but still satisfies both conditions.Let me try to solve the equation:4/√3 = 1/sin α + 2/(√3 cos α - sin α )Let me set t = tan α, so that sin α = t / sqrt(1 + t^2) and cos α = 1 / sqrt(1 + t^2).Substituting into the equation:4/√3 = sqrt(1 + t^2)/t + 2/(√3 * (1 / sqrt(1 + t^2)) - t / sqrt(1 + t^2))Simplify the second term:Denominator: [√3 - t] / sqrt(1 + t^2)So, 2 divided by that is 2 sqrt(1 + t^2) / (√3 - t)Therefore, the equation becomes:4/√3 = sqrt(1 + t^2)/t + 2 sqrt(1 + t^2)/(√3 - t)Let me factor out sqrt(1 + t^2):4/√3 = sqrt(1 + t^2) [1/t + 2/(√3 - t)]Let me compute the expression inside the brackets:1/t + 2/(√3 - t) = [ (√3 - t) + 2t ] / [ t (√3 - t) ] = (√3 + t ) / [ t (√3 - t) ]So, the equation becomes:4/√3 = sqrt(1 + t^2) * (√3 + t ) / [ t (√3 - t) ]This is getting quite involved. Maybe I can square both sides to eliminate the square root.But before that, let me denote s = t for simplicity.So,4/√3 = sqrt(1 + s^2) * (√3 + s ) / [ s (√3 - s) ]Square both sides:(16)/(3) = (1 + s^2) * (√3 + s )^2 / [ s^2 (√3 - s)^2 ]Let me compute (√3 + s)^2 and (√3 - s)^2:(√3 + s)^2 = 3 + 2√3 s + s^2(√3 - s)^2 = 3 - 2√3 s + s^2So, the equation becomes:16/3 = (1 + s^2)(3 + 2√3 s + s^2) / [ s^2 (3 - 2√3 s + s^2) ]Let me denote numerator as N and denominator as D:N = (1 + s^2)(3 + 2√3 s + s^2)D = s^2 (3 - 2√3 s + s^2)So, 16/3 = N/DCross-multiplying:16 D = 3 NCompute N:(1 + s^2)(3 + 2√3 s + s^2) = 1*(3 + 2√3 s + s^2) + s^2*(3 + 2√3 s + s^2)= 3 + 2√3 s + s^2 + 3 s^2 + 2√3 s^3 + s^4= 3 + 2√3 s + 4 s^2 + 2√3 s^3 + s^4Similarly, compute D:s^2 (3 - 2√3 s + s^2) = 3 s^2 - 2√3 s^3 + s^4So, 16 D = 16*(3 s^2 - 2√3 s^3 + s^4) = 48 s^2 - 32√3 s^3 + 16 s^43 N = 3*(3 + 2√3 s + 4 s^2 + 2√3 s^3 + s^4) = 9 + 6√3 s + 12 s^2 + 6√3 s^3 + 3 s^4So, setting 16 D = 3 N:48 s^2 - 32√3 s^3 + 16 s^4 = 9 + 6√3 s + 12 s^2 + 6√3 s^3 + 3 s^4Bring all terms to left side:48 s^2 - 32√3 s^3 + 16 s^4 - 9 - 6√3 s - 12 s^2 - 6√3 s^3 - 3 s^4 = 0Simplify term by term:s^4: 16 s^4 - 3 s^4 = 13 s^4s^3: -32√3 s^3 - 6√3 s^3 = -38√3 s^3s^2: 48 s^2 - 12 s^2 = 36 s^2s: -6√3 sconstants: -9So, the equation becomes:13 s^4 - 38√3 s^3 + 36 s^2 - 6√3 s - 9 = 0This is a quartic equation in s, which seems quite complicated. Maybe I can factor it or find rational roots.Let me try to see if s = √3 is a root:Plug s = √3:13*(√3)^4 - 38√3*(√3)^3 + 36*(√3)^2 - 6√3*(√3) - 9Compute each term:(√3)^4 = (3)^2 = 9So, 13*9 = 117(√3)^3 = 3√3So, -38√3*(3√3) = -38*3*3 = -342(√3)^2 = 3So, 36*3 = 108-6√3*(√3) = -6*3 = -18-9So, summing up:117 - 342 + 108 - 18 - 9 = (117 + 108) + (-342 -18 -9) = 225 - 369 = -144 ≠ 0Not a root.Try s = 1:13*1 - 38√3*1 + 36*1 - 6√3*1 -9 = 13 - 38√3 + 36 -6√3 -9 = (13 + 36 -9) + (-38√3 -6√3) = 40 - 44√3 ≈ 40 - 76.21 ≈ -36.21 ≠ 0Not a root.Try s = 3:13*81 - 38√3*27 + 36*9 - 6√3*3 -9= 1053 - 1026√3 + 324 - 18√3 -9= (1053 + 324 -9) + (-1026√3 -18√3)= 1368 - 1044√3 ≈ 1368 - 1808.3 ≈ -440.3 ≠ 0Not a root.This seems too complicated. Maybe I made a mistake earlier in the algebra.Alternatively, perhaps there's a simpler approach.Wait, going back to the equation:2/sin B = 1/sin A + 1/sin CWe know B = π/3, so sin B = √3/2, so 2/(√3/2) = 4/√3Thus, 4/√3 = 1/sin A + 1/sin CBut A + C = 2π/3, since A + B + C = π, and B = π/3.So, A + C = 2π/3Let me denote A = π/3 - x and C = π/3 + x, so that A + C = 2π/3.So, A = π/3 - x, C = π/3 + xTherefore, sin A = sin(π/3 - x) = sin π/3 cos x - cos π/3 sin x = (√3/2) cos x - (1/2) sin xSimilarly, sin C = sin(π/3 + x) = sin π/3 cos x + cos π/3 sin x = (√3/2) cos x + (1/2) sin xSo, 1/sin A + 1/sin C = 1/[ (√3/2 cos x - 1/2 sin x ) ] + 1/[ (√3/2 cos x + 1/2 sin x ) ]Let me compute this sum:Let me denote u = √3/2 cos x and v = 1/2 sin xSo, 1/(u - v) + 1/(u + v) = [ (u + v) + (u - v) ] / [ (u - v)(u + v) ] = (2u) / (u^2 - v^2)So, 1/sin A + 1/sin C = 2u / (u^2 - v^2)Compute u and v:u = √3/2 cos xv = 1/2 sin xSo, u^2 = (3/4) cos^2 xv^2 = (1/4) sin^2 xThus, u^2 - v^2 = (3/4 cos^2 x - 1/4 sin^2 x ) = (3 cos^2 x - sin^2 x)/4So, 1/sin A + 1/sin C = 2*(√3/2 cos x ) / [ (3 cos^2 x - sin^2 x)/4 ] = (√3 cos x ) / [ (3 cos^2 x - sin^2 x)/4 ] = 4√3 cos x / (3 cos^2 x - sin^2 x )Set this equal to 4/√3:4√3 cos x / (3 cos^2 x - sin^2 x ) = 4/√3Divide both sides by 4:√3 cos x / (3 cos^2 x - sin^2 x ) = 1/√3Multiply both sides by denominator:√3 cos x = (3 cos^2 x - sin^2 x ) / √3Multiply both sides by √3:3 cos x = 3 cos^2 x - sin^2 xBring all terms to one side:3 cos^2 x - sin^2 x - 3 cos x = 0Express sin^2 x in terms of cos^2 x:sin^2 x = 1 - cos^2 xSo,3 cos^2 x - (1 - cos^2 x) - 3 cos x = 0Simplify:3 cos^2 x -1 + cos^2 x - 3 cos x = 0Combine like terms:4 cos^2 x - 3 cos x -1 = 0So, we have a quadratic in cos x:4 cos^2 x - 3 cos x -1 = 0Let me solve for cos x:Let y = cos xThen, 4 y^2 - 3 y -1 = 0Using quadratic formula:y = [3 ± sqrt(9 + 16)] / 8 = [3 ± sqrt(25)] / 8 = [3 ±5]/8So, y = (3 +5)/8 = 8/8 =1 or y=(3 -5)/8= (-2)/8= -1/4So, cos x =1 or cos x= -1/4If cos x =1, then x=0, so A=π/3 -0=π/3, C=π/3 +0=π/3. So, all angles are π/3, which is the equilateral case.If cos x= -1/4, then x= arccos(-1/4). Let's compute x:x ≈ 104.48 degrees, but since x is an angle in a triangle, it must be less than π/2? Wait, no, because A and C are angles in the triangle, so A=π/3 -x and C=π/3 +x must be positive and less than π.So, π/3 -x >0 => x < π/3 ≈60 degreesBut arccos(-1/4)≈104.48 degrees, which is greater than π/3, so this would make A=π/3 -x negative, which is impossible.Therefore, cos x= -1/4 is not acceptable.Thus, the only solution is cos x=1, which gives x=0, leading to A=B=C=π/3.Therefore, the triangle must be equilateral, with all angles equal to 60 degrees.**Final Answer**The angles of triangle ( triangle ABC ) are each (boxed{dfrac{pi}{3}}).