answer：Okay, so I have this problem: Does there exist a natural number which, when divided by the sum of its digits, gives 2011 both as the quotient and the remainder? Hmm, that sounds a bit tricky, but let's try to break it down step by step.First, let me understand what the problem is asking. We need to find a natural number, let's call it ( n ), such that when you divide ( n ) by the sum of its digits, which I'll denote as ( s ), you get 2011 as both the quotient and the remainder. So, in mathematical terms, this can be written as:[ n = 2011 times s + 2011 ]Wait, is that right? Let me think. If you divide ( n ) by ( s ), the quotient is 2011 and the remainder is also 2011. So, according to the division algorithm, we have:[ n = q times s + r ]Where ( q ) is the quotient and ( r ) is the remainder. In this case, both ( q ) and ( r ) are 2011. So, substituting those values in, we get:[ n = 2011 times s + 2011 ]Yes, that seems correct. So, ( n ) is equal to 2011 times the sum of its digits plus 2011.Now, let's simplify this equation a bit. If I factor out 2011 from both terms on the right side, I get:[ n = 2011 times (s + 1) ]So, ( n ) is 2011 multiplied by one more than the sum of its digits. Interesting.Now, I need to find if such a natural number ( n ) exists. To do that, I should probably explore the relationship between ( n ) and ( s ), the sum of its digits.I recall that there's a property in number theory that says a number is congruent to the sum of its digits modulo 9. In other words:[ n equiv s pmod{9} ]This is because 10 is congruent to 1 modulo 9, so each digit contributes itself times 1 to the total modulo 9. So, the entire number is congruent to the sum of its digits modulo 9.Given that, let's apply this property to our equation. We have:[ n = 2011 times (s + 1) ]Taking both sides modulo 9, we get:[ n equiv 2011 times (s + 1) pmod{9} ]But we also know that:[ n equiv s pmod{9} ]So, setting these two congruences equal to each other:[ 2011 times (s + 1) equiv s pmod{9} ]Now, let's compute 2011 modulo 9 to simplify this equation. To find 2011 mod 9, I can sum the digits of 2011:2 + 0 + 1 + 1 = 4So, 2011 ≡ 4 mod 9.Substituting that back into the equation:[ 4 times (s + 1) equiv s pmod{9} ]Let's distribute the 4:[ 4s + 4 equiv s pmod{9} ]Now, subtract ( s ) from both sides:[ 3s + 4 equiv 0 pmod{9} ]So, we have:[ 3s + 4 equiv 0 pmod{9} ]Which simplifies to:[ 3s equiv -4 pmod{9} ]But -4 mod 9 is the same as 5 mod 9 (since -4 + 9 = 5). So:[ 3s equiv 5 pmod{9} ]Now, we need to solve for ( s ) in this congruence. To do that, we can find the multiplicative inverse of 3 modulo 9. However, 3 and 9 are not coprime (their greatest common divisor is 3), so 3 does not have a multiplicative inverse modulo 9. This means that the equation ( 3s equiv 5 pmod{9} ) has no solution because 5 is not divisible by 3.Wait, let me think about that again. If 3 and 9 are not coprime, then the equation ( 3s equiv 5 pmod{9} ) has a solution only if 5 is divisible by the gcd of 3 and 9, which is 3. But 5 is not divisible by 3, so there is no solution.Therefore, there is no natural number ( s ) that satisfies this congruence, which means there is no natural number ( n ) that satisfies the original condition.But just to be thorough, let me double-check my steps.1. I started by expressing ( n ) in terms of ( s ): ( n = 2011s + 2011 ).2. Then, I used the property that ( n equiv s pmod{9} ).3. Substituted ( n ) into the congruence and simplified to get ( 3s equiv 5 pmod{9} ).4. Concluded that since 3 and 9 are not coprime and 5 is not divisible by 3, there's no solution.Yes, that seems correct. Therefore, there does not exist a natural number ( n ) that satisfies the given condition.

answer：Okay, so I need to show that for any two real numbers a and b, the inequality a² + b² is greater than or equal to 2ab holds true. Hmm, let's see. I remember that there are some fundamental inequalities in algebra, and this one seems familiar. Maybe it's related to the AM-GM inequality or something like that? Wait, no, that's for the arithmetic mean being greater than or equal to the geometric mean. This seems a bit different.Let me think about how to approach this. Maybe I can start by rearranging the inequality to see if it simplifies into something I recognize. If I subtract 2ab from both sides, the inequality becomes a² + b² - 2ab ≥ 0. Hmm, that looks like a perfect square. Is that right? Let me check.Yes, I recall that (a - b)² equals a² - 2ab + b². So, a² + b² - 2ab is indeed equal to (a - b)². That means the inequality simplifies to (a - b)² ≥ 0. Now, since any real number squared is always non-negative, this inequality must hold true for all real numbers a and b. Wait, let me make sure I'm not missing anything here. Is there a case where (a - b)² could be negative? Well, squares of real numbers are always non-negative, so even if a and b are negative, their squares will still be positive or zero. So, no matter what values a and b take, (a - b)² will never be negative. That means the original inequality a² + b² ≥ 2ab is always true.But just to be thorough, let me test this with some specific numbers. Suppose a = 3 and b = 2. Then a² + b² is 9 + 4, which is 13, and 2ab is 2*3*2, which is 12. So, 13 ≥ 12, which is true. What if a and b are negative? Let's say a = -4 and b = -5. Then a² + b² is 16 + 25, which is 41, and 2ab is 2*(-4)*(-5) = 40. So, 41 ≥ 40, which is also true.What if a and b are equal? Let's take a = b = 7. Then a² + b² is 49 + 49 = 98, and 2ab is 2*7*7 = 98. So, 98 ≥ 98, which is still true. Okay, so in all these cases, the inequality holds.Is there a way this could fail? Let me think. If a and b are complex numbers, then maybe, but the problem specifies that a and b are real numbers. So, in the realm of real numbers, this should always hold.Another thought: maybe I can use the Cauchy-Schwarz inequality or something else, but I think the approach I took with completing the square is the most straightforward. It directly shows that the difference between a² + b² and 2ab is a perfect square, which is always non-negative.Just to recap, the steps are:1. Start with the inequality a² + b² ≥ 2ab.2. Subtract 2ab from both sides to get a² + b² - 2ab ≥ 0.3. Recognize that a² + b² - 2ab is equal to (a - b)².4. Since (a - b)² is always greater than or equal to zero, the inequality holds.This seems solid. I don't see any gaps in the logic. Maybe I can also think about the geometric interpretation. If I consider a and b as the lengths of the sides of a right triangle, then a² + b² would be the square of the hypotenuse. But I'm not sure if that's directly relevant here.Alternatively, thinking about graphs, if I plot the equation a² + b² = 2ab, it would represent a hyperbola or some conic section, but again, I'm not sure if that's necessary for this proof.I think the key insight here is recognizing that the expression a² + b² - 2ab can be rewritten as a perfect square, which simplifies the problem significantly. It's a common technique in algebra to complete the square or factor expressions to make inequalities or equations easier to handle.So, in conclusion, by rewriting the left-hand side of the inequality as a perfect square, we can clearly see that it is always non-negative, which proves that a² + b² is indeed greater than or equal to 2ab for all real numbers a and b.

answer：First, calculate the original cost per folder, which is 2.50. Then, determine the discount rate of 20%, which is 0.20 in decimal form. Next, calculate the discount amount per folder by multiplying the original price by the discount rate: 2.50 × 0.20 = 0.50. Subtract this discount from the original price to find the discounted price per folder: 2.50 - 0.50 = 2.00.To find the total cost without the discount, multiply the original price by the number of folders: 5 × 2.50 = 12.50. For the total cost with the discount, multiply the discounted price by the number of folders: 5 × 2.00 = 10.00. Finally, subtract the total cost with the discount from the total cost without the discount to determine the savings: 12.50 - 10.00 = 2.50.Thus, Karl could have saved 2.50 by waiting for the sale.

answer：Okay, so I have this trigonometry problem here, and I need to figure out the value of tan left(theta - frac{pi}{4}right) given that sin theta + cos theta = -frac{sqrt{10}}{5} and theta is in the interval left(frac{pi}{2}, piright). Hmm, let me try to break this down step by step.First, I remember that there are some trigonometric identities that relate sin theta and cos theta to tan theta or other trigonometric functions. Maybe I can square both sides of the given equation to use the Pythagorean identity. Let me try that.So, if I square both sides of sin theta + cos theta = -frac{sqrt{10}}{5}, I get:(sin theta + cos theta)^2 = left(-frac{sqrt{10}}{5}right)^2Expanding the left side, I have:sin^2 theta + 2sin theta cos theta + cos^2 theta = frac{10}{25}Simplifying the right side:sin^2 theta + 2sin theta cos theta + cos^2 theta = frac{2}{5}I know that sin^2 theta + cos^2 theta = 1, so substituting that in:1 + 2sin theta cos theta = frac{2}{5}Subtracting 1 from both sides:2sin theta cos theta = frac{2}{5} - 1Calculating the right side:2sin theta cos theta = frac{2}{5} - frac{5}{5} = -frac{3}{5}So, 2sin theta cos theta = -frac{3}{5}. I remember that 2sin theta cos theta is equal to sin 2theta, so:sin 2theta = -frac{3}{5}Alright, so I have sin 2theta = -frac{3}{5}. Now, I need to find tan left(theta - frac{pi}{4}right). I recall the tangent subtraction formula:tan(A - B) = frac{tan A - tan B}{1 + tan A tan B}In this case, A = theta and B = frac{pi}{4}. So, substituting:tan left(theta - frac{pi}{4}right) = frac{tan theta - tan frac{pi}{4}}{1 + tan theta tan frac{pi}{4}}I know that tan frac{pi}{4} = 1, so this simplifies to:tan left(theta - frac{pi}{4}right) = frac{tan theta - 1}{1 + tan theta}Okay, so I need to find tan theta to compute this. How can I find tan theta given that sin 2theta = -frac{3}{5}?I know that sin 2theta = frac{2tan theta}{1 + tan^2 theta}. Let me set t = tan theta for simplicity. Then:sin 2theta = frac{2t}{1 + t^2} = -frac{3}{5}So, setting up the equation:frac{2t}{1 + t^2} = -frac{3}{5}Cross-multiplying:10t = -3(1 + t^2)Expanding the right side:10t = -3 - 3t^2Bringing all terms to one side:3t^2 + 10t + 3 = 0Now, I have a quadratic equation in terms of t:3t^2 + 10t + 3 = 0Let me try to solve this quadratic. Using the quadratic formula:t = frac{-b pm sqrt{b^2 - 4ac}}{2a}Where a = 3, b = 10, and c = 3:t = frac{-10 pm sqrt{100 - 36}}{6} = frac{-10 pm sqrt{64}}{6} = frac{-10 pm 8}{6}So, two possible solutions:1. t = frac{-10 + 8}{6} = frac{-2}{6} = -frac{1}{3}2. t = frac{-10 - 8}{6} = frac{-18}{6} = -3So, tan theta could be -frac{1}{3} or -3. Hmm, but I need to figure out which one is correct given that theta is in the second quadrant, i.e., between frac{pi}{2} and pi.In the second quadrant, sin theta is positive and cos theta is negative. Therefore, tan theta = frac{sin theta}{cos theta} would be negative, which both solutions satisfy. But I need more information to determine which one is correct.Wait, maybe I can use the original equation sin theta + cos theta = -frac{sqrt{10}}{5} to check which value of tan theta is appropriate.Let me denote tan theta = t, so sin theta = frac{t}{sqrt{1 + t^2}} and cos theta = frac{1}{sqrt{1 + t^2}}. But since theta is in the second quadrant, cos theta is negative, so actually:sin theta = frac{t}{sqrt{1 + t^2}} (positive because sin is positive in the second quadrant)cos theta = -frac{1}{sqrt{1 + t^2}} (negative because cos is negative in the second quadrant)Therefore, sin theta + cos theta = frac{t}{sqrt{1 + t^2}} - frac{1}{sqrt{1 + t^2}} = frac{t - 1}{sqrt{1 + t^2}}We know this equals -frac{sqrt{10}}{5}, so:frac{t - 1}{sqrt{1 + t^2}} = -frac{sqrt{10}}{5}Let me square both sides to eliminate the square root:frac{(t - 1)^2}{1 + t^2} = frac{10}{25} = frac{2}{5}So:5(t - 1)^2 = 2(1 + t^2)Expanding the left side:5(t^2 - 2t + 1) = 2 + 2t^2Which becomes:5t^2 - 10t + 5 = 2 + 2t^2Bringing all terms to one side:5t^2 - 10t + 5 - 2 - 2t^2 = 0Simplify:3t^2 - 10t + 3 = 0Wait, this is the same quadratic equation I had earlier: 3t^2 + 10t + 3 = 0, but with a sign change on the middle term. Wait, no, actually, it's different. Wait, let me check my steps again.Wait, I had:5(t^2 - 2t + 1) = 2 + 2t^2Which is:5t^2 - 10t + 5 = 2 + 2t^2Subtracting 2 + 2t^2 from both sides:5t^2 - 10t + 5 - 2 - 2t^2 = 0Which simplifies to:3t^2 - 10t + 3 = 0So, the quadratic is 3t^2 - 10t + 3 = 0, which is different from the previous one. So, perhaps I made a mistake earlier.Wait, earlier I had:sin 2theta = -frac{3}{5}, which led to 2sin theta cos theta = -frac{3}{5}, which is correct.Then, I used the identity sin 2theta = frac{2t}{1 + t^2}, which is correct.So, setting that equal to -frac{3}{5}, I had:frac{2t}{1 + t^2} = -frac{3}{5}Cross-multiplying:10t = -3(1 + t^2)Which is:10t = -3 - 3t^2Bringing all terms to one side:3t^2 + 10t + 3 = 0Which is correct. So, why is the quadratic equation different when I used the other approach?Wait, in the second approach, I used the original equation sin theta + cos theta = -frac{sqrt{10}}{5} and expressed sin theta and cos theta in terms of t = tan theta, considering the quadrant.So, perhaps I made a mistake in the second approach.Wait, let's see:sin theta + cos theta = frac{t - 1}{sqrt{1 + t^2}} = -frac{sqrt{10}}{5}Then, squaring both sides:frac{(t - 1)^2}{1 + t^2} = frac{10}{25} = frac{2}{5}So, 5(t - 1)^2 = 2(1 + t^2)Expanding:5t^2 - 10t + 5 = 2 + 2t^2Bringing all terms to left:5t^2 - 10t + 5 - 2 - 2t^2 = 0Simplify:3t^2 - 10t + 3 = 0So, this is a different quadratic equation than the previous one. So, which one is correct?Wait, perhaps I made a mistake in the first approach when I set up the equation.Wait, in the first approach, I had:sin 2theta = frac{2t}{1 + t^2} = -frac{3}{5}Which led to 3t^2 + 10t + 3 = 0But in the second approach, I ended up with 3t^2 - 10t + 3 = 0So, these are two different quadratics. That suggests that I made a mistake somewhere.Wait, let me check the second approach again.I had:sin theta + cos theta = frac{t - 1}{sqrt{1 + t^2}} = -frac{sqrt{10}}{5}Wait, is that correct?Wait, sin theta = frac{t}{sqrt{1 + t^2}} and cos theta = -frac{1}{sqrt{1 + t^2}} because theta is in the second quadrant.So, sin theta + cos theta = frac{t}{sqrt{1 + t^2}} - frac{1}{sqrt{1 + t^2}} = frac{t - 1}{sqrt{1 + t^2}}Yes, that seems correct.So, setting that equal to -frac{sqrt{10}}{5}:frac{t - 1}{sqrt{1 + t^2}} = -frac{sqrt{10}}{5}Then, squaring both sides:frac{(t - 1)^2}{1 + t^2} = frac{10}{25} = frac{2}{5}So, 5(t - 1)^2 = 2(1 + t^2)Expanding:5t^2 - 10t + 5 = 2 + 2t^2Bringing all terms to left:5t^2 - 10t + 5 - 2 - 2t^2 = 0Simplify:3t^2 - 10t + 3 = 0So, that's correct. So, in the first approach, I had 3t^2 + 10t + 3 = 0, but in the second approach, I have 3t^2 - 10t + 3 = 0Hmm, that suggests that I made a mistake in the first approach.Wait, in the first approach, I had:sin 2theta = frac{2t}{1 + t^2} = -frac{3}{5}Which is correct.So, cross-multiplying:10t = -3(1 + t^2)Which is:10t = -3 - 3t^2Bringing all terms to one side:3t^2 + 10t + 3 = 0Yes, that's correct.But in the second approach, I have 3t^2 - 10t + 3 = 0So, which one is correct?Wait, perhaps I made a mistake in the second approach when squaring both sides.Wait, let me check:frac{t - 1}{sqrt{1 + t^2}} = -frac{sqrt{10}}{5}Squaring both sides:frac{(t - 1)^2}{1 + t^2} = frac{10}{25} = frac{2}{5}Yes, that's correct.So, 5(t - 1)^2 = 2(1 + t^2)Expanding:5t^2 - 10t + 5 = 2 + 2t^2Bringing all terms to left:5t^2 - 10t + 5 - 2 - 2t^2 = 0Simplify:3t^2 - 10t + 3 = 0So, that's correct.So, in the first approach, I had 3t^2 + 10t + 3 = 0, and in the second approach, I have 3t^2 - 10t + 3 = 0This is confusing. How can both be correct?Wait, perhaps I made a mistake in the first approach when I set up the equation.Wait, in the first approach, I had:sin 2theta = frac{2t}{1 + t^2} = -frac{3}{5}But wait, sin 2theta is negative, which is consistent with 2theta being in the third or fourth quadrant.But since theta is in the second quadrant, 2theta would be in the third or fourth quadrant.Wait, theta is between frac{pi}{2} and pi, so 2theta is between pi and 2pi, which is the third and fourth quadrants.So, sin 2theta is negative, which is consistent with the given value of -frac{3}{5}.So, that's correct.But then, in the first approach, solving 3t^2 + 10t + 3 = 0 gives t = -frac{1}{3} and t = -3In the second approach, solving 3t^2 - 10t + 3 = 0 gives different roots.Wait, let me solve 3t^2 - 10t + 3 = 0.Using quadratic formula:t = frac{10 pm sqrt{100 - 36}}{6} = frac{10 pm sqrt{64}}{6} = frac{10 pm 8}{6}So, two solutions:1. t = frac{10 + 8}{6} = frac{18}{6} = 32. t = frac{10 - 8}{6} = frac{2}{6} = frac{1}{3}So, in the second approach, t is either 3 or frac{1}{3}.But in the first approach, t is either -frac{1}{3} or -3.So, which one is correct?Wait, since theta is in the second quadrant, tan theta is negative, as sin is positive and cos is negative.Therefore, in the second approach, the solutions t = 3 and t = frac{1}{3} are positive, which would not be possible because tan theta should be negative.Therefore, the second approach must have an error.Wait, but in the second approach, I had:sin theta + cos theta = frac{t - 1}{sqrt{1 + t^2}} = -frac{sqrt{10}}{5}But since sin theta + cos theta is negative, and frac{t - 1}{sqrt{1 + t^2}} is negative, then t - 1 must be negative because the denominator is always positive.So, t - 1 < 0 implies t < 1. So, t is less than 1.But in the second approach, the solutions are t = 3 and t = frac{1}{3}, which are both greater than 0, but t = 3 is greater than 1, which contradicts t < 1.Wait, t = frac{1}{3} is less than 1, so that's possible.But in the first approach, t = -frac{1}{3} and t = -3, both less than 1, which is consistent.So, perhaps the second approach is flawed because when I squared both sides, I introduced extraneous solutions.Therefore, the correct solutions are from the first approach: t = -frac{1}{3} and t = -3.But now, how do I determine which one is correct?I need to check which value of t satisfies the original equation sin theta + cos theta = -frac{sqrt{10}}{5}.Let me test t = -frac{1}{3}.If t = -frac{1}{3}, then:sin theta = frac{t}{sqrt{1 + t^2}} = frac{-frac{1}{3}}{sqrt{1 + left(-frac{1}{3}right)^2}} = frac{-frac{1}{3}}{sqrt{1 + frac{1}{9}}} = frac{-frac{1}{3}}{sqrt{frac{10}{9}}} = frac{-frac{1}{3}}{frac{sqrt{10}}{3}} = -frac{1}{sqrt{10}}But wait, sin theta should be positive in the second quadrant. So, this is a problem.Wait, no, because I had sin theta = frac{t}{sqrt{1 + t^2}}, but since theta is in the second quadrant, sin theta is positive, so actually, sin theta = frac{|t|}{sqrt{1 + t^2}}, but since t is negative, it's frac{-t}{sqrt{1 + t^2}}.Wait, no, that's not correct. Let me think again.If t = tan theta = frac{sin theta}{cos theta}, and since sin theta is positive and cos theta is negative in the second quadrant, t must be negative.So, sin theta = frac{t}{sqrt{1 + t^2}}, but since t is negative, sin theta would be negative, which contradicts the fact that sin theta is positive in the second quadrant.Wait, that can't be right. So, perhaps I need to adjust the expressions for sin theta and cos theta when dealing with negative t.Wait, let me recall that sin theta = frac{t}{sqrt{1 + t^2}} is valid when theta is in the first quadrant. But in the second quadrant, sin theta is positive and cos theta is negative, so perhaps the correct expressions are:sin theta = frac{t}{sqrt{1 + t^2}} if t is positive, but since t is negative, we need to adjust.Wait, actually, no. The standard identity is sin theta = frac{tan theta}{sqrt{1 + tan^2 theta}} regardless of the quadrant, but we have to consider the sign based on the quadrant.So, in the second quadrant, sin theta is positive, so:sin theta = frac{|tan theta|}{sqrt{1 + tan^2 theta}}But since tan theta is negative, |tan theta| = -tan thetaSo, sin theta = frac{-tan theta}{sqrt{1 + tan^2 theta}}Similarly, cos theta = -frac{1}{sqrt{1 + tan^2 theta}}So, with t = tan theta, which is negative, we have:sin theta = frac{-t}{sqrt{1 + t^2}}cos theta = -frac{1}{sqrt{1 + t^2}}Therefore, sin theta + cos theta = frac{-t}{sqrt{1 + t^2}} - frac{1}{sqrt{1 + t^2}} = frac{-t - 1}{sqrt{1 + t^2}}Which is equal to -frac{sqrt{10}}{5}So, setting up the equation:frac{-t - 1}{sqrt{1 + t^2}} = -frac{sqrt{10}}{5}Multiply both sides by -1:frac{t + 1}{sqrt{1 + t^2}} = frac{sqrt{10}}{5}Now, squaring both sides:frac{(t + 1)^2}{1 + t^2} = frac{10}{25} = frac{2}{5}So, 5(t + 1)^2 = 2(1 + t^2)Expanding:5t^2 + 10t + 5 = 2 + 2t^2Bringing all terms to left:5t^2 + 10t + 5 - 2 - 2t^2 = 0Simplify:3t^2 + 10t + 3 = 0Which is the same quadratic as in the first approach.So, solving 3t^2 + 10t + 3 = 0, we get:t = frac{-10 pm sqrt{100 - 36}}{6} = frac{-10 pm 8}{6}So, t = frac{-10 + 8}{6} = -frac{2}{6} = -frac{1}{3}Or t = frac{-10 - 8}{6} = -frac{18}{6} = -3So, t = -frac{1}{3} or t = -3Now, let's test these values in the original equation.First, let's test t = -frac{1}{3}.Compute sin theta + cos theta:sin theta = frac{-t}{sqrt{1 + t^2}} = frac{-(-frac{1}{3})}{sqrt{1 + (frac{1}{3})^2}} = frac{frac{1}{3}}{sqrt{1 + frac{1}{9}}} = frac{frac{1}{3}}{sqrt{frac{10}{9}}} = frac{frac{1}{3}}{frac{sqrt{10}}{3}} = frac{1}{sqrt{10}}cos theta = -frac{1}{sqrt{1 + t^2}} = -frac{1}{sqrt{1 + frac{1}{9}}} = -frac{1}{frac{sqrt{10}}{3}} = -frac{3}{sqrt{10}}So, sin theta + cos theta = frac{1}{sqrt{10}} - frac{3}{sqrt{10}} = -frac{2}{sqrt{10}} = -frac{sqrt{10}}{5}Which matches the given equation. So, t = -frac{1}{3} is a valid solution.Now, let's test t = -3.Compute sin theta + cos theta:sin theta = frac{-t}{sqrt{1 + t^2}} = frac{-(-3)}{sqrt{1 + 9}} = frac{3}{sqrt{10}}cos theta = -frac{1}{sqrt{1 + t^2}} = -frac{1}{sqrt{10}}So, sin theta + cos theta = frac{3}{sqrt{10}} - frac{1}{sqrt{10}} = frac{2}{sqrt{10}} = frac{sqrt{10}}{5}But the given equation is sin theta + cos theta = -frac{sqrt{10}}{5}, so this is not valid.Therefore, t = -3 is an extraneous solution introduced when we squared both sides.So, the correct value is t = -frac{1}{3}.Therefore, tan theta = -frac{1}{3}.Now, going back to the tangent subtraction formula:tan left(theta - frac{pi}{4}right) = frac{tan theta - 1}{1 + tan theta}Substituting tan theta = -frac{1}{3}:tan left(theta - frac{pi}{4}right) = frac{-frac{1}{3} - 1}{1 + (-frac{1}{3})} = frac{-frac{4}{3}}{frac{2}{3}} = frac{-4}{3} times frac{3}{2} = -2So, tan left(theta - frac{pi}{4}right) = -2Therefore, the correct answer is D: -2

answer：First, I need to calculate John's total weekly income before the raise. He earns 40 from his job, 10 from his side gig, and 5 from dividends. Adding these together gives a total income of 55 per week.Next, I'll determine his total weekly income after the raise. His job income increased to 80, while his side gig and dividends remain unchanged at 10 and 5, respectively. Adding these amounts results in a total income of 95 per week.To find the percentage increase in his income, I'll use the formula for percentage increase: ((New Income - Old Income) / Old Income) × 100. Plugging in the numbers, the calculation becomes ((95 - 55) / 55) × 100, which simplifies to (40 / 55) × 100. This results in approximately 72.73%.Therefore, John's total income increased by approximately 72.73% after the raise.

answer：Okay, so I have this problem about a sequence where each term is one more than the previous term, starting from 1. I need to find the sum of the reciprocals of the products of consecutive terms, from the first term up to the 99th and 100th terms. Hmm, let me try to break this down.First, the sequence is defined by a_{n+1} = a_n + 1, and a_1 = 1. So, that means each term increases by 1. Let me write out the first few terms to see the pattern:a_1 = 1a_2 = a_1 + 1 = 1 + 1 = 2a_3 = a_2 + 1 = 2 + 1 = 3a_4 = a_3 + 1 = 3 + 1 = 4Wait, so it's just the sequence of natural numbers: 1, 2, 3, 4, ..., right? So, in general, a_n = n. That makes things simpler.Now, the problem asks for the sum: 1/(a_1 a_2) + 1/(a_2 a_3) + ... + 1/(a_{99} a_{100}). Since a_n = n, this becomes:1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*100)Okay, so each term is of the form 1/(k(k+1)) where k goes from 1 to 99. I remember that fractions like 1/(k(k+1)) can be broken down using partial fractions. Let me recall how that works.Partial fraction decomposition for 1/(k(k+1)) would be A/k + B/(k+1). Let's solve for A and B:1/(k(k+1)) = A/k + B/(k+1)Multiplying both sides by k(k+1):1 = A(k+1) + BkExpanding the right side:1 = Ak + A + BkCombine like terms:1 = (A + B)k + ANow, since this must hold for all k, the coefficients of like terms on both sides must be equal. On the left side, the coefficient of k is 0, and the constant term is 1. On the right side, the coefficient of k is (A + B), and the constant term is A.So, setting up the equations:A + B = 0 (coefficient of k)A = 1 (constant term)From the second equation, A = 1. Plugging that into the first equation:1 + B = 0 => B = -1So, the partial fractions are:1/(k(k+1)) = 1/k - 1/(k+1)Great, so each term in the sum can be written as the difference of two reciprocals. Let me rewrite the entire sum using this decomposition:Sum = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/99 - 1/100)Now, let's write out the terms to see if there's a pattern or cancellation:First term: 1/1 - 1/2Second term: 1/2 - 1/3Third term: 1/3 - 1/4...99th term: 1/99 - 1/100When we add all these up, let's see what happens:1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100Notice that most of the middle terms cancel out. The -1/2 cancels with the +1/2, the -1/3 cancels with the +1/3, and so on, all the way up to -1/99 canceling with +1/99. So, what's left is just the first term of the first expression and the last term of the last expression:Sum = 1/1 - 1/100Simplifying that:Sum = 1 - 1/100 = 99/100So, the value of the sum is 99/100.Wait, let me double-check to make sure I didn't make a mistake. Each term was correctly decomposed into partial fractions, and the cancellation seems to work as expected. The telescoping nature of the series means that all intermediate terms cancel out, leaving only the first and the last term. Yeah, that makes sense.I think that's the correct answer. It's always good to verify, especially when dealing with telescoping series, but I feel confident about this one.