💡First, I need to understand the given sequence. The initial terms are ( a_1 = 2 ) and ( a_2 = 5 ). The recursive formula provided is ( a_{n+1} = a_{n+2} + a_n ).To find ( a_6 ), I'll calculate each subsequent term step by step using the recursive formula.1. **Calculate ( a_3 ):** [ a_3 = a_2 - a_1 = 5 - 2 = 3 ]2. **Calculate ( a_4 ):** [ a_4 = a_3 - a_2 = 3 - 5 = -2 ]3. **Calculate ( a_5 ):** [ a_5 = a_4 - a_3 = -2 - 3 = -5 ]4. **Calculate ( a_6 ):** [ a_6 = a_5 - a_4 = -5 - (-2) = -5 + 2 = -3 ]After computing each term, I find that ( a_6 = -3 ).

💡Okay, so I've got these two math problems to solve. Let me take them one at a time because I want to make sure I understand each step properly. Starting with the first problem: Calculate (40 + left(frac{1}{6} - frac{2}{3} + frac{3}{4}right) times 12). Hmm, this looks like it involves fractions and order of operations. I remember that according to PEMDAS, I should handle the operations inside the parentheses first, then multiplication and division, and finally addition and subtraction. Alright, so the expression inside the parentheses is (frac{1}{6} - frac{2}{3} + frac{3}{4}). I need to simplify this part before multiplying by 12. To add or subtract fractions, they need a common denominator. Let me find the least common denominator (LCD) for 6, 3, and 4. The denominators are 6, 3, and 4. The multiples of 6 are 6, 12, 18, etc. Checking if 12 is divisible by 3 and 4: yes, because 12 divided by 3 is 4 and 12 divided by 4 is 3. So, 12 is the LCD.Now, I'll convert each fraction to have 12 as the denominator. For (frac{1}{6}): Multiply numerator and denominator by 2 to get (frac{2}{12}).For (frac{2}{3}): Multiply numerator and denominator by 4 to get (frac{8}{12}).For (frac{3}{4}): Multiply numerator and denominator by 3 to get (frac{9}{12}).So, substituting back into the expression: (frac{2}{12} - frac{8}{12} + frac{9}{12}).Now, let's perform the operations from left to right.First, (frac{2}{12} - frac{8}{12}) equals (-frac{6}{12}).Then, (-frac{6}{12} + frac{9}{12}) equals (frac{3}{12}).Simplifying (frac{3}{12}) gives (frac{1}{4}).So, the expression inside the parentheses simplifies to (frac{1}{4}).Now, multiply this by 12: (frac{1}{4} times 12 = 3).Finally, add this result to 40: (40 + 3 = 43).Okay, that seems straightforward. Let me double-check my steps to make sure I didn't make a mistake. I converted all fractions to have a common denominator, performed the addition and subtraction correctly, and then multiplied by 12 before adding to 40. It all adds up, so I think the answer is 43.Moving on to the second problem: Calculate (-1^{2} times (-5) + [(-3)^{2} + 2 times (-5)]). This one looks a bit trickier because of the exponents and the order of operations. I need to be careful with the signs here.First, let's break it down. There are two main parts: (-1^{2} times (-5)) and ([(-3)^{2} + 2 times (-5)]). I'll handle each part separately and then combine them.Starting with (-1^{2} times (-5)). Wait, is the exponentiation applied before the negative sign? I think so, because exponentiation has higher precedence than the unary minus. So, (1^{2}) is 1, and then the negative sign makes it (-1). So, (-1 times (-5)) equals 5. Okay, that part is done.Now, the second part: ([(-3)^{2} + 2 times (-5)]). Let's handle the exponent first. ((-3)^{2}) is positive 9 because a negative times a negative is positive. Then, (2 times (-5)) is (-10). So, adding those together: (9 + (-10)) equals (-1).Now, putting it all together: the first part was 5, and the second part is (-1). So, (5 + (-1)) equals 4.Wait, let me make sure I did that correctly. The first part was (-1^{2} times (-5)). If I interpret it as (- (1^{2}) times (-5)), that would be (-1 times (-5) = 5). Alternatively, if it were ((-1)^{2} times (-5)), that would be (1 times (-5) = -5). Which interpretation is correct? I think it's the first one because exponentiation precedes the unary minus unless there are parentheses. So, since it's (-1^{2}), it's (- (1^{2})), which is (-1). So, (-1 times (-5) = 5). That seems right.Then, in the second part, ((-3)^{2}) is definitely 9, and (2 times (-5)) is (-10). So, 9 plus (-10) is indeed (-1). Adding that to the first part: 5 + (-1) = 4. Yeah, that checks out.I think I've got both problems solved correctly. Let me just recap:Problem 1: Simplified the fractions, did the arithmetic step by step, ended up with 43.Problem 2: Handled the exponents carefully, made sure about the order of operations, ended up with 4.I feel confident about these answers now.

💡Okay, so I have this inequality to prove: given real numbers a, b, c, d with a > b > c > d, I need to show that 1/(a - b) + 1/(b - c) + 1/(c - d) is greater than or equal to 9/(a - d). Hmm, interesting. Let me try to figure out how to approach this.First, I notice that the denominators in the left-hand side are the differences between consecutive variables, and the right-hand side has the difference between the first and last variables. Maybe there's a way to relate these differences somehow. I remember that inequalities often involve techniques like the Cauchy-Schwarz inequality or the AM-HM inequality. Maybe one of those could be useful here. Let me recall what those are.The Cauchy-Schwarz inequality in its basic form states that for any real numbers, (sum of a_i*b_i)^2 ≤ (sum of a_i^2)*(sum of b_i^2). But I'm not sure how to apply that directly here. Maybe if I set up the terms appropriately.Alternatively, the AM-HM inequality says that for positive real numbers, the arithmetic mean is always greater than or equal to the harmonic mean. The harmonic mean is the reciprocal of the average of the reciprocals. So, if I can express the left-hand side as a harmonic mean, maybe that could help.Wait, let's see. The left-hand side is 1/(a - b) + 1/(b - c) + 1/(c - d). If I consider these three terms, maybe I can use the AM-HM inequality on them. The harmonic mean of three numbers x, y, z is 3/(1/x + 1/y + 1/z). So, the harmonic mean here would be 3 divided by the left-hand side. But the right-hand side is 9/(a - d). Hmm, how does that relate? Let me think. The sum (a - b) + (b - c) + (c - d) is equal to a - d. So, the denominators in the left-hand side add up to a - d. Maybe that's a clue.Let me denote x = a - b, y = b - c, z = c - d. Then, x, y, z are all positive real numbers because a > b > c > d. So, the left-hand side becomes 1/x + 1/y + 1/z, and the right-hand side becomes 9/(x + y + z). So, the inequality is 1/x + 1/y + 1/z ≥ 9/(x + y + z).Oh! That looks familiar. That's exactly the statement of the AM-HM inequality for three variables. The AM-HM inequality states that (x + y + z)/3 ≥ 3/(1/x + 1/y + 1/z). If I rearrange that, it becomes 1/x + 1/y + 1/z ≥ 9/(x + y + z). So, that's exactly the inequality I need to prove. Therefore, by the AM-HM inequality, the given inequality holds. Equality occurs when x = y = z, which translates back to a - b = b - c = c - d. So, when the differences between consecutive terms are equal, the equality holds.Wait, let me make sure I didn't skip any steps. I defined x, y, z as the differences, which are positive. Then, the inequality is in terms of these x, y, z. Since they are positive, the AM-HM inequality applies. So, yes, it seems correct.Alternatively, could I have used Cauchy-Schwarz here? Let me see. If I consider the Cauchy-Schwarz inequality in the form (1/x + 1/y + 1/z)(x + y + z) ≥ (1 + 1 + 1)^2 = 9. That's another way to write it. So, (1/x + 1/y + 1/z) ≥ 9/(x + y + z). So, that's another way to see it using Cauchy-Schwarz.Either way, whether I think of it as AM-HM or Cauchy-Schwarz, the inequality holds. So, that's the proof.But just to be thorough, let me write it out step by step.1. Let x = a - b, y = b - c, z = c - d. Since a > b > c > d, x, y, z are positive real numbers.2. The left-hand side of the inequality is 1/x + 1/y + 1/z.3. The right-hand side is 9/(a - d) = 9/(x + y + z).4. By the AM-HM inequality, (x + y + z)/3 ≥ 3/(1/x + 1/y + 1/z).5. Rearranging this gives 1/x + 1/y + 1/z ≥ 9/(x + y + z).6. Therefore, 1/(a - b) + 1/(b - c) + 1/(c - d) ≥ 9/(a - d).Yes, that seems solid. I think I've got it.**Final Answer**The inequality is proven, and the result is boxed{frac{1}{a-b} + frac{1}{b-c} + frac{1}{c-d} geqslant frac{9}{a-d}}.

💡First, identify the coordinates of Point ( C ), which are ( (3, k) ).When Point ( C ) is reflected over the ( y )-axis, the ( x )-coordinate changes sign, resulting in the coordinates of Point ( D ) being ( (-3, k) ).Next, increase the ( y )-coordinate of Point ( D ) by 4 units. This changes the coordinates of Point ( D ) to ( (-3, k + 4) ).Now, sum all four coordinate values of Points ( C ) and ( D ):[3 + k + (-3) + (k + 4)]Simplify the expression:[3 - 3 + k + k + 4 = 2k + 4]Thus, the sum of the four coordinate values is ( 2k + 4 ).

💡Okay, so I have this problem where I need to prove that if a six-digit number, written as (overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}}), is divisible by 7, then another six-digit number formed by moving the last digit to the front, which is (overline{a_{6} a_{1} a_{2} a_{3} a_{4} a_{5}}), is also divisible by 7. Hmm, interesting.First, let me try to understand what it means for a number to be divisible by 7. I know that a number is divisible by 7 if, when you divide it by 7, there's no remainder. So, if (overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}}) is divisible by 7, then (overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}} equiv 0 mod 7).Now, I need to see how moving the last digit to the front affects the divisibility by 7. Let me denote the original number as (N = overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}}). This can be written in terms of its digits as:[ N = 10^5 a_1 + 10^4 a_2 + 10^3 a_3 + 10^2 a_4 + 10 a_5 + a_6 ]The new number, which I'll call (M), is (overline{a_{6} a_{1} a_{2} a_{3} a_{4} a_{5}}), so it can be written as:[ M = 10^5 a_6 + 10^4 a_1 + 10^3 a_2 + 10^2 a_3 + 10 a_4 + a_5 ]I need to show that if (N) is divisible by 7, then (M) is also divisible by 7. That is, if (N equiv 0 mod 7), then (M equiv 0 mod 7).Let me think about how (N) and (M) are related. If I can express (M) in terms of (N), that might help. Let me see:If I take (N), which is (10^5 a_1 + 10^4 a_2 + 10^3 a_3 + 10^2 a_4 + 10 a_5 + a_6), and move the last digit (a_6) to the front, I get (M = a_6 times 10^5 + (N - a_6)/10). Wait, is that right?Let me check. If I remove the last digit (a_6) from (N), I get the number (A = 10^4 a_1 + 10^3 a_2 + 10^2 a_3 + 10 a_4 + a_5). So, (N = 10A + a_6). Then, (M) can be written as (a_6 times 10^5 + A).So, (M = 10^5 a_6 + A).Since (N = 10A + a_6), and (N) is divisible by 7, we have:[ 10A + a_6 equiv 0 mod 7 ]Which implies:[ 10A equiv -a_6 mod 7 ]Now, I need to relate this to (M). Let's express (M) modulo 7:[ M = 10^5 a_6 + A ]I need to find (10^5 mod 7). Let me compute that. Since (10 equiv 3 mod 7), then:[ 10^2 equiv 3^2 = 9 equiv 2 mod 7 ][ 10^3 equiv 10^2 times 10 equiv 2 times 3 = 6 mod 7 ][ 10^4 equiv 10^3 times 10 equiv 6 times 3 = 18 equiv 4 mod 7 ][ 10^5 equiv 10^4 times 10 equiv 4 times 3 = 12 equiv 5 mod 7 ]So, (10^5 equiv 5 mod 7). Therefore:[ M equiv 5a_6 + A mod 7 ]But from earlier, we have (10A equiv -a_6 mod 7). Let me solve for (A):[ 10A equiv -a_6 mod 7 ][ A equiv -a_6 times 10^{-1} mod 7 ]I need to find the modular inverse of 10 modulo 7. Since (10 equiv 3 mod 7), I need to find (x) such that (3x equiv 1 mod 7). Trying (x=5), since (3 times 5 = 15 equiv 1 mod 7). So, (10^{-1} equiv 5 mod 7).Therefore:[ A equiv -a_6 times 5 mod 7 ][ A equiv -5a_6 mod 7 ]Now, substitute this back into (M):[ M equiv 5a_6 + A mod 7 ][ M equiv 5a_6 + (-5a_6) mod 7 ][ M equiv 0 mod 7 ]So, (M) is also divisible by 7. That proves the statement.Wait, let me double-check my steps to make sure I didn't make a mistake. I expressed (N) as (10A + a_6), then found (M = 10^5 a_6 + A). I computed (10^5 mod 7) correctly as 5. Then, I used the fact that (10A equiv -a_6 mod 7) to solve for (A), finding (A equiv -5a_6 mod 7). Substituting back into (M), I indeed get (M equiv 0 mod 7). So, it seems correct.Another way to think about it is that moving the last digit to the front is equivalent to multiplying the last digit by (10^5) and adding the rest of the number. Since (10^5 equiv 5 mod 7), and we already have a relationship from the original number's divisibility, it all ties together neatly.I think this makes sense. The key was recognizing the modular relationships and using the fact that (10) has an inverse modulo (7). This allowed me to express (A) in terms of (a_6) and substitute back into (M), showing its divisibility by 7.

💡First, I need to determine the relative speed between the two whales since they are moving in the same direction. The faster whale is moving at 18 meters per second, and the slower whale is moving at 15 meters per second. The relative speed is calculated by subtracting the speed of the slower whale from the speed of the faster whale:18 mps - 15 mps = 3 mpsNext, I'll calculate the distance the faster whale travels relative to the slower whale during the time it takes to cross it, which is 15 seconds. Using the formula:Distance = Speed × TimeDistance = 3 mps × 15 s = 45 metersThis distance represents the length of the slower whale because it is the distance the faster whale covers to completely pass the slower whale.