💡Okay, so I need to find the greatest common divisor (GCD) of three numbers: 16434, 24651, and 43002. I remember that the GCD is the largest number that divides all the given numbers without leaving a remainder. Since there are three numbers, I think I can find the GCD of two numbers first and then find the GCD of that result with the third number. That makes sense because the GCD of multiple numbers can be found step by step.First, I'll start by finding the GCD of 16434 and 24651. To do this, I can use the Euclidean algorithm, which is a reliable method for finding the GCD of two numbers. The Euclidean algorithm involves dividing the larger number by the smaller one and then replacing the larger number with the smaller number and the smaller number with the remainder from the division. I repeat this process until the remainder is zero, and the last non-zero remainder is the GCD.So, let's apply the Euclidean algorithm to 16434 and 24651.First, divide 24651 by 16434. 24651 ÷ 16434 = 1 with a remainder. To find the remainder:24651 - (16434 × 1) = 24651 - 16434 = 8217.Now, replace the numbers: GCD(16434, 24651) becomes GCD(16434, 8217).Next, divide 16434 by 8217.16434 ÷ 8217 = 2 with a remainder.16434 - (8217 × 2) = 16434 - 16434 = 0.Since the remainder is 0, the last non-zero remainder is 8217. So, the GCD of 16434 and 24651 is 8217.Now, I need to find the GCD of 8217 and the third number, 43002.Again, I'll use the Euclidean algorithm.First, divide 43002 by 8217.43002 ÷ 8217 ≈ 5.23. So, 5 times 8217 is 41085.Subtract that from 43002:43002 - 41085 = 1917.Now, replace the numbers: GCD(8217, 43002) becomes GCD(8217, 1917).Next, divide 8217 by 1917.8217 ÷ 1917 ≈ 4.28. So, 4 times 1917 is 7668.Subtract that from 8217:8217 - 7668 = 549.Now, replace the numbers: GCD(1917, 8217) becomes GCD(1917, 549).Next, divide 1917 by 549.1917 ÷ 549 = 3 with a remainder.1917 - (549 × 3) = 1917 - 1647 = 270.Replace the numbers: GCD(549, 1917) becomes GCD(549, 270).Next, divide 549 by 270.549 ÷ 270 = 2 with a remainder.549 - (270 × 2) = 549 - 540 = 9.Replace the numbers: GCD(270, 549) becomes GCD(270, 9).Next, divide 270 by 9.270 ÷ 9 = 30 with no remainder.Since the remainder is 0, the last non-zero remainder is 9. So, the GCD of 8217 and 43002 is 9.Wait a minute, earlier I thought the GCD was 3, but now I'm getting 9. Did I make a mistake somewhere?Let me double-check my calculations.Starting again with GCD(8217, 43002):43002 ÷ 8217 = 5 with a remainder of 1917.Then GCD(8217, 1917):8217 ÷ 1917 = 4 with a remainder of 549.GCD(1917, 549):1917 ÷ 549 = 3 with a remainder of 270.GCD(549, 270):549 ÷ 270 = 2 with a remainder of 9.GCD(270, 9):270 ÷ 9 = 30 with a remainder of 0.So, the GCD is indeed 9.Wait, but earlier I thought it was 3. Maybe I confused the steps. Let me verify by checking if 9 divides all three original numbers.16434 ÷ 9 = 1826, which is an integer.24651 ÷ 9 = 2739, which is an integer.43002 ÷ 9 = 4778, which is an integer.So, 9 does divide all three numbers. Therefore, the GCD should be 9, not 3.I think I made a mistake in my initial calculation where I thought the GCD was 3. It should actually be 9.So, the greatest common divisor of 16434, 24651, and 43002 is 9.

💡First, I need to determine the total amount of ribbon tape Soeun used to wrap the gifts. Since each gift requires 0.84 meters of ribbon and she prepared gifts for 10 friends, I multiply 0.84 meters by 10.Next, I need to account for the leftover ribbon tape, which is 50 centimeters. To ensure consistency in units, I convert 50 centimeters to meters by dividing by 100, resulting in 0.50 meters.Finally, I add the total ribbon tape used for the gifts (8.4 meters) to the leftover ribbon tape (0.50 meters) to find the original length of the ribbon tape Soeun had at first.

💡Okay, so I have this problem about a hyperbola with two foci at F₁(-√5, 0) and F₂(√5, 0). There's a point P on the hyperbola such that the lines PF₁ and PF₂ are perpendicular to each other, and the product of their lengths is 2. I need to find the equation of the hyperbola from the given options.First, I remember that for a hyperbola, the standard form is either (x²/a²) - (y²/b²) = 1 or (y²/a²) - (x²/b²) = 1, depending on whether it's opening horizontally or vertically. Since the foci are on the x-axis, this hyperbola opens left and right, so the standard form should be (x²/a²) - (y²/b²) = 1.The distance between the foci is 2c, so in this case, the distance between F₁ and F₂ is 2√5, which means c = √5. I also recall that for hyperbolas, c² = a² + b². So, once I find a² and b², I can write the equation.Now, the point P is on the hyperbola, and PF₁ is perpendicular to PF₂. That means the vectors PF₁ and PF₂ are perpendicular, so their dot product is zero. Also, the product of their lengths is 2.Let me denote the coordinates of point P as (x, y). Then, the vectors PF₁ and PF₂ can be written as:PF₁ = (x - (-√5), y - 0) = (x + √5, y)PF₂ = (x - √5, y - 0) = (x - √5, y)Since PF₁ is perpendicular to PF₂, their dot product should be zero:(x + √5)(x - √5) + y * y = 0Simplifying this:(x² - (√5)²) + y² = 0x² - 5 + y² = 0x² + y² = 5So, point P lies on the circle centered at the origin with radius √5. Interesting.Also, the product of the lengths |PF₁| and |PF₂| is 2. Let's calculate |PF₁| and |PF₂|.|PF₁| = √[(x + √5)² + y²]|PF₂| = √[(x - √5)² + y²]So, |PF₁| * |PF₂| = 2.Let me compute |PF₁|² and |PF₂|² first:|PF₁|² = (x + √5)² + y² = x² + 2x√5 + 5 + y²|PF₂|² = (x - √5)² + y² = x² - 2x√5 + 5 + y²I can also compute |PF₁|² * |PF₂|², but that might be complicated. Alternatively, since I know that |PF₁| * |PF₂| = 2, maybe I can find |PF₁|² + |PF₂|².Wait, from the dot product condition, I have x² + y² = 5. Let me see:From |PF₁|² + |PF₂|²:= (x² + 2x√5 + 5 + y²) + (x² - 2x√5 + 5 + y²)= 2x² + 2y² + 10But since x² + y² = 5, this becomes:= 2*5 + 10 = 10 + 10 = 20So, |PF₁|² + |PF₂|² = 20.Also, since |PF₁| * |PF₂| = 2, I can use the identity:(|PF₁| + |PF₂|)² = |PF₁|² + 2|PF₁||PF₂| + |PF₂|²But I don't know |PF₁| + |PF₂|.Alternatively, I can use the identity:(|PF₁| - |PF₂|)² = |PF₁|² - 2|PF₁||PF₂| + |PF₂|²But for hyperbola, I know that | |PF₁| - |PF₂| | = 2a. So, let's denote D = |PF₁| - |PF₂|, then |D| = 2a.But I don't know D yet. However, I can relate it to the expressions I have.I have |PF₁|² + |PF₂|² = 20 and |PF₁| * |PF₂| = 2.Let me denote |PF₁| = m and |PF₂| = n. Then:m² + n² = 20m * n = 2I can also find (m - n)² = m² - 2mn + n² = 20 - 4 = 16So, (m - n)² = 16, which means |m - n| = 4. Since for hyperbola, | |PF₁| - |PF₂| | = 2a, so 2a = 4, which gives a = 2.So, a² = 4.Now, since c² = a² + b², and c = √5, so c² = 5.Therefore, 5 = 4 + b² => b² = 1.So, the equation of the hyperbola is (x²)/4 - y² = 1.Looking at the options, that's option C.Wait, let me double-check.Given a² = 4 and b² = 1, so the equation is x²/4 - y²/1 = 1, which is x²/4 - y² = 1. Yes, that's option C.I think that makes sense. Let me recap:1. Foci at (-√5, 0) and (√5, 0) => c = √5.2. Point P on hyperbola with PF₁ ⊥ PF₂ and |PF₁|*|PF₂|=2.3. Derived that x² + y² = 5 (circle with radius √5).4. Calculated |PF₁|² + |PF₂|² = 20.5. Used the product |PF₁|*|PF₂|=2 to find (|PF₁| - |PF₂|)² = 16 => 2a=4 => a=2.6. Then, c² = a² + b² => 5=4 + b² => b²=1.7. Therefore, equation is x²/4 - y² =1, which is option C.Yes, that seems correct. I don't see any mistakes in the steps.**Final Answer**boxed{C}

💡First, I note that Jacob is currently 11 years old. In 2 years, Jacob will be 13 years old.At that time, Jacob will be exactly half of Peter's age. Therefore, Peter will be 26 years old in 2 years.This means Peter is currently 24 years old.Since Peter is 4 years older than Drew, Drew must be 20 years old.Drew is 5 years older than Maya, so Maya is 15 years old.Finally, John is twice as old as Maya, making John 30 years old.

💡First, I need to determine the total distance the train must travel to completely cross the bridge. This includes both the length of the train and the length of the bridge.Next, I will convert the train's speed from kilometers per hour to meters per second to ensure the units are consistent for the calculation.Finally, I will use the formula for time, which is distance divided by speed, to find out how many seconds it takes for the train to cross the bridge.

💡Okay, so I have this problem where Jane is currently 30 years old, and Tom is older than Jane. In 'n' years, both of their ages will be two-digit numbers, and Jane's age will be the reverse of Tom's age. I need to find how many ordered pairs (t, n) are possible, where 't' is Tom's current age and 'n' is the number of years into the future when this digit reversal happens.First, let me break down the problem. Jane is 30 now, so in 'n' years, she will be 30 + n years old. Tom is older than Jane now, so his current age 't' is greater than 30. In 'n' years, Tom's age will be t + n.The key condition is that in 'n' years, Jane's age is the reverse of Tom's age. So, if Jane's age is a two-digit number, say 10a + b, then Tom's age will be 10b + a. Since Tom is older, 10b + a must be greater than 10a + b. That gives me an inequality: 10b + a > 10a + b. Let me simplify that:10b + a > 10a + b 10b - b > 10a - a 9b > 9a b > aSo, the digit in the tens place of Jane's future age must be less than the digit in the ones place of Tom's future age. That makes sense because Tom is older, so his age should be a larger number.Now, Jane's future age is 30 + n, which is equal to 10a + b. Similarly, Tom's future age is t + n, which is equal to 10b + a. So, I can write two equations:1. 30 + n = 10a + b2. t + n = 10b + aFrom the first equation, I can solve for n: n = 10a + b - 30. Since n must be a positive integer, 10a + b must be greater than 30. Also, since both a and b are digits, they can only be integers from 0 to 9. However, since 10a + b is a two-digit number, a must be at least 1.But wait, Jane's future age is 30 + n, which is at least 31 because n is a positive integer. So, 10a + b must be at least 31. That means a can be from 3 to 9, but let's see:If a is 3, then b must be greater than a, so b can be 4 to 9. If a is 4, b can be 5 to 9, and so on. Let me list all possible pairs (a, b) where b > a and 10a + b >= 31.Starting with a = 3:- b can be 4,5,6,7,8,9 => pairs (3,4), (3,5), (3,6), (3,7), (3,8), (3,9)a = 4:- b can be 5,6,7,8,9 => pairs (4,5), (4,6), (4,7), (4,8), (4,9)a = 5:- b can be 6,7,8,9 => pairs (5,6), (5,7), (5,8), (5,9)a = 6:- b can be 7,8,9 => pairs (6,7), (6,8), (6,9)a = 7:- b can be 8,9 => pairs (7,8), (7,9)a = 8:- b can be 9 => pair (8,9)a = 9:- b has to be greater than 9, but since b is a digit, it can't be. So, no pairs here.So, let's count these pairs:- a=3: 6 pairs- a=4: 5 pairs- a=5: 4 pairs- a=6: 3 pairs- a=7: 2 pairs- a=8: 1 pairTotal pairs: 6+5+4+3+2+1 = 21 pairs.Each of these pairs corresponds to a unique (a, b), which gives us a unique n and t.From the first equation, n = 10a + b - 30.From the second equation, t = 10b + a - n. But since n = 10a + b - 30, substitute that in:t = 10b + a - (10a + b - 30) t = 10b + a -10a - b + 30 t = 9b -9a +30 t = 9(b - a) +30Since t must be greater than 30 (Tom is older than Jane), and b > a, so (b - a) is at least 1, making t at least 39. That seems reasonable.So, each pair (a, b) gives a unique n and t. Therefore, there are 21 possible ordered pairs (t, n).Wait, let me double-check if all these pairs result in valid ages. For example, take a=3, b=4:n = 10*3 +4 -30 = 34 -30=4 t=9*(4-3)+30=9+30=39 So, in 4 years, Jane will be 34, Tom will be 39+4=43. 34 and 43 are reverses, and 43>34, which is correct.Another example: a=8, b=9:n=10*8 +9 -30=89-30=59 t=9*(9-8)+30=9+30=39 Wait, in 59 years, Jane will be 30+59=89, Tom will be 39+59=98. 89 and 98 are reverses, and 98>89. That works.Wait, but Tom is currently 39, which is older than Jane's current age of 30. That's fine.Another edge case: a=3, b=9:n=10*3 +9 -30=39-30=9 t=9*(9-3)+30=54+30=84 In 9 years, Jane will be 39, Tom will be 84+9=93. 39 and 93 are reverses, and 93>39. Correct.What about a=7, b=8:n=78-30=48 t=9*(8-7)+30=9+30=39 In 48 years, Jane will be 78, Tom will be 39+48=87. 78 and 87 are reverses, correct.Wait, but Tom is currently 39, which is older than Jane's 30. That's okay.Wait, but in the case where a=3, b=4, n=4, t=39. So, Tom is 39 now, Jane is 30. In 4 years, Jane is 34, Tom is 43. That works.I think all these cases are valid. So, the total number of ordered pairs is 21.**Final Answer**The number of possible ordered pairs (t, n) is boxed{21}.