💡First, I recognize that there are a total of 20 red and yellow balls in the bag.Xiao Ming observed that the frequency of picking red balls stabilizes around 0.25, which means that the probability of drawing a red ball is 25%.To find the number of red balls, I can set up the equation:Number of red balls divided by total number of balls equals the probability of picking a red ball.So, ( frac{x}{20} = 0.25 ).Solving for ( x ), I multiply both sides by 20:( x = 0.25 times 20 ).Calculating this gives ( x = 5 ).Therefore, there are 5 red balls in the bag.

💡First, substitute the given values ( a = 3 ) and ( b = 2 ) into the expression ( (a^b)^a + (b^a)^b ).Calculate ( (3^2)^3 ):[(3^2)^3 = 9^3 = 729]Next, calculate ( (2^3)^2 ):[(2^3)^2 = 8^2 = 64]Finally, add the two results together:[729 + 64 = 793]

💡Okay, so I have this problem where there are numbers from 1 to 100 on a board. Each student who enters the classroom circles the numbers that are not divisible by any of the other numbers listed, and then erases all the circled numbers. The first student erases only the number 1 because it's not divisible by any other number. Then, when Carl erased some numbers, no numbers remained on the board. I need to figure out which numbers Carl erased.Hmm, let me break this down. So, each student is supposed to circle numbers that are not divisible by any other numbers on the board. That means they are looking for numbers that are prime or maybe some composite numbers that aren't factors of any other numbers on the board.Wait, actually, if a number is not divisible by any other number on the board, it means it's a prime number because primes are only divisible by 1 and themselves. But 1 is already erased by the first student. So, the first student erases 1, then the next student would erase all the prime numbers because they are only divisible by 1 and themselves, and since 1 is already gone, they can't be divided by any other number on the board.But wait, is that right? Let me think again. If a number is not divisible by any other number on the board, it doesn't necessarily have to be prime. For example, 4 is not divisible by any other number except 1 and 2, but 2 is still on the board. So, 4 would not be erased in the second step because it is divisible by 2.So, maybe the second student would erase only the prime numbers because they are not divisible by any other numbers except 1, which is already erased. So, primes would be the next to go.Then, the third student would come in and erase numbers that are not divisible by any remaining numbers. So, after primes are erased, the next set of numbers to be erased would be composite numbers that are not divisible by any remaining composite numbers.Wait, this is getting a bit confusing. Maybe I need to think about the process step by step.First, the board has numbers 1 to 100.1. First student erases 1 because it's not divisible by any other number.2. Second student comes in and looks for numbers not divisible by any other numbers. Since 1 is gone, the primes are now the numbers not divisible by any other numbers because their only divisors are 1 and themselves, and 1 is already erased. So, the second student erases all prime numbers.3. Third student comes in and now looks for numbers not divisible by any remaining numbers. The remaining numbers are composite numbers. But some composite numbers are still divisible by others. For example, 4 is divisible by 2, but 2 was erased in the second step. Wait, no, 2 was erased, so 4 is not divisible by any remaining numbers because 2 is gone. Hmm, so 4 would be erased in the third step.Wait, so after the primes are erased, the next numbers to be erased are the composite numbers that are not divisible by any remaining numbers. But since the primes are gone, the only divisors left are composite numbers. So, a composite number like 4 is only divisible by 2, which is gone, so 4 is not divisible by any remaining numbers. Similarly, 6 is divisible by 2 and 3, both of which are gone, so 6 would also be erased in the third step.Wait, so in the third step, all composite numbers that are products of two primes would be erased because their prime factors are already gone. So, numbers like 4, 6, 8, 9, 10, etc., would be erased in the third step.Then, the fourth student comes in and looks for numbers not divisible by any remaining numbers. The remaining numbers would be composite numbers with more prime factors. For example, 12 is divisible by 2, 3, 4, 6, which are all gone, so 12 would be erased in the fourth step.Wait, no, if 4 and 6 are gone, then 12 is not divisible by any remaining numbers because its divisors are all gone. So, 12 would be erased in the fourth step.This seems like a process where each step erases numbers with a certain number of prime factors. The first step erases 1, the second step erases primes (which have one prime factor), the third step erases numbers with two prime factors, the fourth step erases numbers with three prime factors, and so on.But wait, 1 is not a prime, so it's a special case. Then primes are numbers with one prime factor, composites with two prime factors, three, etc.But in reality, the number of prime factors isn't the only thing that matters. For example, 8 is 2^3, which has only one distinct prime factor but multiple prime factors when considering multiplicity.So, maybe the process is based on the number of prime factors with multiplicity.Let me think again. After 1 is erased, primes are erased next because they are only divisible by 1 and themselves, and 1 is gone. Then, numbers that are products of two primes (like 4=2*2, 6=2*3, 8=2*2*2, 9=3*3, etc.) would be erased next because their prime factors are gone.Wait, but 8 is 2^3, which is a product of three primes, but since 2 is gone, 8 is not divisible by any remaining numbers. So, maybe it's not about the number of prime factors but about the smallest prime factor.Wait, perhaps it's about the number of prime factors in terms of exponents. For example, 4 is 2^2, which has two prime factors with multiplicity. 6 is 2*3, which has two distinct prime factors. So, maybe the third step erases numbers with two prime factors, whether distinct or not.But then, 8 would be erased in the third step as well because it has three prime factors with multiplicity, but since 2 is already gone, it's not divisible by any remaining numbers.Wait, this is getting confusing. Maybe I need to think about it differently.Each time a student enters, they erase numbers that are not divisible by any other numbers on the board. So, the first student erases 1 because it's not divisible by any other number. Then, the second student erases all primes because their only divisors are 1 and themselves, and 1 is gone. Then, the third student erases numbers that are not divisible by any remaining numbers, which would be numbers that are products of two primes, because their prime factors have been erased.Wait, no, if the primes are erased, then numbers like 4, which is 2^2, would have 2 erased, so 4 is not divisible by any remaining numbers. Similarly, 6 is 2*3, both erased, so 6 is not divisible by any remaining numbers. So, the third student would erase all composite numbers that are products of two primes, whether distinct or not.Then, the fourth student would come in and erase numbers that are not divisible by any remaining numbers. At this point, the remaining numbers would be composite numbers with three prime factors, like 8=2^3, 12=2^2*3, etc. But wait, 8 is 2^3, and 2 is already erased, so 8 is not divisible by any remaining numbers. Similarly, 12 is 2^2*3, both 2 and 3 are erased, so 12 is not divisible by any remaining numbers. So, the fourth student would erase numbers with three prime factors.Wait, but 12 is 2^2*3, which is three prime factors with multiplicity. So, maybe the number of prime factors with multiplicity determines the step in which they are erased.So, the first student erases 1 (which has zero prime factors). The second student erases primes (one prime factor). The third student erases numbers with two prime factors. The fourth student erases numbers with three prime factors, and so on.But wait, 1 is a special case. So, the process is:1. Erase 1 (step 1).2. Erase primes (step 2).3. Erase numbers with two prime factors (step 3).4. Erase numbers with three prime factors (step 4).5. Continue until all numbers are erased.But the problem says that when Carl erased some numbers, no numbers remained on the board. So, Carl must have been the last student to erase numbers, meaning he erased the last set of numbers that were left.So, I need to figure out which numbers are erased last in this process.To do that, I need to determine how many prime factors (with multiplicity) each number has and see which numbers have the highest number of prime factors within 1 to 100.The number with the highest number of prime factors (with multiplicity) less than or equal to 100 would be the one that is erased last.Let me think about the maximum number of prime factors a number can have within 1 to 100.Starting with the smallest prime, 2:2^6 = 642^7 = 128 > 100So, 2^6 = 64 is the highest power of 2 within 100, which has 6 prime factors (all 2s).Similarly, 3^4 = 813^5 = 243 > 100So, 3^4 = 81 has 4 prime factors.5^3 = 125 > 100So, 5^2 = 25 has 2 prime factors.7^2 = 49 has 2 prime factors.So, the number with the highest number of prime factors (with multiplicity) is 64, which has 6 prime factors.But wait, are there other numbers with 6 prime factors?Let's see:- 64 = 2^6 (6 factors)- 96 = 2^5 * 3 (5 + 1 = 6 factors)- 81 = 3^4 (4 factors)- 48 = 2^4 * 3 (4 + 1 = 5 factors)- 32 = 2^5 (5 factors)- 16 = 2^4 (4 factors)- etc.So, 64 and 96 both have 6 prime factors with multiplicity.Therefore, these would be the last numbers to be erased because they have the highest number of prime factors.So, when Carl erased some numbers, he must have erased 64 and 96, which were the last remaining numbers on the board.Therefore, the numbers Carl erased are 64 and 96.

💡Okay, so I have this problem where x is a positive real number, and it's given that x plus 1 over x equals 100. I need to find the square root of x plus 1 over the square root of x, and also verify if x squared plus 1 equals 10,002. Hmm, let's break this down step by step.First, let me write down what's given:x + (1/x) = 100.I need to find sqrt(x) + (1/sqrt(x)). Let me denote sqrt(x) as y. So, y = sqrt(x), which means that y squared is x. Similarly, 1/sqrt(x) would be 1/y. So, the expression I need to find is y + 1/y.Wait, that looks similar to the given equation, but instead of x and 1/x, it's y and 1/y. Maybe I can relate these two expressions somehow.Let me square the expression I need to find. If I square y + 1/y, I get:(y + 1/y)^2 = y^2 + 2*(y)*(1/y) + (1/y)^2 = y^2 + 2 + 1/y^2.But y^2 is x, and 1/y^2 is 1/x. So, substituting back, this becomes:x + 2 + 1/x.But from the given equation, x + 1/x is 100. So, substituting that in, we have:(y + 1/y)^2 = 100 + 2 = 102.Therefore, y + 1/y is the square root of 102. Since x is positive, sqrt(x) is also positive, so y + 1/y must be positive. Therefore, the value is sqrt(102).Alright, that seems straightforward. So, the first part is done.Now, the second part is to verify if x^2 + 1 equals 10,002. Let me see.Given that x + 1/x = 100, I can try to find x^2 + 1/x^2, which might help me relate it to x^2 + 1.Wait, let's square the given equation:(x + 1/x)^2 = x^2 + 2*(x)*(1/x) + (1/x)^2 = x^2 + 2 + 1/x^2.So, (x + 1/x)^2 = x^2 + 2 + 1/x^2.Given that x + 1/x is 100, squaring both sides gives:100^2 = x^2 + 2 + 1/x^2 => 10,000 = x^2 + 2 + 1/x^2.So, x^2 + 1/x^2 = 10,000 - 2 = 9,998.Hmm, so x^2 + 1/x^2 is 9,998. But the question is about x^2 + 1. How can I find x^2 + 1?Well, if I have x^2 + 1/x^2 = 9,998, maybe I can find x^2 + 1 by manipulating this equation.Let me denote z = x^2. Then, 1/x^2 = 1/z. So, z + 1/z = 9,998.But I need z + 1, not z + 1/z. Hmm, is there a way to relate z + 1 to z + 1/z?Alternatively, maybe I can find x^2 + 1 by using the original equation.Wait, from x + 1/x = 100, can I find x - 1/x? Let me see.If I square x - 1/x, I get x^2 - 2 + 1/x^2. So, (x - 1/x)^2 = x^2 - 2 + 1/x^2.But I already know that x^2 + 1/x^2 = 9,998. So, (x - 1/x)^2 = 9,998 - 2 = 9,996.Therefore, x - 1/x is sqrt(9,996) or negative sqrt(9,996). But since x is positive, and if x > 1, then x - 1/x is positive. If x < 1, then x - 1/x is negative. Hmm, but I don't know if x is greater than or less than 1.Wait, from x + 1/x = 100, which is a large number, x must be either very large or very small. If x is very large, then 1/x is negligible, so x is approximately 100. If x is very small, then 1/x is approximately 100, so x is approximately 1/100.So, x is either approximately 100 or approximately 0.01.Let me check both possibilities.Case 1: x ≈ 100.Then, x^2 + 1 ≈ 100^2 + 1 = 10,000 + 1 = 10,001.Case 2: x ≈ 0.01.Then, x^2 + 1 ≈ (0.01)^2 + 1 = 0.0001 + 1 = 1.0001.Neither of these is 10,002. So, x^2 + 1 is approximately 10,001 or 1.0001, depending on whether x is large or small.But wait, maybe I can calculate it more precisely.Let me solve for x from the equation x + 1/x = 100.Multiplying both sides by x, we get x^2 + 1 = 100x.So, x^2 - 100x + 1 = 0.This is a quadratic equation in x. Using the quadratic formula:x = [100 ± sqrt(100^2 - 4*1*1)] / 2 = [100 ± sqrt(10,000 - 4)] / 2 = [100 ± sqrt(9,996)] / 2.Calculating sqrt(9,996). Let's see, 100^2 is 10,000, so sqrt(9,996) is slightly less than 100. Let me approximate it.Let me compute 99.98^2: 99.98^2 = (100 - 0.02)^2 = 100^2 - 2*100*0.02 + (0.02)^2 = 10,000 - 4 + 0.0004 = 9,996.0004.So, sqrt(9,996) ≈ 99.98.Therefore, x = [100 ± 99.98]/2.So, two solutions:x = (100 + 99.98)/2 = 199.98/2 = 99.99.Or, x = (100 - 99.98)/2 = 0.02/2 = 0.01.So, x is either approximately 99.99 or 0.01.Now, let's compute x^2 + 1 for both cases.Case 1: x ≈ 99.99.x^2 ≈ (99.99)^2 = (100 - 0.01)^2 = 100^2 - 2*100*0.01 + (0.01)^2 = 10,000 - 2 + 0.0001 = 9,998.0001.So, x^2 + 1 ≈ 9,998.0001 + 1 = 9,999.0001.Case 2: x ≈ 0.01.x^2 ≈ (0.01)^2 = 0.0001.So, x^2 + 1 ≈ 0.0001 + 1 = 1.0001.Neither of these is 10,002. Therefore, x^2 + 1 is not equal to 10,002.Wait, but in the quadratic equation, we had x^2 - 100x + 1 = 0, which can be rearranged as x^2 + 1 = 100x.So, x^2 + 1 = 100x.Therefore, if x is 99.99, then 100x is 9,999, which is approximately equal to x^2 + 1, which we calculated as 9,999.0001.Similarly, if x is 0.01, then 100x is 1, which is approximately equal to x^2 + 1, which is 1.0001.So, in both cases, x^2 + 1 is approximately equal to 100x, but not equal to 10,002.Therefore, the statement x^2 + 1 = 10,002 is false.Wait, but let me double-check. Maybe I made a mistake in my calculations.Let me compute x^2 + 1 for x = 99.99.x = 99.99.x^2 = (99.99)^2 = 99.99*99.99.Let me compute 100*100 = 10,000.But 99.99 is 100 - 0.01.So, (100 - 0.01)^2 = 100^2 - 2*100*0.01 + (0.01)^2 = 10,000 - 2 + 0.0001 = 9,998.0001.So, x^2 + 1 = 9,998.0001 + 1 = 9,999.0001.Similarly, for x = 0.01, x^2 + 1 = 0.0001 + 1 = 1.0001.So, neither is 10,002. Therefore, the verification shows that x^2 + 1 is not equal to 10,002.Wait, but maybe I can find x^2 + 1 in terms of the given equation.From x + 1/x = 100, we can find x^2 + 1/x^2 = (x + 1/x)^2 - 2 = 100^2 - 2 = 10,000 - 2 = 9,998.So, x^2 + 1/x^2 = 9,998.But the question is about x^2 + 1, not x^2 + 1/x^2.Hmm, so x^2 + 1 is not directly related to x^2 + 1/x^2. Unless we can express x^2 + 1 in terms of x^2 + 1/x^2.Wait, let me think.We have x^2 + 1/x^2 = 9,998.We also have x + 1/x = 100.If I multiply x + 1/x by x - 1/x, I get x^2 - 1/x^2.So, (x + 1/x)(x - 1/x) = x^2 - 1/x^2.We know x + 1/x = 100, and x^2 - 1/x^2 = (x^2 + 1/x^2) - 2*(1/x^2) = 9,998 - 2*(1/x^2). Wait, that might not help.Alternatively, since we have x^2 + 1/x^2 = 9,998, and we can write x^2 + 1 = (x^2 + 1/x^2) + (1 - 1/x^2).But that seems convoluted.Alternatively, maybe we can express x^2 + 1 in terms of x.From x + 1/x = 100, we can write x^2 + 1 = 100x, as I did earlier.So, x^2 + 1 = 100x.Therefore, x^2 + 1 is equal to 100x, which is either approximately 9,999 or 1, depending on the value of x.So, it's not 10,002.Therefore, the verification shows that x^2 + 1 is not equal to 10,002.Wait, but let me check if 100x is equal to 10,002.If x is 99.99, then 100x is 9,999, which is less than 10,002.If x is 0.01, then 100x is 1, which is way less than 10,002.Therefore, x^2 + 1 cannot be 10,002.So, in conclusion, sqrt(x) + 1/sqrt(x) is sqrt(102), and x^2 + 1 is not equal to 10,002.

💡Alright, so I have this geometry problem here. It says: "A chord which is the perpendicular bisector of a radius of length 12 in a circle, has length..." and then it gives me multiple choices. Hmm, okay, let's break this down step by step.First, I need to visualize the problem. There's a circle with a radius of length 12 units. Let me draw that in my mind: a circle, center O, and a radius OA, where A is a point on the circumference. Now, there's a chord CD that is the perpendicular bisector of this radius OA. So, CD intersects OA at a right angle and cuts it exactly in half.Okay, so if OA is 12 units long, then the point where CD bisects OA must be at 6 units from the center O. Let's call that point M. So, OM is 6 units, and MA is also 6 units. Since CD is perpendicular to OA at M, triangle OMC is a right-angled triangle, right? With OM = 6, OC being the radius of the circle, which is 12 units, and MC being half the length of the chord CD.Wait, hold on, OC is the radius, so OC = 12. OM is 6, and MC is the other leg of the right triangle. So, using the Pythagorean theorem, I can find MC.Let me write that down:In triangle OMC,( OC^2 = OM^2 + MC^2 )Plugging in the known values:( 12^2 = 6^2 + MC^2 )( 144 = 36 + MC^2 )Subtract 36 from both sides:( MC^2 = 108 )Taking the square root:( MC = sqrt{108} )Simplify that:( sqrt{108} = sqrt{36 times 3} = 6sqrt{3} )Okay, so MC is ( 6sqrt{3} ). But wait, MC is just half of the chord CD, right? Because the chord is bisected by the radius OA. So, the full length of CD is twice that.So, CD = 2 * MC = 2 * 6√3 = 12√3.Let me check the answer choices: (A) 3√3, (B) 27, (C) 6√3, (D) 12√3, (E) none of these.So, 12√3 is option D. That seems to be the answer.But wait, let me make sure I didn't make any mistakes. So, OA is 12, CD is the perpendicular bisector, so it cuts OA at M, which is 6 units from O. Triangle OMC is right-angled at M, with OC = 12, OM = 6, so MC = sqrt(12^2 - 6^2) = sqrt(144 - 36) = sqrt(108) = 6√3. Then, CD is twice that, so 12√3. Yeah, that seems correct.I think I got it. So, the length of the chord CD is 12√3, which is option D.**Final Answer**boxed{D}

💡First, I need to find the complement of set ( B ) with respect to the universal set ( U ). This means identifying all elements in ( U ) that are not in ( B ).Given:[U = {1, 2, 3, 4, 5}][B = {1, 2, 4}]The complement of ( B ) in ( U ) is:[complement_{U}B = U - B = {3, 5}]Next, I need to find the union of ( complement_{U}B ) and set ( A ). Set ( A ) is given as:[A = {1, 3}]So, the union is:[(complement_{U}B) cup A = {3, 5} cup {1, 3} = {1, 3, 5}]The resulting set ( {1, 3, 5} ) has 3 elements. The number of subsets of a set with ( n ) elements is ( 2^n ). Therefore, the number of subsets is:[2^3 = 8]