💡Okay, so I need to find the twentieth prime number. Hmm, prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. Let me start listing them one by one to make sure I don't miss any.First, I know that 2 is the first prime number. It's the only even prime number. Then, 3 is the next prime. After that, 5, because 4 is even and divisible by 2, so it's not prime. Then 7, since it's not divisible by any number other than 1 and 7. Continuing, 11 is next because 9 is divisible by 3 and 10 is even. So, 11 is prime. Then 13, since 12 is even and divisible by 3, but 13 isn't. Next, 17 comes after 15, which is divisible by 3 and 5, and 16 is even. So, 17 is prime. Then 19, because 18 is even and divisible by 3, but 19 isn't. Now, moving on, 23 is next. 20 is even, 21 is divisible by 3 and 7, and 22 is even, so 23 is prime. Then 29 comes after 25, which is 5 squared, and 26 and 27 are even and divisible by 3 respectively. So, 29 is prime. Next, 31 is prime because 30 is even and divisible by 3 and 5, but 31 isn't. Then 37, since 32 is even, 33 is divisible by 3, 34 is even, 35 is divisible by 5 and 7, so 37 is prime. Continuing, 41 is next. 38 is even, 39 is divisible by 3, 40 is even, so 41 is prime. Then 43, because 42 is even and divisible by 3, but 43 isn't. Next, 47, since 44 is even, 45 is divisible by 5 and 9, and 46 is even, so 47 is prime. Now, 53 comes after 50, which is even and divisible by 5, 51 is divisible by 3 and 17, and 52 is even. So, 53 is prime. Then 59, because 54 is even and divisible by 3, 55 is divisible by 5 and 11, 56 is even, 57 is divisible by 3 and 19, 58 is even, so 59 is prime. Next, 61 is prime. 60 is even and divisible by 3 and 5, but 61 isn't. Then 67 comes after 62, which is even, 63 is divisible by 3 and 7, 64 is even, 65 is divisible by 5 and 13, 66 is even, so 67 is prime. Finally, 71 is next. 68 is even, 69 is divisible by 3, 70 is even, so 71 is prime. Let me count them to make sure I have twenty: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71. Yep, that's twenty primes. So, the twentieth prime number is 71.

💡First, I need to calculate the binomial coefficient (dbinom{60}{3}). This is done using the formula:[dbinom{60}{3} = frac{60!}{3!(60-3)!} = frac{60 times 59 times 58}{3 times 2 times 1} = frac{205740}{6} = 34290]Next, I calculate (10!):[10! = 10 times 9 times 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 = 3628800]Finally, I multiply the two results together:[34290 times 3628800 = 124467072000]

💡Okay, so I have this system of inequalities to solve. It looks like there are two inequalities here, and I need to find the values of x that satisfy both of them at the same time. Let me write them down again to make sure I have them right:First inequality: (-x + 3)/2 < xSecond inequality: 2(x + 6) ≥ 5xAlright, let me start with the first inequality. It's a bit tricky because there's a fraction involved, but I think I can handle that. So, the inequality is (-x + 3)/2 < x. Hmm, maybe I should get rid of the denominator to make it simpler. To do that, I can multiply both sides of the inequality by 2. That should eliminate the fraction.So, multiplying both sides by 2 gives me: -x + 3 < 2x.Now, I have -x + 3 < 2x. I need to solve for x. Let me get all the x terms on one side. I'll add x to both sides to move the -x to the right. That gives me: 3 < 3x.Wait, that seems a bit off. Let me double-check my steps. Starting from (-x + 3)/2 < x, multiplying both sides by 2 gives -x + 3 < 2x. Then, adding x to both sides: 3 < 3x. Yeah, that seems right. So, now I have 3 < 3x. To solve for x, I can divide both sides by 3. That gives me 1 < x, or x > 1.Okay, so the solution to the first inequality is x > 1. Got it.Now, moving on to the second inequality: 2(x + 6) ≥ 5x. Let me expand the left side first. Distribute the 2 into the parentheses: 2*x + 2*6 = 2x + 12. So, the inequality becomes 2x + 12 ≥ 5x.Now, I need to solve for x. Let me get all the x terms on one side. I'll subtract 2x from both sides: 12 ≥ 3x.Hmm, that's 12 ≥ 3x. To solve for x, I can divide both sides by 3. That gives me 4 ≥ x, or x ≤ 4.Alright, so the solution to the second inequality is x ≤ 4.Now, I need to find the values of x that satisfy both inequalities simultaneously. So, from the first inequality, x has to be greater than 1, and from the second inequality, x has to be less than or equal to 4.Putting these together, x must be greater than 1 and less than or equal to 4. So, the solution set is all real numbers between 1 and 4, not including 1 but including 4.Let me write that in interval notation to make sure I have it right. It would be (1, 4]. Yeah, that makes sense.Just to double-check, let me pick a number in this interval and see if it satisfies both inequalities. Let's choose x = 2.For the first inequality: (-2 + 3)/2 = 1/2 < 2. That's true.For the second inequality: 2(2 + 6) = 2*8 = 16 ≥ 5*2 = 10. 16 ≥ 10 is true.Okay, that works. Let me try the boundary points too. x = 1: For the first inequality, (-1 + 3)/2 = 2/2 = 1 < 1? No, it's equal, so it doesn't satisfy the strict inequality. That's why x has to be greater than 1, not equal to.x = 4: For the first inequality, (-4 + 3)/2 = (-1)/2 = -0.5 < 4. That's true.For the second inequality, 2(4 + 6) = 2*10 = 20 ≥ 5*4 = 20. 20 ≥ 20 is true, so x = 4 is included.Alright, everything checks out. So, the solution is x > 1 and x ≤ 4, which can be written as 1 < x ≤ 4.

💡Okay, so I have this quadratic equation: (x^2 + bx + 25). The problem says that it has at least one real root, and I need to find all possible values of (b). Hmm, okay. I remember that for a quadratic equation (ax^2 + bx + c = 0), the discriminant determines the nature of the roots. The discriminant is given by (D = b^2 - 4ac). If (D) is positive, there are two distinct real roots; if (D = 0), there's exactly one real root (a repeated root); and if (D) is negative, there are no real roots, just complex ones.So, in this case, since the equation has at least one real root, the discriminant must be greater than or equal to zero. That means (D geq 0). Let me write that down:(D = b^2 - 4ac geq 0)Given the quadratic equation (x^2 + bx + 25), the coefficients are (a = 1), (b = b), and (c = 25). Plugging these into the discriminant formula:(D = b^2 - 4(1)(25))Simplifying that:(D = b^2 - 100)So, for the equation to have at least one real root, we need:(b^2 - 100 geq 0)Hmm, okay. Let me solve this inequality for (b). So, (b^2 - 100 geq 0) can be rewritten as:(b^2 geq 100)To solve for (b), I can take the square root of both sides. Remember, when taking the square root in an inequality, I have to consider both the positive and negative roots. So:(|b| geq sqrt{100})Which simplifies to:(|b| geq 10)This absolute value inequality means that (b) is either greater than or equal to 10 or less than or equal to -10. In other words:(b geq 10) or (b leq -10)So, in interval notation, this would be the union of two intervals: from negative infinity to -10, and from 10 to positive infinity. Therefore, the possible values of (b) are:((-infty, -10] cup [10, infty))Let me double-check to make sure I didn't make any mistakes. If (b = 10), the equation becomes (x^2 + 10x + 25), which factors to ((x + 5)^2 = 0), so it has a repeated real root at (x = -5). Similarly, if (b = -10), the equation becomes (x^2 - 10x + 25), which factors to ((x - 5)^2 = 0), giving a repeated real root at (x = 5). If (b) is greater than 10 or less than -10, the discriminant will be positive, resulting in two distinct real roots. If (b) is between -10 and 10, the discriminant becomes negative, meaning there are no real roots. So, my solution seems correct.I think I've covered all the necessary steps and checked my work. The final answer in interval notation is ((-infty, -10] cup [10, infty)).

💡First, I need to determine the value of ( N ) based on the given information that 50% of ( N ) is 16.1. **Expressing 50% as a fraction:** [ 50% = frac{1}{2} ]2. **Setting up the equation:** [ frac{1}{2} times N = 16 ]3. **Solving for ( N ):** [ N = 16 times 2 = 32 ]Now that I have found ( N = 32 ), I need to calculate 75% of ( N ).4. **Expressing 75% as a fraction:** [ 75% = frac{3}{4} ]5. **Calculating 75% of ( N ):** [ frac{3}{4} times 32 = frac{96}{4} = 24 ]Therefore, 75% of ( N ) is 24.

💡Okay, so I have this problem here with vectors a and b, and I need to find the value of k such that the vector k*a + b is perpendicular to the vector 2*a - b. Hmm, let me try to break this down step by step.First, I remember that if two vectors are perpendicular, their dot product is zero. So, that must mean that the dot product of (k*a + b) and (2*a - b) is zero. That gives me an equation to work with. Let me write that down:(k*a + b) · (2*a - b) = 0Alright, now I need to compute this dot product. But before I do that, maybe I should figure out what the vectors k*a + b and 2*a - b actually are. Let me compute each component.Given that vector a is (1, 1, 0) and vector b is (-1, 0, 2), let's compute k*a + b first.Multiplying vector a by k gives me (k*1, k*1, k*0) which is (k, k, 0). Then adding vector b, which is (-1, 0, 2), to that:k*a + b = (k + (-1), k + 0, 0 + 2) = (k - 1, k, 2)Okay, so k*a + b is (k - 1, k, 2). Now, let's compute 2*a - b.Multiplying vector a by 2 gives me (2*1, 2*1, 2*0) which is (2, 2, 0). Then subtracting vector b, which is (-1, 0, 2), from that:2*a - b = (2 - (-1), 2 - 0, 0 - 2) = (3, 2, -2)So, 2*a - b is (3, 2, -2). Now, I need to compute the dot product of (k - 1, k, 2) and (3, 2, -2) and set it equal to zero.Let me write that out:(k - 1)*3 + k*2 + 2*(-2) = 0Now, let me compute each term:First term: (k - 1)*3 = 3k - 3Second term: k*2 = 2kThird term: 2*(-2) = -4Adding all these together:3k - 3 + 2k - 4 = 0Combine like terms:3k + 2k = 5k-3 - 4 = -7So, the equation becomes:5k - 7 = 0Now, solving for k:5k = 7k = 7/5Hmm, 7/5 is equal to 1.4. Let me check if that makes sense. Maybe I made a mistake in my calculations.Wait, let me double-check the dot product:(k - 1)*3 + k*2 + 2*(-2) = 0Yes, that's correct. So, 3(k - 1) + 2k - 4 = 0.Expanding that:3k - 3 + 2k - 4 = 0Combining like terms:5k - 7 = 0Yes, that's right. So, k = 7/5. Hmm, 7/5 is 1.4, which is 1 and 2/5. Let me see if that's one of the options. Wait, the options given are A:1, and then B, C, D are missing. Maybe I need to express 7/5 as a fraction or something else.Wait, 7/5 is equal to 1.4, which is 1 and 2/5. But since the options aren't fully provided, maybe I should check my steps again to make sure I didn't make a mistake.Let me go back to the beginning. Vector a is (1,1,0), vector b is (-1,0,2). So, k*a is (k, k, 0), and adding b gives (k -1, k, 2). Correct.2*a is (2,2,0), subtracting b gives (2 - (-1), 2 - 0, 0 - 2) which is (3,2,-2). Correct.Dot product: (k -1)*3 + k*2 + 2*(-2) = 0.So, 3(k -1) + 2k -4 = 0.3k -3 + 2k -4 = 0.5k -7 = 0.5k =7.k=7/5.Hmm, that seems correct. Maybe the answer is 7/5, which is 1.4, but in the options, A is 1, and then B, C, D are missing. Perhaps the correct option is D, which might be 7/5 or 1.4. Since the user didn't provide the full options, but in the initial problem, the answer was given as D, so maybe D is 7/5.Alternatively, maybe I made a mistake in computing the vectors. Let me check again.Wait, when I computed 2*a - b, I had 2*a as (2,2,0), and then subtracting b which is (-1,0,2). So, 2 - (-1) is 3, 2 - 0 is 2, and 0 - 2 is -2. That's correct.Similarly, k*a + b is (k, k, 0) + (-1, 0, 2) which is (k -1, k, 2). Correct.Dot product: (k -1)*3 + k*2 + 2*(-2) = 0.Yes, that's correct. So, 3k -3 + 2k -4 = 0.5k -7 =0.k=7/5.So, unless I made a mistake in the arithmetic, which I don't see, the answer should be 7/5, which is 1.4. Since the options are A:1, and then B, C, D missing, but the initial problem said the correct answer is D, so maybe D is 7/5.Alternatively, maybe I made a mistake in the initial step. Let me think again.Wait, another way to approach this is to compute the dot product directly using the components of a and b without expanding the vectors. Let me try that.Given that (k*a + b) · (2*a - b) = 0.We can expand this dot product as:k*a · 2*a + k*a · (-b) + b · 2*a + b · (-b) = 0Which simplifies to:2k*(a · a) -k*(a · b) + 2*(b · a) - (b · b) = 0Now, let's compute each dot product separately.First, a · a: (1,1,0) · (1,1,0) = 1*1 + 1*1 + 0*0 = 1 +1 +0 =2Next, a · b: (1,1,0) · (-1,0,2) =1*(-1) +1*0 +0*2 = -1 +0 +0 = -1Similarly, b · a is the same as a · b, which is -1.And b · b: (-1,0,2) · (-1,0,2) = (-1)*(-1) +0*0 +2*2 =1 +0 +4=5So, plugging these back into the equation:2k*(2) -k*(-1) + 2*(-1) -5 =0Compute each term:2k*2=4k- k*(-1)=k2*(-1)=-2So, putting it all together:4k +k -2 -5=0Combine like terms:5k -7=0So, 5k=7 => k=7/5Same result as before. So, that confirms it. So, k=7/5.Therefore, the correct answer is 7/5, which is 1.4. Since the options are A:1, and then B, C, D missing, but the initial problem said the answer is D, so I think D is 7/5.I think that's solid. I don't see any mistakes in my calculations. Both methods gave me the same result, so I'm confident that k=7/5 is correct.