💡Okay, so I've got this math problem here: 100 divided by (6 divided by 2). Hmm, let me see. I remember that when there are parentheses, I should do what's inside them first. So, I need to figure out what 6 divided by 2 is. That seems straightforward enough. 6 divided by 2 is 3, right? Yeah, because 2 times 3 is 6. So, now the problem simplifies to 100 divided by 3. Wait, is that all? I mean, I know that dividing by a number is the same as multiplying by its reciprocal. So, 100 divided by 3 is the same as 100 multiplied by 1/3. Let me write that down: 100 × (1/3). Hmm, 100 times 1 is 100, and then divided by 3. So, that would be 100/3. But wait, can I simplify that further? 100 divided by 3 doesn't give me a whole number. It's a fraction. So, maybe I can leave it as 100/3 or convert it to a decimal. Let me try that. If I divide 100 by 3, it goes like this: 3 goes into 10 three times (which is 9), subtract 9 from 10, you get 1. Bring down the next 0, making it 10 again. So, 3 goes into 10 three times again, and this pattern repeats. So, 100 divided by 3 is approximately 33.333... with the 3 repeating indefinitely. But since the problem didn't specify whether to leave it as a fraction or a decimal, I think it's safer to present it as a fraction, 100/3. That way, it's exact and doesn't involve rounding. Let me double-check my steps to make sure I didn't make any mistakes. First, I solved the parentheses: 6 divided by 2 is 3. Then, I took 100 divided by that result, which is 3, giving me 100/3. Yeah, that seems right. I don't think I missed anything here. I wonder if there's another way to approach this problem. Maybe by using the order of operations differently? But no, according to PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction), I did the right thing by solving the parentheses first. Another thought: sometimes, people get confused with the order of division, thinking that division is associative, but it's not. So, 100 divided by (6 divided by 2) is not the same as (100 divided by 6) divided by 2. Let me check that. If I do (100 divided by 6) first, that's approximately 16.666..., and then dividing that by 2 would give me about 8.333..., which is definitely different from 33.333... So, it's important to follow the order of operations correctly here. Alright, I think I've covered all the bases. I solved the problem step by step, checked my work, and even considered alternative approaches to ensure I didn't make any mistakes. I'm pretty confident that the answer is 100/3.

💡First, I need to evaluate the absolute value of (13 - 3). Calculating the difference inside the absolute value, 13 minus 3 equals 10. The absolute value of 10 is simply 10.Next, I evaluate the absolute value of (4 - 10). Calculating the difference inside the absolute value, 4 minus 10 equals -6. The absolute value of -6 is 6.Finally, I subtract the second absolute value from the first one: 10 minus 6 equals 4.Therefore, the result of |13 - 3| - |4 - 10| is 4.

💡Okay, so I have this trigonometry problem here. It says that the angle θ satisfies two conditions: sin(2θ) < 0 and cosθ - sinθ < 0. I need to figure out which quadrant θ is in, and the options are the first, second, third, or fourth quadrant. Alright, let's break this down step by step.First, I remember that sin(2θ) can be expressed using the double-angle identity. So, sin(2θ) is equal to 2sinθcosθ. The problem states that sin(2θ) is less than zero, which means 2sinθcosθ < 0. Since 2 is a positive number, this simplifies to sinθcosθ < 0. Hmm, okay. So, sinθ times cosθ is negative. That tells me something about the signs of sinθ and cosθ individually. Remember, in different quadrants, sine and cosine have different signs:- In the first quadrant, both sinθ and cosθ are positive.- In the second quadrant, sinθ is positive and cosθ is negative.- In the third quadrant, both sinθ and cosθ are negative.- In the fourth quadrant, sinθ is negative and cosθ is positive.So, if sinθcosθ is negative, that means one of them is positive and the other is negative. That can only happen in the second or fourth quadrants because in the first and third quadrants, both are positive or both are negative, respectively. So, θ must be in either the second or fourth quadrant.Alright, so that narrows it down to two options: B (second quadrant) or D (fourth quadrant). Now, let's look at the second condition: cosθ - sinθ < 0. That simplifies to cosθ < sinθ. So, we need to figure out in which quadrants cosθ is less than sinθ.Let me think about the unit circle. In the second quadrant, sinθ is positive and cosθ is negative. So, in this case, cosθ is definitely less than sinθ because a negative number is always less than a positive number. So, in the second quadrant, cosθ - sinθ will definitely be negative.In the fourth quadrant, sinθ is negative and cosθ is positive. So, cosθ is positive and sinθ is negative. Therefore, cosθ - sinθ would be positive minus negative, which is like adding two positive numbers. So, cosθ - sinθ would be positive in the fourth quadrant, which doesn't satisfy the condition cosθ - sinθ < 0.Wait, hold on. Let me make sure I'm not making a mistake here. If θ is in the fourth quadrant, cosθ is positive and sinθ is negative. So, cosθ - sinθ would be positive minus negative, which is positive plus positive, so it's definitely positive. Therefore, in the fourth quadrant, cosθ - sinθ is positive, which doesn't satisfy the inequality. So, the fourth quadrant is out.Therefore, θ must be in the second quadrant because that's the only quadrant left where both conditions are satisfied: sin(2θ) < 0 and cosθ - sinθ < 0.Let me just recap to make sure I didn't miss anything. The first condition tells me θ is in either the second or fourth quadrant. The second condition tells me that cosθ is less than sinθ. In the second quadrant, sinθ is positive and cosθ is negative, so cosθ is definitely less than sinθ. In the fourth quadrant, sinθ is negative and cosθ is positive, so cosθ is greater than sinθ. Therefore, only the second quadrant satisfies both conditions.I think that makes sense. I don't see any mistakes in my reasoning. So, θ must be in the second quadrant.**Final Answer**boxed{B}

💡Okay, so I have this problem: I need to find the value of ( b ) such that the quadratic expression ( bx^2 + 20x + 9 ) is the square of a binomial. Hmm, that sounds like it should be a perfect square trinomial. I remember that a perfect square trinomial has the form ( (px + q)^2 ), which expands to ( p^2x^2 + 2pqx + q^2 ). So, I guess I need to match the given quadratic to this form.Let me write down the given quadratic: ( bx^2 + 20x + 9 ). And the general form of a perfect square trinomial is ( p^2x^2 + 2pqx + q^2 ). So, if I can find values of ( p ) and ( q ) such that these two expressions are equal, then I can find ( b ).First, let's match the coefficients term by term. The coefficient of ( x^2 ) in the given quadratic is ( b ), and in the perfect square trinomial, it's ( p^2 ). So, I can write:1. ( p^2 = b )Next, the coefficient of ( x ) in the given quadratic is 20, and in the perfect square trinomial, it's ( 2pq ). So, that gives me:2. ( 2pq = 20 )Lastly, the constant term in the given quadratic is 9, and in the perfect square trinomial, it's ( q^2 ). So, that gives me:3. ( q^2 = 9 )Alright, so now I have three equations. Let's solve them step by step.Starting with equation 3: ( q^2 = 9 ). To find ( q ), I take the square root of both sides. Remember, when I take the square root, there are both positive and negative solutions. So, ( q = sqrt{9} ) or ( q = -sqrt{9} ), which simplifies to ( q = 3 ) or ( q = -3 ).Okay, so ( q ) can be either 3 or -3. Let's consider both cases.**Case 1: ( q = 3 )**Now, substitute ( q = 3 ) into equation 2: ( 2pq = 20 ). Plugging in ( q = 3 ), we get:( 2p times 3 = 20 )Simplify that:( 6p = 20 )To solve for ( p ), divide both sides by 6:( p = frac{20}{6} )Simplify the fraction by dividing numerator and denominator by 2:( p = frac{10}{3} )So, ( p = frac{10}{3} ).Now, substitute ( p = frac{10}{3} ) into equation 1: ( p^2 = b ).Calculating ( p^2 ):( left( frac{10}{3} right)^2 = frac{100}{9} )So, ( b = frac{100}{9} ).**Case 2: ( q = -3 )**Let's check the other possibility where ( q = -3 ). Substitute ( q = -3 ) into equation 2: ( 2pq = 20 ).So,( 2p times (-3) = 20 )Simplify:( -6p = 20 )Solve for ( p ):( p = frac{20}{-6} )Simplify the fraction:( p = -frac{10}{3} )Now, substitute ( p = -frac{10}{3} ) into equation 1: ( p^2 = b ).Calculating ( p^2 ):( left( -frac{10}{3} right)^2 = frac{100}{9} )So, ( b = frac{100}{9} ) again.Wait, that's interesting. Whether ( q ) is 3 or -3, the value of ( b ) comes out to be the same, ( frac{100}{9} ). So, regardless of the sign of ( q ), ( b ) remains ( frac{100}{9} ).Let me just verify this result to make sure I didn't make any mistakes. If ( b = frac{100}{9} ), then the quadratic expression becomes ( frac{100}{9}x^2 + 20x + 9 ). Let's see if this is indeed a perfect square.Assume it's equal to ( (px + q)^2 ). We found ( p = frac{10}{3} ) and ( q = 3 ). Let's square ( left( frac{10}{3}x + 3 right) ):( left( frac{10}{3}x + 3 right)^2 = left( frac{10}{3}x right)^2 + 2 times frac{10}{3}x times 3 + 3^2 )Calculating each term:1. ( left( frac{10}{3}x right)^2 = frac{100}{9}x^2 )2. ( 2 times frac{10}{3}x times 3 = 2 times 10x = 20x )3. ( 3^2 = 9 )Adding them up:( frac{100}{9}x^2 + 20x + 9 )Which matches the given quadratic expression. So, yes, it checks out.Alternatively, if we take ( p = -frac{10}{3} ) and ( q = -3 ), squaring ( left( -frac{10}{3}x - 3 right) ):( left( -frac{10}{3}x - 3 right)^2 = left( -frac{10}{3}x right)^2 + 2 times -frac{10}{3}x times -3 + (-3)^2 )Calculating each term:1. ( left( -frac{10}{3}x right)^2 = frac{100}{9}x^2 )2. ( 2 times -frac{10}{3}x times -3 = 2 times 10x = 20x )3. ( (-3)^2 = 9 )Adding them up:( frac{100}{9}x^2 + 20x + 9 )Again, it matches the given quadratic. So, whether ( p ) and ( q ) are positive or negative, the result is the same.Therefore, the value of ( b ) that makes ( bx^2 + 20x + 9 ) a perfect square trinomial is ( frac{100}{9} ).I think that's solid. I don't see any mistakes in my calculations, and I verified both cases where ( q ) is positive and negative, which both led to the same value of ( b ). So, I'm confident that ( b = frac{100}{9} ) is the correct answer.**Final Answer**The value of ( b ) is boxed{dfrac{100}{9}}.

💡Okay, so I have this problem here where I need to compute the value of this fraction involving binomial coefficients. Let me try to understand what it's asking for. The expression is:[frac{binom{-1/2}{2015} cdot 4^{2015}}{binom{4030}{2015}}]Alright, so I know that binomial coefficients are usually defined for integers, but here they're using a generalized version where the top can be any real number. The definition is given as:[binom{x}{k} = frac{x(x - 1)(x - 2) dots (x - k + 1)}{k!}]So, for this problem, x is -1/2 and k is 2015. I need to compute this generalized binomial coefficient, multiply it by 4 raised to the 2015th power, and then divide by the binomial coefficient of 4030 choose 2015.Let me break this down step by step.First, let's compute (binom{-1/2}{2015}). Using the definition, that would be:[binom{-1/2}{2015} = frac{(-1/2)(-1/2 - 1)(-1/2 - 2) dots (-1/2 - 2015 + 1)}{2015!}]Simplifying the terms in the numerator:Each term is of the form (-1/2 - n) where n starts at 0 and goes up to 2014. So, the first term is (-1/2), the next is (-3/2), then (-5/2), and so on, up to (-4029/2). So, the numerator is:[(-1/2)(-3/2)(-5/2) dots (-4029/2)]I notice that each term is negative and has a denominator of 2. There are 2015 terms in the numerator since k is 2015. So, I can factor out the negative signs and the denominators:[(-1)^{2015} times frac{1}{2^{2015}} times (1 times 3 times 5 times dots times 4029)]Wait, the product (1 times 3 times 5 times dots times 4029) is the double factorial of 4029, denoted as (4029!!). So, putting it all together, the numerator becomes:[(-1)^{2015} times frac{4029!!}{2^{2015}}]Therefore, the binomial coefficient (binom{-1/2}{2015}) is:[frac{(-1)^{2015} times 4029!!}{2^{2015} times 2015!}]Okay, that's the first part. Now, the next step is to multiply this by (4^{2015}). Let's write that out:[binom{-1/2}{2015} times 4^{2015} = frac{(-1)^{2015} times 4029!!}{2^{2015} times 2015!} times 4^{2015}]Simplify the constants. Since (4^{2015} = (2^2)^{2015} = 2^{4030}), we can substitute that in:[frac{(-1)^{2015} times 4029!!}{2^{2015} times 2015!} times 2^{4030} = frac{(-1)^{2015} times 4029!! times 2^{4030}}{2^{2015} times 2015!}]Simplify the exponents of 2:[2^{4030} / 2^{2015} = 2^{2015}]So, now we have:[frac{(-1)^{2015} times 4029!! times 2^{2015}}{2015!}]Alright, so that's the numerator of our original expression. Now, the denominator is (binom{4030}{2015}). Let's recall that:[binom{4030}{2015} = frac{4030!}{2015! times 2015!}]So, putting it all together, our expression becomes:[frac{frac{(-1)^{2015} times 4029!! times 2^{2015}}{2015!}}{frac{4030!}{2015! times 2015!}} = frac{(-1)^{2015} times 4029!! times 2^{2015} times 2015! times 2015!}{2015! times 4030!}]Simplify the 2015! terms:The numerator has (2015! times 2015!) and the denominator has (2015!), so they cancel out one 2015!:[frac{(-1)^{2015} times 4029!! times 2^{2015} times 2015!}{4030!}]Now, let's think about the relationship between 4030! and 4029!!. Remember that 4030! is the product of all integers from 1 to 4030, while 4029!! is the product of all odd integers from 1 to 4029. So, perhaps we can express 4030! in terms of 4029!! and some other terms.Let me recall that:[4030! = 4030 times 4029 times 4028 times dots times 2 times 1]But 4029!! is:[4029!! = 4029 times 4027 times dots times 3 times 1]So, 4030! can be written as:[4030! = 4030 times (4029 times 4028 times dots times 2 times 1) = 4030 times (4029!! times 2^{2015} times 2015!)]Wait, let me explain that step. The product of all even numbers from 2 to 4030 is (2^{2015} times 2015!), because each even number can be written as 2 times an integer from 1 to 2015. So, the product of even numbers is (2^{2015} times 2015!). Then, the product of the odd numbers is 4029!!. Therefore, 4030! is:[4030! = 4030 times (4029!!) times (2^{2015} times 2015!)]So, substituting back into our expression:[frac{(-1)^{2015} times 4029!! times 2^{2015} times 2015!}{4030!} = frac{(-1)^{2015} times 4029!! times 2^{2015} times 2015!}{4030 times 4029!! times 2^{2015} times 2015!}]Now, I can cancel out the common terms in the numerator and denominator:- (4029!!) cancels out- (2^{2015}) cancels out- (2015!) cancels outSo, we're left with:[frac{(-1)^{2015}}{4030}]Since 2015 is an odd number, ( (-1)^{2015} = -1 ). Therefore, the expression simplifies to:[-frac{1}{4030}]So, the value of the given expression is (-frac{1}{4030}).**Final Answer**boxed{-dfrac{1}{4030}}

💡Alright, so I've got this problem here about sampling methods in a high school setting. Let me try to wrap my head around it step by step. First, the problem states that there are 35 classes in the freshman year, each with 56 students numbered from 1 to 56. The requirement is that the student with the number 14 in each class stays for an exchange. I need to figure out what kind of sampling method this is. The options are Stratified Sampling, Lottery Sampling, Random Sampling, or Systematic Sampling.Okay, let's break this down. So, there are 35 classes, each with 56 students. Each class has a student numbered 14. The school wants these specific students to stay for the exchange. So, they're selecting the 14th student from every class.Hmm, what do I know about these sampling methods?1. **Stratified Sampling**: This is when you divide the population into subgroups or strata that share similar characteristics, and then randomly select from each stratum. For example, if you wanted to ensure representation from different classes, you might use stratified sampling.2. **Lottery Sampling**: This is a simple random sampling method where each individual has an equal chance of being selected, usually through some random process like drawing names out of a hat.3. **Random Sampling**: Similar to lottery sampling, this is a method where each member of the population has an equal chance of being selected, without any specific structure or pattern.4. **Systematic Sampling**: This involves selecting individuals from a population at regular intervals. For example, every 10th person in a list.In this problem, the selection isn't random because they're specifically choosing the 14th student from each class. It's not like they're randomly picking students or ensuring proportional representation from different strata. Instead, they're following a fixed rule: pick the 14th student in every class.So, it seems like they're using a systematic approach because they're selecting every 14th student in each class. But wait, is that exactly how systematic sampling works? I thought systematic sampling usually involves selecting every nth individual from a list. In this case, it's selecting a specific position within each subgroup (class).Maybe I need to think about it differently. If they were selecting every 14th student from the entire population of 35 classes, that would be systematic. But here, they're selecting the 14th student from each individual class. Is that still considered systematic?I guess it depends on how you look at it. If you consider each class as a separate list and then select the 14th student from each list, it's still following a systematic rule. You're not randomly selecting, but rather following a specific interval or position within each subgroup.On the other hand, stratified sampling would involve dividing the population into classes (strata) and then randomly selecting from each stratum. But in this case, they're not randomly selecting; they're specifically choosing the 14th student.Lottery sampling and random sampling both involve randomness, which isn't the case here since the selection is deterministic—always the 14th student.So, even though it's a specific position within each class, the method still follows a systematic rule. Therefore, I think the correct answer is systematic sampling.