💡First, identify the number of hockey cards, which is given as 200.Next, calculate the number of football cards by multiplying the number of hockey cards by 4:4 × 200 = 800.Now, determine the number of baseball cards by subtracting the total number of hockey and football cards from the overall total:1750 - (800 + 200) = 750.Finally, find the difference between the number of football cards and baseball cards:800 - 750 = 50.

💡To determine the domain of the function ( y = sqrt{1 - x} ), we need to ensure that the expression inside the square root is non-negative because the square root of a negative number is not defined in the set of real numbers.1. **Set up the inequality**: [ 1 - x geq 0 ]2. **Solve for ( x )**: [ 1 - x geq 0 implies -x geq -1 implies x leq 1 ]3. **Interpret the result**: The solution ( x leq 1 ) means that the function ( y = sqrt{1 - x} ) is defined for all real numbers ( x ) that are less than or equal to 1.4. **Select the correct option**: The domain ( {x | x leq 1} ) corresponds to option D.Therefore, the correct answer is (boxed{D}).

💡Let me try to solve this problem step by step. First, let's understand what the problem is saying. We have a sixth grade at Dongfang Elementary School, and they selected some girls and boys to participate in a math competition. Specifically, they selected 1/11 of the girls and 22 boys. After this selection, the remaining number of girls is twice the remaining number of boys. Additionally, we know that there are 2 more girls than boys in total. We need to find the total number of sixth graders in the school.Okay, so let's break this down. Let me denote the number of girls as ( x ) and the number of boys as ( y ). From the problem, we know two main things:1. The number of girls is 2 more than the number of boys. So, ( x = y + 2 ).2. After selecting 1/11 of the girls and 22 boys, the remaining girls are twice the remaining boys. Let's translate this into equations. First, the number of girls selected is ( frac{1}{11}x ), so the remaining girls are ( x - frac{1}{11}x = frac{10}{11}x ).Similarly, the number of boys selected is 22, so the remaining boys are ( y - 22 ).According to the problem, the remaining girls are twice the remaining boys. So, we can write:[frac{10}{11}x = 2(y - 22)]Now, we have two equations:1. ( x = y + 2 )2. ( frac{10}{11}x = 2(y - 22) )Our goal is to find ( x ) and ( y ), and then add them together to get the total number of students.Let's substitute the first equation into the second equation. Since ( x = y + 2 ), we can replace ( x ) in the second equation with ( y + 2 ):[frac{10}{11}(y + 2) = 2(y - 22)]Now, let's solve for ( y ). First, multiply both sides by 11 to eliminate the denominator:[10(y + 2) = 22(y - 22)]Expanding both sides:[10y + 20 = 22y - 484]Now, let's bring all terms to one side to solve for ( y ):[10y + 20 - 22y + 484 = 0][-12y + 504 = 0][-12y = -504][y = frac{504}{12}][y = 42]So, the number of boys is 42. Now, using the first equation ( x = y + 2 ):[x = 42 + 2 = 44]Therefore, the number of girls is 44.To find the total number of sixth graders, we add the number of girls and boys:[x + y = 44 + 42 = 86]So, the total number of sixth graders in the elementary school is 86.

💡Okay, so I have this problem here with two vectors, a and b. Vector a is given as 3e₁ minus 4e₂, and vector b is (1 - n)e₁ plus 3n e₂. I need to find the value of n such that vector a is parallel to vector b. Hmm, okay, let me think about how to approach this.First, I remember that two vectors are parallel if one is a scalar multiple of the other. That means there must be some scalar k such that vector a equals k times vector b. So, I can write that as:a = k * bSubstituting the given vectors into this equation, we get:3e₁ - 4e₂ = k * [(1 - n)e₁ + 3n e₂]Let me distribute the k on the right side:3e₁ - 4e₂ = k(1 - n)e₁ + k(3n)e₂Now, since e₁ and e₂ are basis vectors, they are linearly independent, right? That means the coefficients of e₁ and e₂ on both sides must be equal. So, I can set up two equations:1. For the e₁ component: 3 = k(1 - n)2. For the e₂ component: -4 = k(3n)Okay, so now I have a system of two equations with two unknowns: k and n. I need to solve for n. Let me write these equations down again:Equation 1: 3 = k(1 - n)Equation 2: -4 = k(3n)Hmm, maybe I can solve for k from one equation and substitute into the other. Let's solve Equation 1 for k:From Equation 1: k = 3 / (1 - n)Now, plug this expression for k into Equation 2:-4 = (3 / (1 - n)) * (3n)Let me simplify this:-4 = (9n) / (1 - n)Now, I can cross-multiply to solve for n:-4(1 - n) = 9nLet me distribute the -4:-4 + 4n = 9nNow, subtract 4n from both sides:-4 = 5nSo, n = -4 / 5Wait, that seems straightforward. Let me just double-check my steps to make sure I didn't make any mistakes.Starting from the beginning: a = k * b, which gives 3e₁ - 4e₂ = k[(1 - n)e₁ + 3n e₂]. Then equating components:3 = k(1 - n) and -4 = k(3n). Solving for k in the first equation gives k = 3 / (1 - n). Plugging into the second equation: -4 = (3 / (1 - n))(3n) => -4 = 9n / (1 - n). Cross-multiplying: -4(1 - n) = 9n => -4 + 4n = 9n => -4 = 5n => n = -4/5.Looks correct. So, n is -4/5.But wait, just to be thorough, let me check if this value of n actually makes the vectors parallel. Let's substitute n = -4/5 into vector b and see if it's a scalar multiple of vector a.Vector b with n = -4/5:b = (1 - (-4/5))e₁ + 3*(-4/5)e₂= (1 + 4/5)e₁ - 12/5 e₂= (9/5)e₁ - (12/5)e₂Now, vector a is 3e₁ - 4e₂. Let me see if b is a scalar multiple of a. Let's factor out 3/5 from b:b = (9/5)e₁ - (12/5)e₂ = (3/5)(3e₁ - 4e₂) = (3/5)aYes, so b is indeed (3/5) times a, which means they are parallel. So, n = -4/5 is correct.I think that's solid. I don't see any mistakes in my reasoning. So, n is -4/5.**Final Answer**The value of ( n ) is boxed{-dfrac{4}{5}}.

💡Okay, so I have this equation: 2x² + 8y² = 17xy - 423. I need to find integer solutions for x and y. Hmm, that looks a bit complicated, but maybe I can rearrange it to make it easier to handle.First, let me bring all the terms to one side to see if I can factor it or something. So, subtracting 17xy and adding 423 to both sides gives me:2x² - 17xy + 8y² + 423 = 0.Hmm, that's a quadratic in terms of x and y. Maybe I can treat this as a quadratic equation in x, with y as a parameter. Let me write it as:2x² - 17y x + (8y² + 423) = 0.Yeah, that looks like a quadratic in x. So, using the quadratic formula, x would be:x = [17y ± sqrt((17y)² - 4*2*(8y² + 423))]/(2*2).Let me calculate the discriminant D inside the square root:D = (17y)² - 4*2*(8y² + 423)D = 289y² - 8*(8y² + 423)D = 289y² - 64y² - 3384D = 225y² - 3384.Okay, so for x to be an integer, D must be a perfect square. Let me denote D = z², so:225y² - z² = 3384.Hmm, this looks like a difference of squares. Maybe I can factor it:(15y - z)(15y + z) = 3384.So, 15y - z and 15y + z are two factors of 3384. Also, since 15y - z < 15y + z, and both are positive because y and z are positive integers (I think), I can list the factor pairs of 3384 and solve for y and z.First, let me factorize 3384 to find all possible pairs. Let's see:3384 ÷ 2 = 16921692 ÷ 2 = 846846 ÷ 2 = 423423 ÷ 3 = 141141 ÷ 3 = 4747 is a prime number.So, the prime factorization of 3384 is 2³ * 3² * 47. Now, let's list all the factor pairs:1 and 33842 and 16923 and 11284 and 8466 and 5648 and 4239 and 37612 and 28218 and 18824 and 14136 and 9447 and 72...Wait, but since 15y - z and 15y + z are both even or both odd. Let me check: 15y is divisible by 15, which is odd. So, if z is even, 15y - z and 15y + z are both odd or both even? Wait, 15y is odd times y, so if y is even, 15y is even; if y is odd, 15y is odd.But 3384 is even, so both factors must be even. Therefore, both 15y - z and 15y + z must be even. So, let me consider only the factor pairs where both factors are even.Looking back at the factor pairs:2 and 16924 and 8466 and 5648 and 423 (423 is odd, so discard)12 and 28218 and 18824 and 141 (141 is odd, discard)36 and 9447 and 72 (47 is odd, discard)...So, possible pairs are:(2, 1692), (4, 846), (6, 564), (12, 282), (18, 188), (36, 94).Now, let me take each pair and solve for y and z.First pair: 15y - z = 2 and 15y + z = 1692.Adding both equations: 30y = 1694 ⇒ y = 1694 / 30 ≈ 56.466... Not an integer. So discard this pair.Second pair: 15y - z = 4 and 15y + z = 846.Adding: 30y = 850 ⇒ y = 850 / 30 ≈ 28.333... Not integer. Discard.Third pair: 15y - z = 6 and 15y + z = 564.Adding: 30y = 570 ⇒ y = 570 / 30 = 19. Okay, integer y = 19.Now, subtracting the equations: (15y + z) - (15y - z) = 564 - 6 ⇒ 2z = 558 ⇒ z = 279.So, z = 279. Let me check if 225y² - z² = 3384.225*(19)² = 225*361 = 81225.z² = 279² = 77841.81225 - 77841 = 3384. Yes, that works.So, y = 19, z = 279.Now, going back to the quadratic formula for x:x = [17y ± z]/4.So, x = [17*19 ± 279]/4.Calculate 17*19: 17*20=340, minus 17=323.So, x = (323 ± 279)/4.First, 323 + 279 = 602 ⇒ 602/4 = 150.5. Not integer.Second, 323 - 279 = 44 ⇒ 44/4 = 11. So, x = 11.So, one solution is x = 11, y = 19.Now, let's check the next factor pair.Fourth pair: 15y - z = 12 and 15y + z = 282.Adding: 30y = 294 ⇒ y = 294 / 30 = 9.8. Not integer. Discard.Fifth pair: 15y - z = 18 and 15y + z = 188.Adding: 30y = 206 ⇒ y = 206 / 30 ≈ 6.866... Not integer. Discard.Sixth pair: 15y - z = 36 and 15y + z = 94.Adding: 30y = 130 ⇒ y = 130 / 30 ≈ 4.333... Not integer. Discard.So, the only valid factor pair is (6, 564), giving y = 19 and z = 279, leading to x = 11.But wait, since the equation is symmetric in a way, if (x, y) is a solution, then (-x, -y) should also be a solution. Let me check that.If x = -11 and y = -19, plug into the original equation:2*(-11)² + 8*(-19)² = 2*121 + 8*361 = 242 + 2888 = 3130.17*(-11)*(-19) - 423 = 17*209 - 423 = 3553 - 423 = 3130.Yes, that works. So, another solution is x = -11, y = -19.Are there any other solutions? Let me think. Maybe I missed some factor pairs.Wait, earlier I considered only the factor pairs where both factors are even. But maybe I should also consider if both factors are odd. But 3384 is even, so both factors must be even. So, I think I covered all possible pairs.Alternatively, maybe y can be negative? But since y is squared, it doesn't matter. So, the only integer solutions are (11, 19) and (-11, -19).Let me double-check with another approach. Maybe rearranging the original equation.Original equation: 2x² + 8y² = 17xy - 423.Let me bring all terms to one side:2x² - 17xy + 8y² + 423 = 0.Alternatively, maybe I can write this as:2x² - 17xy + 8y² = -423.Hmm, not sure if that helps. Maybe factor the quadratic in x:2x² -17xy +8y². Let me see if this factors.Looking for factors of 2x² -17xy +8y².Multiply 2*8 = 16. Looking for two numbers that multiply to 16 and add to -17. Hmm, -16 and -1.So, 2x² -16xy -xy +8y².Grouping:(2x² -16xy) + (-xy +8y²) = 2x(x -8y) - y(x -8y) = (2x - y)(x -8y).So, 2x² -17xy +8y² = (2x - y)(x -8y).So, the equation becomes:(2x - y)(x -8y) = -423.Okay, so now I have a product of two integers equal to -423. So, I can list all pairs of integers (a, b) such that a*b = -423, and solve for x and y.First, factorize 423. 423 ÷ 3 = 141, 141 ÷ 3 = 47. So, 423 = 3² *47.Thus, the positive divisors are 1, 3, 9, 47, 141, 423.So, the factor pairs of -423 are:(1, -423), (-1, 423), (3, -141), (-3, 141), (9, -47), (-9, 47), (47, -9), (-47, 9), (141, -3), (-141, 3), (423, -1), (-423, 1).So, for each pair (a, b), set:2x - y = a,x -8y = b.Then solve for x and y.Let me go through each pair.First pair: a =1, b = -423.So,2x - y =1,x -8y = -423.Let me solve this system.From the first equation: y = 2x -1.Substitute into the second equation:x -8*(2x -1) = -423x -16x +8 = -423-15x +8 = -423-15x = -431x = 431/15 ≈28.733... Not integer. Discard.Second pair: a = -1, b =423.So,2x - y = -1,x -8y =423.From first equation: y = 2x +1.Substitute into second equation:x -8*(2x +1) =423x -16x -8 =423-15x -8 =423-15x =431x = -431/15 ≈-28.733... Not integer. Discard.Third pair: a=3, b=-141.So,2x - y =3,x -8y =-141.From first equation: y =2x -3.Substitute into second equation:x -8*(2x -3) =-141x -16x +24 =-141-15x +24 =-141-15x =-165x=11.Then y=2*11 -3=22-3=19.So, x=11, y=19. Which is the solution we found earlier.Fourth pair: a=-3, b=141.So,2x - y =-3,x -8y =141.From first equation: y=2x +3.Substitute into second equation:x -8*(2x +3)=141x -16x -24=141-15x -24=141-15x=165x= -11.Then y=2*(-11)+3= -22+3= -19.So, x=-11, y=-19. Another solution.Fifth pair: a=9, b=-47.So,2x - y=9,x -8y=-47.From first equation: y=2x -9.Substitute into second equation:x -8*(2x -9) =-47x -16x +72 =-47-15x +72 =-47-15x= -119x=119/15≈7.933... Not integer. Discard.Sixth pair: a=-9, b=47.So,2x - y=-9,x -8y=47.From first equation: y=2x +9.Substitute into second equation:x -8*(2x +9)=47x -16x -72=47-15x -72=47-15x=119x= -119/15≈-7.933... Not integer. Discard.Seventh pair: a=47, b=-9.So,2x - y=47,x -8y=-9.From first equation: y=2x -47.Substitute into second equation:x -8*(2x -47)=-9x -16x +376=-9-15x +376=-9-15x= -385x=385/15≈25.666... Not integer. Discard.Eighth pair: a=-47, b=9.So,2x - y=-47,x -8y=9.From first equation: y=2x +47.Substitute into second equation:x -8*(2x +47)=9x -16x -376=9-15x -376=9-15x=385x= -385/15≈-25.666... Not integer. Discard.Ninth pair: a=141, b=-3.So,2x - y=141,x -8y=-3.From first equation: y=2x -141.Substitute into second equation:x -8*(2x -141)=-3x -16x +1128=-3-15x +1128=-3-15x= -1131x=1131/15=75.4. Not integer. Discard.Tenth pair: a=-141, b=3.So,2x - y=-141,x -8y=3.From first equation: y=2x +141.Substitute into second equation:x -8*(2x +141)=3x -16x -1128=3-15x -1128=3-15x=1131x= -1131/15≈-75.4. Not integer. Discard.Eleventh pair: a=423, b=-1.So,2x - y=423,x -8y=-1.From first equation: y=2x -423.Substitute into second equation:x -8*(2x -423)=-1x -16x +3384=-1-15x +3384=-1-15x= -3385x=3385/15≈225.666... Not integer. Discard.Twelfth pair: a=-423, b=1.So,2x - y=-423,x -8y=1.From first equation: y=2x +423.Substitute into second equation:x -8*(2x +423)=1x -16x -3384=1-15x -3384=1-15x=3385x= -3385/15≈-225.666... Not integer. Discard.So, after checking all factor pairs, the only integer solutions are (11,19) and (-11,-19).I think that's all. So, the integer solutions are x=11, y=19 and x=-11, y=-19.

💡Alright, so I've got this problem here about an arithmetic sequence. Let me try to figure it out step by step. First, the problem says that ( S_n ) is the sum of the first ( n ) terms of the arithmetic sequence ( {a_n} ). We're given that the first term ( a_1 = -2014 ), and there's this equation involving the sums: ( dfrac{S_{2014}}{2014} - dfrac{S_{2008}}{2008} = 6 ). We need to find ( S_{2017} ).Okay, so let's recall what we know about arithmetic sequences. An arithmetic sequence has a common difference ( d ) between consecutive terms. The sum of the first ( n ) terms, ( S_n ), can be calculated using the formula:[S_n = dfrac{n}{2} times [2a_1 + (n - 1)d]]Alternatively, this can also be written as:[S_n = n times a_1 + dfrac{n(n - 1)}{2} times d]So, if I use the second formula, ( S_n = n a_1 + dfrac{n(n - 1)}{2} d ), that might be helpful here.The problem gives us ( a_1 = -2014 ), so let's plug that into the formula:[S_n = n(-2014) + dfrac{n(n - 1)}{2} d]Simplify that a bit:[S_n = -2014n + dfrac{n(n - 1)}{2} d]Now, the problem also gives us an equation involving ( S_{2014} ) and ( S_{2008} ):[dfrac{S_{2014}}{2014} - dfrac{S_{2008}}{2008} = 6]Let me compute ( dfrac{S_n}{n} ) using the formula for ( S_n ):[dfrac{S_n}{n} = dfrac{-2014n + dfrac{n(n - 1)}{2} d}{n} = -2014 + dfrac{(n - 1)}{2} d]So, ( dfrac{S_n}{n} = -2014 + dfrac{(n - 1)}{2} d ).Now, let's plug in ( n = 2014 ) and ( n = 2008 ) into this expression:For ( n = 2014 ):[dfrac{S_{2014}}{2014} = -2014 + dfrac{2013}{2} d]For ( n = 2008 ):[dfrac{S_{2008}}{2008} = -2014 + dfrac{2007}{2} d]Now, subtract these two expressions as given in the problem:[dfrac{S_{2014}}{2014} - dfrac{S_{2008}}{2008} = left( -2014 + dfrac{2013}{2} d right) - left( -2014 + dfrac{2007}{2} d right)]Simplify the right-hand side:The ( -2014 ) and ( +2014 ) cancel out, so we have:[dfrac{2013}{2} d - dfrac{2007}{2} d = 6]Factor out ( dfrac{d}{2} ):[dfrac{d}{2} (2013 - 2007) = 6]Calculate ( 2013 - 2007 ):[2013 - 2007 = 6]So, the equation becomes:[dfrac{d}{2} times 6 = 6]Simplify:[3d = 6]Divide both sides by 3:[d = 2]Alright, so the common difference ( d ) is 2. Now, we need to find ( S_{2017} ).Using the sum formula again:[S_n = n a_1 + dfrac{n(n - 1)}{2} d]Plug in ( n = 2017 ), ( a_1 = -2014 ), and ( d = 2 ):[S_{2017} = 2017 times (-2014) + dfrac{2017 times 2016}{2} times 2]Let me compute each term step by step.First term: ( 2017 times (-2014) )Hmm, that's a big number. Let me compute it:( 2017 times 2014 ). Wait, but it's negative, so it's ( -2017 times 2014 ).But maybe I can compute it as:( 2017 times 2014 = (2000 + 17)(2000 + 14) )But that might not be the quickest way. Alternatively, perhaps I can note that ( 2017 times 2014 = 2014 times 2017 ), which is symmetric.But regardless, let's compute it:( 2017 times 2014 )Let me compute 2017 * 2000 = 4,034,000Then, 2017 * 14 = 28,238So, total is 4,034,000 + 28,238 = 4,062,238Therefore, ( 2017 times (-2014) = -4,062,238 )Now, the second term:( dfrac{2017 times 2016}{2} times 2 )Simplify this:The 2 in the denominator and the 2 outside cancel out, so it's just ( 2017 times 2016 )Compute ( 2017 times 2016 ):Again, let's break it down:2017 * 2000 = 4,034,0002017 * 16 = 32,272So, total is 4,034,000 + 32,272 = 4,066,272Therefore, the second term is 4,066,272Now, add the two terms together:( S_{2017} = -4,062,238 + 4,066,272 )Compute that:4,066,272 - 4,062,238 = 4,034So, ( S_{2017} = 4,034 )Looking back at the options, D is 4034, so that's the answer.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, computing ( 2017 times 2014 ):2017 * 2000 = 4,034,0002017 * 14 = 28,238Adding them gives 4,034,000 + 28,238 = 4,062,238. That seems correct.Then, ( 2017 times 2016 ):2017 * 2000 = 4,034,0002017 * 16 = 32,272Adding them gives 4,034,000 + 32,272 = 4,066,272. That also seems correct.Subtracting: 4,066,272 - 4,062,238 = 4,034. Yes, that's correct.So, the sum ( S_{2017} = 4034 ), which is option D.I think that makes sense. Let me see if there's another way to approach this problem to confirm.Alternatively, since ( dfrac{S_n}{n} = a_1 + dfrac{(n - 1)}{2} d ), which we used earlier, we found that ( d = 2 ). So, knowing that, we can compute ( S_{2017} ) as:[S_{2017} = 2017 times a_1 + dfrac{2017 times 2016}{2} times d]Which is exactly what we did. So, that confirms the approach.Another way to think about it is that ( dfrac{S_n}{n} ) is the average of the first ( n ) terms. So, the difference between the average of the first 2014 terms and the average of the first 2008 terms is 6. Since each term increases by ( d = 2 ), the average increases by ( d/2 ) each time. So, over 6 terms (from 2008 to 2014), the average increases by 6 * (2/2) = 6, which matches the given condition. So, that makes sense.Therefore, I'm confident that the answer is D: 4034.**Final Answer**boxed{D}