💡Okay, so I'm trying to solve this problem about sets. Let me read it again carefully to make sure I understand what's being asked.The universal set I is the set of natural numbers, which I think means all positive integers like 1, 2, 3, and so on. Set A is defined as all x in I such that 2 ≤ x ≤ 10. So, A should be the numbers from 2 to 10, inclusive. That would be {2, 3, 4, 5, 6, 7, 8, 9, 10}.Set B is the set of all prime numbers. Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. So, the prime numbers less than or equal to 10 are 2, 3, 5, and 7. Therefore, B should be {2, 3, 5, 7}.Now, the problem is asking for A ∩ complement of B with respect to I, which is written as A ∩ ∁ᵢB. I think the complement of B in I would be all the natural numbers that are not in B. Since B is the set of primes, the complement would be all natural numbers that are not prime. But since we're dealing with set A, which only goes up to 10, I guess we'll focus on the numbers from 2 to 10.So, the complement of B in I, ∁ᵢB, would include numbers like 1, 4, 6, 8, 9, 10, and so on. But since we're looking at the intersection with A, which is from 2 to 10, we can ignore numbers outside that range. Therefore, ∁ᵢB within the context of A would be {4, 6, 8, 9, 10}.Now, to find A ∩ ∁ᵢB, we need to find the elements that are in both A and ∁ᵢB. Since A is {2, 3, 4, 5, 6, 7, 8, 9, 10} and ∁ᵢB is {4, 6, 8, 9, 10}, the intersection would be the numbers that appear in both sets. Looking at both sets, the common elements are 4, 6, 8, 9, and 10.So, A ∩ ∁ᵢB should be {4, 6, 8, 9, 10}. Let me check the options given:A: {4,6,8,10}B: {4,5,6,8,9}C: {2,4,6,8,10}D: {4,6,8,9,10}Option D matches what I got, which is {4,6,8,9,10}. I think that's the correct answer.But wait, let me double-check to make sure I didn't make a mistake. Maybe I missed something.First, set A is definitely {2,3,4,5,6,7,8,9,10}. Set B is {2,3,5,7}. The complement of B in I would exclude these primes, so in the range of A, the non-primes are 4,6,8,9,10. So, intersecting A with these non-primes gives exactly those numbers. Yeah, that seems right.I think I'm confident that D is the correct answer.

💡First, I need to determine the total amount paid by the customer through the payment plan. This includes the down payment plus the sum of the 12 monthly payments of 10 each.Next, I'll calculate the interest paid by subtracting the purchase price from the total amount paid.Given that the interest paid is 15.254237288135593% of the purchase price, I'll convert this percentage to a decimal to make the calculations easier.Using the purchase price of 118, I'll compute the exact amount of interest paid.With the interest amount known, I can then find the total amount paid by adding the purchase price and the interest.Finally, by subtracting the total of the monthly payments from the total amount paid, I'll determine the down payment amount.

💡Alright, so I've got this system of equations here:begin{cases}y - 2x - a = 0, y^2 - xy + x^2 - b = 0.end{cases}And I need to prove that if there's a set of rational numbers that satisfy this system, then those numbers must actually be integers. Okay, let's break this down step by step.First, let's look at the first equation: ( y - 2x - a = 0 ). That seems straightforward. I can solve for one variable in terms of the other. Let me solve for ( y ) because it might be simpler.So, from the first equation:y = 2x + aAlright, now I can substitute this expression for ( y ) into the second equation to eliminate ( y ) and have an equation in terms of ( x ) only. Let's do that.Substituting ( y = 2x + a ) into the second equation:(2x + a)^2 - x(2x + a) + x^2 - b = 0Now, let's expand and simplify this equation step by step.First, expand ( (2x + a)^2 ):(2x)^2 + 2 cdot 2x cdot a + a^2 = 4x^2 + 4ax + a^2Next, expand ( x(2x + a) ):2x^2 + axNow, substitute these back into the equation:4x^2 + 4ax + a^2 - (2x^2 + ax) + x^2 - b = 0Let's distribute the negative sign:4x^2 + 4ax + a^2 - 2x^2 - ax + x^2 - b = 0Now, combine like terms:- ( 4x^2 - 2x^2 + x^2 = 3x^2 )- ( 4ax - ax = 3ax )- ( a^2 - b ) remains as is.So, the equation simplifies to:3x^2 + 3ax + (a^2 - b) = 0Hmm, okay, so now we have a quadratic equation in terms of ( x ):3x^2 + 3ax + (a^2 - b) = 0Since ( x ) is rational, the solutions to this quadratic equation must be rational. For a quadratic equation ( Ax^2 + Bx + C = 0 ), the solutions are rational if and only if the discriminant ( B^2 - 4AC ) is a perfect square.Let's compute the discriminant for our equation:D = (3a)^2 - 4 cdot 3 cdot (a^2 - b) = 9a^2 - 12(a^2 - b)Simplify the discriminant:D = 9a^2 - 12a^2 + 12b = -3a^2 + 12b = 12b - 3a^2Factor out a 3:D = 3(4b - a^2)So, the discriminant is ( 3(4b - a^2) ). For the solutions ( x ) to be rational, ( D ) must be a perfect square. Let's denote ( D = k^2 ) where ( k ) is an integer because ( a ) and ( b ) are integers.Therefore:3(4b - a^2) = k^2This implies that ( k^2 ) is divisible by 3. Since 3 is a prime number, ( k ) must be divisible by 3. Let's write ( k = 3m ) where ( m ) is an integer.Substituting back:3(4b - a^2) = (3m)^2 = 9m^2Divide both sides by 3:4b - a^2 = 3m^2So, we have:4b = a^2 + 3m^2Now, let's recall that ( x ) is rational. The solutions to the quadratic equation are:x = frac{-3a pm sqrt{D}}{2 cdot 3} = frac{-3a pm k}{6}But since ( k = 3m ), this becomes:x = frac{-3a pm 3m}{6} = frac{-a pm m}{2}So, ( x = frac{-a + m}{2} ) or ( x = frac{-a - m}{2} ).Since ( x ) is rational, and ( a ) and ( m ) are integers, ( x ) must be a rational number. But we need to show that ( x ) is actually an integer.Looking at ( x = frac{-a pm m}{2} ), for ( x ) to be an integer, ( -a pm m ) must be even. That is, ( -a pm m ) must be divisible by 2.Let's analyze this. Since ( a ) and ( m ) are integers, their sum or difference will also be an integer. For ( frac{-a pm m}{2} ) to be an integer, ( -a pm m ) must be even. This implies that ( a ) and ( m ) must have the same parity (both even or both odd).But let's see if we can derive more information from the equation ( 4b = a^2 + 3m^2 ). Since ( 4b ) is divisible by 4, the right-hand side must also be divisible by 4.So, ( a^2 + 3m^2 equiv 0 mod 4 ).Let's consider the possible residues of squares modulo 4.Squares modulo 4 can be 0 or 1:- If ( a ) is even, ( a^2 equiv 0 mod 4 ).- If ( a ) is odd, ( a^2 equiv 1 mod 4 ).Similarly for ( m ):- If ( m ) is even, ( m^2 equiv 0 mod 4 ).- If ( m ) is odd, ( m^2 equiv 1 mod 4 ).So, let's consider all possibilities for ( a ) and ( m ):1. Both ( a ) and ( m ) are even: - ( a^2 equiv 0 mod 4 ) - ( 3m^2 equiv 0 mod 4 ) - Sum: ( 0 + 0 = 0 mod 4 ) → satisfies ( a^2 + 3m^2 equiv 0 mod 4 )2. ( a ) even, ( m ) odd: - ( a^2 equiv 0 mod 4 ) - ( 3m^2 equiv 3 cdot 1 = 3 mod 4 ) - Sum: ( 0 + 3 = 3 mod 4 ) → Not congruent to 03. ( a ) odd, ( m ) even: - ( a^2 equiv 1 mod 4 ) - ( 3m^2 equiv 0 mod 4 ) - Sum: ( 1 + 0 = 1 mod 4 ) → Not congruent to 04. Both ( a ) and ( m ) are odd: - ( a^2 equiv 1 mod 4 ) - ( 3m^2 equiv 3 cdot 1 = 3 mod 4 ) - Sum: ( 1 + 3 = 4 equiv 0 mod 4 ) → SatisfiesSo, the only possibilities where ( a^2 + 3m^2 equiv 0 mod 4 ) are when both ( a ) and ( m ) are even or both are odd.This means that ( a ) and ( m ) have the same parity, which is exactly what we needed for ( x = frac{-a pm m}{2} ) to be an integer.Therefore, ( x ) must be an integer.Now, since ( x ) is an integer and from the first equation ( y = 2x + a ), and ( a ) is an integer, ( y ) must also be an integer.So, both ( x ) and ( y ) are integers.**Final Answer**boxed{text{Both } x text{ and } y text{ are integers.}}

💡First, I need to determine the total amount Chris spent on the two hockey sticks. I know that he spent a total of 68 and bought a helmet for 25. By subtracting the cost of the helmet from the total amount spent, I can find out how much was spent on the hockey sticks.So, 68 (total) minus 25 (helmet) equals 43 spent on the two hockey sticks.Now, since Chris bought two hockey sticks at the same price, I can find the cost of one hockey stick by dividing the total amount spent on the hockey sticks by 2.Therefore, 43 divided by 2 equals 21.50 per hockey stick.

💡Okay, so I need to find the smallest integer that's greater than -7/3. Hmm, let me think about how to approach this. First, I remember that integers are whole numbers, both positive and negative, without any fractions or decimals. So, I need to figure out where -7/3 falls on the number line and then identify the next integer to the right of it.Let me start by converting -7/3 into a decimal to get a better sense of its value. Dividing 7 by 3 gives me approximately 2.333... So, -7/3 is the same as -2.333... That means it's somewhere between -3 and -2 on the number line. Specifically, it's two and a third units to the left of zero.Now, I need to find the smallest integer that's greater than this value. Since -7/3 is approximately -2.333, the next integer to the right would be -2. That makes sense because -2 is greater than -2.333, and there's no integer between -2.333 and -2. So, -2 is the smallest integer that is still greater than -7/3.Wait, let me double-check to make sure I'm not making a mistake here. If I think about the number line, moving from left to right, the numbers increase. So, starting from -3, the next integer is -2, then -1, and so on. Since -7/3 is between -3 and -2, the next integer greater than -7/3 should indeed be -2.Is there any chance I could have confused this with the ceiling function? I think the ceiling function gives the smallest integer greater than or equal to a given number. In this case, since -7/3 is not an integer, the ceiling would be the next integer up, which is -2. That aligns with what I thought earlier.Just to be thorough, let me consider some examples. If I have a number like 1.2, the smallest integer greater than that is 2. Similarly, for -1.2, the smallest integer greater than that would be -1. Applying the same logic, for -2.333, the smallest integer greater than that should be -2.I think I'm confident now that -2 is the correct answer. It's the immediate next integer to the right of -7/3 on the number line, making it the smallest integer that is still greater than -7/3.

💡First, I need to find ( f(1) ) using the given equation ( f(e^x) = x + 2 ). To do this, I'll set ( e^x = 1 ), which implies ( x = 0 ). Substituting ( x = 0 ) into the equation gives ( f(1) = 0 + 2 = 2 ).Next, to find the analytical expression of ( f(x) ), I'll introduce a substitution. Let ( t = e^x ), which means ( x = ln{t} ). Substituting ( x = ln{t} ) into the original equation ( f(e^x) = x + 2 ) gives ( f(t) = ln{t} + 2 ). Therefore, the function ( f(x) ) can be expressed as ( f(x) = ln{x} + 2 ).