💡Okay, so I have this problem about painting three faces of a die red, and I need to make sure that none of the sums of any two red faces equal 7. Hmm, let me think about how to approach this.First, I know that a standard die has six faces, numbered from 1 to 6. Each face is opposite another face, and the numbers on opposite faces add up to 7. So, the pairs are (1,6), (2,5), and (3,4). That means if I paint one face red, I can't paint its opposite face red because their sum would be 7, which is not allowed.Now, the problem is asking how many ways I can paint three faces red without any two of them adding up to 7. So, essentially, I need to choose three numbers from 1 to 6 such that none of these numbers are opposites of each other.Let me start by calculating the total number of ways to choose any three faces out of six, without any restrictions. That should be a combination problem, right? So, the formula for combinations is:[text{Total ways} = binom{6}{3} = frac{6!}{3!(6-3)!} = frac{6 times 5 times 4}{3 times 2 times 1} = 20]So, there are 20 possible ways to paint any three faces red. But now I need to subtract the number of ways that include at least one pair of opposite faces because those would sum to 7, which is not allowed.Let's think about how many such invalid combinations there are. For each pair of opposite faces, if I choose both of them, then I have to choose one more face from the remaining four faces. But wait, the remaining four faces also include the opposites of the other two pairs. So, I need to make sure that the third face I choose doesn't form another pair that sums to 7.Wait, no, actually, if I choose one pair, say (1,6), then the remaining faces are 2, 3, 4, and 5. None of these pairs sum to 7 because 2 and 5 are opposites, and 3 and 4 are opposites. So, if I choose (1,6) and then choose any one of the remaining four faces, none of those will form a sum of 7 with either 1 or 6. So, for each pair, there are 4 ways to choose the third face.Since there are three such pairs: (1,6), (2,5), and (3,4), each contributing 4 invalid combinations, that gives:[3 times 4 = 12]So, there are 12 invalid combinations where at least one pair of opposite faces is chosen. Therefore, the number of valid combinations should be:[20 - 12 = 8]Wait, but let me double-check to make sure I didn't miss anything. Is there any overlap where choosing two pairs might have been subtracted twice? For example, could there be a combination where both (1,6) and (2,5) are chosen? But if I choose both (1,6) and (2,5), that would require choosing four faces, which is more than three. So, in this case, since we're only choosing three faces, it's impossible to have two pairs. Therefore, there's no overlap, and I don't need to adjust for that.So, my final answer should be 8 ways.Just to be thorough, let me list out all the possible valid combinations to make sure there are indeed 8.The pairs that sum to 7 are (1,6), (2,5), and (3,4). So, to avoid any of these pairs, I need to choose numbers such that none of these pairs are included.Let's consider each number and see which numbers they can be paired with:- If I choose 1, I can't choose 6.- If I choose 2, I can't choose 5.- If I choose 3, I can't choose 4.So, let's try to list all possible combinations:1. Choose 1, 2, 32. Choose 1, 2, 43. Choose 1, 3, 54. Choose 1, 4, 55. Choose 2, 3, 66. Choose 2, 4, 67. Choose 3, 5, 68. Choose 4, 5, 6Wait, hold on. Let me check each of these to ensure none of them include a pair that sums to 7.1. 1, 2, 3: No pairs sum to 7. Good.2. 1, 2, 4: No pairs sum to 7. Good.3. 1, 3, 5: No pairs sum to 7. Good.4. 1, 4, 5: No pairs sum to 7. Good.5. 2, 3, 6: No pairs sum to 7. Good.6. 2, 4, 6: No pairs sum to 7. Good.7. 3, 5, 6: No pairs sum to 7. Good.8. 4, 5, 6: No pairs sum to 7. Good.So, all eight combinations are valid. That confirms my earlier calculation.Therefore, the number of ways to paint three faces red without any two of them summing to 7 is indeed 8.

💡First, I need to find the base length of the Great Pyramid using the given height and slope angle. I know that the pyramid can be divided into right-angled triangles by cutting along its height. Each of these triangles has a height of 520 feet and a slope angle of 52 degrees.I can use the tangent function to relate the height and half of the base length. The tangent of the slope angle is equal to the height divided by half of the base length.Next, I'll set up the equation using the tangent of 52 degrees:[ tan(52^circ) = frac{520}{frac{b}{2}} ]To solve for ( frac{b}{2} ), I'll rearrange the equation:[ frac{b}{2} = frac{520}{tan(52^circ)} ]Using a calculator, I'll find the value of ( tan(52^circ) ) and then compute ( frac{b}{2} ). After obtaining ( frac{b}{2} ), I'll multiply it by 2 to get the full base length ( b ).Finally, I'll add the height and the base length to find their sum.

💡First, I need to determine the total number of books your class received from the library.Initially, your class received 54 books.Then, your class received an additional 23 books.To find the total number of books, I add these two quantities together: 54 + 23.Calculating the sum, 54 plus 23 equals 77.Therefore, your class received a total of 77 books from the library.

💡Okay, so I have this problem where I'm given two equations involving x, y, and a, and I need to find cos(x + 2y). Let me try to unpack this step by step.First, let me write down the given equations:1. ( x^3 + sin x - 2a = 0 )2. ( 4y^3 + sin y cos y + a = 0 )And I know that x and y are both in the interval ([- frac{pi}{4}, frac{pi}{4}]). Hmm, that interval is important because it tells me about the possible values of x and y, which might help later on.I need to find ( cos(x + 2y) ). So, maybe I can find expressions for x and y in terms of a, or find a relationship between x and y that allows me to compute x + 2y directly.Looking at the first equation: ( x^3 + sin x = 2a ). Let me denote this as Equation (1). The second equation is ( 4y^3 + sin y cos y = -a ). Let me denote this as Equation (2).I notice that both equations involve a, so perhaps I can eliminate a by expressing a from one equation and substituting into the other.From Equation (1): ( 2a = x^3 + sin x ) => ( a = frac{x^3 + sin x}{2} ).From Equation (2): ( 4y^3 + sin y cos y = -a ) => ( a = - (4y^3 + sin y cos y) ).So, setting the two expressions for a equal to each other:( frac{x^3 + sin x}{2} = - (4y^3 + sin y cos y) )Multiply both sides by 2 to eliminate the denominator:( x^3 + sin x = -2(4y^3 + sin y cos y) )Simplify the right-hand side:( x^3 + sin x = -8y^3 - 2 sin y cos y )Hmm, this seems a bit complicated. Maybe I can find a way to relate x and y directly. Let me think about the functions involved.I notice that both equations have cubic terms and sine terms. Maybe I can define a function that combines these terms to simplify things.Let me define ( f(t) = t^3 + sin t ). Then, Equation (1) becomes:( f(x) = 2a )And Equation (2) can be rewritten as:( 4y^3 + sin y cos y = -a )Wait, I can express 4y^3 as ( (2y)^3 / 2 ), right? Because ( (2y)^3 = 8y^3 ), so ( 4y^3 = (8y^3)/2 = (2y)^3 / 2 ). Let me try that.So, ( 4y^3 = frac{(2y)^3}{2} ). Therefore, Equation (2) becomes:( frac{(2y)^3}{2} + sin y cos y = -a )Multiply both sides by 2 to eliminate the denominator:( (2y)^3 + 2 sin y cos y = -2a )But ( 2 sin y cos y = sin(2y) ) by the double-angle identity. So, Equation (2) becomes:( (2y)^3 + sin(2y) = -2a )But wait, ( (2y)^3 + sin(2y) ) is exactly the function ( f(2y) ) because ( f(t) = t^3 + sin t ). So, ( f(2y) = -2a ).From Equation (1), we have ( f(x) = 2a ). So, combining these two results:( f(x) = 2a ) and ( f(2y) = -2a )Which implies:( f(x) = -f(2y) )Since ( f(t) = t^3 + sin t ), which is an odd function because both ( t^3 ) and ( sin t ) are odd functions. Therefore, ( f(-t) = -f(t) ).So, if ( f(x) = -f(2y) ), then:( f(x) = f(-2y) )Since ( f(t) ) is an increasing function on the interval ([- frac{pi}{2}, frac{pi}{2}]) (because its derivative ( f'(t) = 3t^2 + cos t ) is always positive in this interval), it is injective (one-to-one) on this interval. Therefore, if ( f(x) = f(-2y) ), then ( x = -2y ).So, ( x = -2y ). Therefore, ( x + 2y = -2y + 2y = 0 ).Thus, ( cos(x + 2y) = cos(0) = 1 ).Wait, that seems straightforward. Let me double-check to make sure I didn't make any mistakes.1. I defined ( f(t) = t^3 + sin t ), which is correct.2. I transformed Equation (2) correctly by expressing ( 4y^3 ) as ( (2y)^3 / 2 ) and then multiplied by 2.3. I used the double-angle identity correctly: ( 2 sin y cos y = sin(2y) ).4. Recognized that ( f(2y) = -2a ) and ( f(x) = 2a ), leading to ( f(x) = -f(2y) ).5. Since ( f ) is odd, ( f(x) = f(-2y) ).6. Since ( f ) is injective on the given interval, ( x = -2y ).7. Therefore, ( x + 2y = 0 ), so ( cos(0) = 1 ).Everything seems to check out. I don't see any errors in the reasoning.**Final Answer**boxed{1}

💡First, I need to understand the conditions given for the amount of change George is trying to get. He is receiving change less than one dollar, and there are two specific scenarios:1. **Quarters Condition**: When George takes the maximum number of quarters possible, he needs 3 pennies to make up the total amount. This means the total amount can be expressed as 25q + 3, where q is the number of quarters. Since the amount is less than a dollar, q can be 0, 1, 2, or 3, leading to possible amounts of 3, 28, 53, or 78 cents.2. **Dimes Condition**: When George takes the maximum number of dimes possible, he needs 8 pennies to make up the total amount. This means the total amount can be expressed as 10d + 8, where d is the number of dimes. The amount must end with 8 cents, so possible amounts are 8, 18, 28, 38, 48, 58, 68, 78, or 88 cents.Next, I need to find the amounts that satisfy both conditions. Comparing the possible amounts from both conditions:- From quarters: 3, 28, 53, 78- From dimes: 8, 18, 28, 38, 48, 58, 68, 78, 88The common amounts that appear in both lists are 28 and 78 cents.Finally, to find the sum of these possible amounts, I add them together:28 + 78 = 106 cents

💡Alright, let's tackle this problem step by step. I need to find the smallest positive integer ( n ) that satisfies three conditions: it's divisible by 36, ( n^2 ) is a perfect cube, and ( n^3 ) is a perfect square. Then, I have to determine the largest prime factor of ( n ).First, let's break down the problem. The number ( n ) must be divisible by 36. I know that 36 factors into primes as ( 2^2 times 3^2 ). So, ( n ) must have at least ( 2^2 ) and ( 3^2 ) in its prime factorization.Next, ( n^2 ) must be a perfect cube. For a number to be a perfect cube, all the exponents in its prime factorization must be multiples of 3. So, if I let ( n = 2^a times 3^b times ) (other primes if necessary), then ( n^2 = 2^{2a} times 3^{2b} times ) ... must have exponents that are multiples of 3. This means ( 2a ) and ( 2b ) must be multiples of 3. Therefore, ( a ) and ( b ) must be multiples of ( frac{3}{2} ). But since ( a ) and ( b ) must be integers, the smallest integers that satisfy this are ( a = 3 ) and ( b = 3 ). Wait, but hold on, ( 2a ) needs to be a multiple of 3, so ( a ) must be a multiple of ( frac{3}{2} ). Since ( a ) has to be an integer, the smallest ( a ) is 3, because ( 3 times 2 = 6 ), which is a multiple of 3. Similarly, ( b ) must also be 3.But wait, ( n ) also needs to be divisible by 36, which is ( 2^2 times 3^2 ). So, if ( a = 3 ) and ( b = 3 ), then ( n = 2^3 times 3^3 ). Is that sufficient? Let me check.Now, the third condition is that ( n^3 ) must be a perfect square. For ( n^3 ) to be a perfect square, all exponents in its prime factorization must be even. So, ( n^3 = 2^{3a} times 3^{3b} times ) ... must have exponents that are even. Therefore, ( 3a ) and ( 3b ) must be even numbers. This means ( a ) and ( b ) must be even numbers because 3 times an even number is even. But earlier, we found that ( a ) and ( b ) must be multiples of ( frac{3}{2} ). So, ( a ) and ( b ) need to satisfy both conditions: they must be multiples of ( frac{3}{2} ) and also even numbers.Let me think about this. If ( a ) and ( b ) must be multiples of ( frac{3}{2} ), that means ( a = frac{3}{2}k ) and ( b = frac{3}{2}m ) where ( k ) and ( m ) are integers. But since ( a ) and ( b ) must also be even, ( frac{3}{2}k ) and ( frac{3}{2}m ) must be even. So, ( k ) and ( m ) must be multiples of 4 because ( frac{3}{2} times 4 = 6 ), which is even. Therefore, the smallest ( k ) and ( m ) are 4, making ( a = 6 ) and ( b = 6 ).So, ( n = 2^6 times 3^6 ). Let me verify if this satisfies all the conditions.1. Divisible by 36: ( 2^6 times 3^6 ) certainly includes ( 2^2 times 3^2 ), so yes.2. ( n^2 ) is a perfect cube: ( n^2 = (2^6 times 3^6)^2 = 2^{12} times 3^{12} ). Both 12 and 12 are multiples of 3, so yes.3. ( n^3 ) is a perfect square: ( n^3 = (2^6 times 3^6)^3 = 2^{18} times 3^{18} ). Both 18 and 18 are even, so yes.Therefore, the smallest ( n ) is ( 2^6 times 3^6 ). The prime factors of ( n ) are 2 and 3. The largest prime factor is 3.Wait, but let me double-check if there's a smaller ( n ) that satisfies all conditions. Suppose I take ( a = 3 ) and ( b = 3 ), making ( n = 2^3 times 3^3 ). Then:1. Divisible by 36: ( 2^3 times 3^3 ) includes ( 2^2 times 3^2 ), so yes.2. ( n^2 = 2^6 times 3^6 ), which is a perfect cube because 6 is a multiple of 3.3. ( n^3 = 2^9 times 3^9 ). But 9 is odd, so ( n^3 ) is not a perfect square. Therefore, ( a = 3 ) and ( b = 3 ) don't work.So, indeed, ( a ) and ( b ) need to be 6. Therefore, the smallest ( n ) is ( 2^6 times 3^6 ), and the largest prime factor is 3.