💡First, I need to determine Sam's weight by subtracting Jack's weight from the total weight of both Jack and Sam.Sam's weight = Total weight - Jack's weightSam's weight = 96 pounds - 52 poundsSam's weight = 44 poundsNow that I know Sam's weight, I can find out how much heavier Jack is than Sam by subtracting Sam's weight from Jack's weight.Jack's weight - Sam's weight = How much heavier Jack is52 pounds - 44 pounds = 8 poundsTherefore, Jack is 8 pounds heavier than Sam.

💡Alright, so I've got this problem here about natural numbers p and q, and two conditions: Condition A is that p cubed minus q cubed is even, and Condition B is that p plus q is even. The question is asking about the relationship between these two conditions—whether A is sufficient, necessary, both, or neither for B.Okay, let's start by recalling what it means for a number to be even or odd. An even number is any integer that's divisible by 2, and an odd number is any integer that's not divisible by 2. So, if p and q are natural numbers, they can either be even or odd.Now, let's think about Condition A: p³ - q³ is even. I remember that the difference of cubes can be factored as (p - q)(p² + pq + q²). Hmm, so p³ - q³ equals (p - q)(p² + pq + q²). For this product to be even, at least one of the factors must be even. So, either (p - q) is even, or (p² + pq + q²) is even, or both.Let me consider the parity of p and q. If both p and q are even, then p - q is even, and p², pq, and q² are all even, so p² + pq + q² is even. Therefore, the entire expression p³ - q³ is even. Similarly, if both p and q are odd, then p - q is even (since odd minus odd is even), and p², pq, and q² are all odd, so p² + pq + q² is odd plus odd plus odd, which is odd. Wait, that would make p² + pq + q² odd, but p - q is even, so the product would still be even. So, in both cases where p and q have the same parity (both even or both odd), p³ - q³ is even.What if p and q have different parities? Let's say p is even and q is odd. Then p - q is odd, and p² is even, pq is even, and q² is odd. So p² + pq + q² is even + even + odd, which is odd. Therefore, the product (p - q)(p² + pq + q²) is odd times odd, which is odd. So, p³ - q³ would be odd in this case. Similarly, if p is odd and q is even, the same thing happens. So, p³ - q³ is odd when p and q have different parities.Okay, so Condition A (p³ - q³ is even) holds if and only if p and q have the same parity. Now, let's look at Condition B: p + q is even. For p + q to be even, p and q must also have the same parity. If both are even or both are odd, their sum is even. If one is even and the other is odd, their sum is odd.So, both Condition A and Condition B depend on p and q having the same parity. That suggests that they are related. Specifically, if p³ - q³ is even, then p + q must be even, and vice versa. Wait, is that true?Let me think again. If p³ - q³ is even, we've established that p and q must have the same parity, which means p + q is even. So, Condition A implies Condition B. Conversely, if p + q is even, then p and q have the same parity, which means p³ - q³ is even. So, Condition B implies Condition A as well.Therefore, Condition A and Condition B are equivalent. They imply each other. So, A is a necessary and sufficient condition for B.But wait, let me double-check. Is there any case where p + q is even, but p³ - q³ is odd? Or vice versa? From the earlier analysis, it seems that whenever p and q have the same parity, both p + q and p³ - q³ are even. And when they have different parities, both are odd. So, they always move together.Therefore, A is both necessary and sufficient for B. So, the correct answer should be option C.**Final Answer**boxed{C}

💡Okay, so I have this problem where a rectangle can be divided into ( n ) equal squares, and the same rectangle can also be divided into ( n + 76 ) equal squares. I need to find ( n ). Hmm, let me try to figure this out step by step.First, let's visualize the problem. If a rectangle can be divided into equal squares, that means the rectangle's sides must be multiples of the square's side length. So, if the rectangle has sides ( a ) and ( b ), then when divided into ( n ) squares, each square has a side length of ( sqrt{frac{ab}{n}} ). Similarly, when divided into ( n + 76 ) squares, each square has a side length of ( sqrt{frac{ab}{n + 76}} ).Since both divisions result in squares that perfectly fit the rectangle, the ratio of the side lengths of these squares must be a rational number. Let me denote this ratio as ( frac{p}{q} ), where ( p ) and ( q ) are coprime integers. So, we can write:[frac{sqrt{frac{ab}{n}}}{sqrt{frac{ab}{n + 76}}} = frac{p}{q}]Simplifying this, we get:[sqrt{frac{n + 76}{n}} = frac{p}{q}]Squaring both sides to eliminate the square roots:[frac{n + 76}{n} = left( frac{p}{q} right)^2]Let me denote ( frac{p^2}{q^2} ) as ( frac{n + 76}{n} ). This implies:[p^2 n = q^2 (n + 76)]Expanding the right side:[p^2 n = q^2 n + 76 q^2]Subtracting ( q^2 n ) from both sides:[n (p^2 - q^2) = 76 q^2]So, solving for ( n ):[n = frac{76 q^2}{p^2 - q^2}]Since ( n ) must be an integer, ( frac{76 q^2}{p^2 - q^2} ) must also be an integer. This means that ( p^2 - q^2 ) must be a divisor of 76. Let's list the positive divisors of 76: 1, 2, 4, 19, 38, and 76.Now, I need to check each divisor to see if it can be expressed as ( p^2 - q^2 ) where ( p ) and ( q ) are coprime integers. Let's go through them one by one.1. **Divisor 1**: ( p^2 - q^2 = 1 ) - This factors into ( (p + q)(p - q) = 1 ). The only solution is ( p + q = 1 ) and ( p - q = 1 ), but this would imply ( q = 0 ), which isn't valid since ( q ) must be a positive integer.2. **Divisor 2**: ( p^2 - q^2 = 2 ) - Factoring gives ( (p + q)(p - q) = 2 ). The possible pairs are (2,1), but solving these gives non-integer values for ( p ) and ( q ).3. **Divisor 4**: ( p^2 - q^2 = 4 ) - Factoring into ( (p + q)(p - q) = 4 ). Possible pairs are (4,1) and (2,2). Checking these: - For (4,1): ( p + q = 4 ), ( p - q = 1 ). Solving gives ( p = 2.5 ), which isn't an integer. - For (2,2): ( p + q = 2 ), ( p - q = 2 ). Solving gives ( q = 0 ), which isn't valid.4. **Divisor 19**: ( p^2 - q^2 = 19 ) - Factoring into ( (p + q)(p - q) = 19 ). Since 19 is prime, the only pair is (19,1). Solving gives non-integer values for ( p ) and ( q ).5. **Divisor 38**: ( p^2 - q^2 = 38 ) - Factoring into ( (p + q)(p - q) = 38 ). The possible pairs are (38,1) and (19,2). Both lead to non-integer solutions for ( p ) and ( q ).6. **Divisor 76**: ( p^2 - q^2 = 76 ) - Factoring into ( (p + q)(p - q) = 76 ). The possible pairs are (38,2) and (19,4). Let's check these: - For (38,2): ( p + q = 38 ), ( p - q = 2 ). Solving these equations: - Adding: ( 2p = 40 ) ⇒ ( p = 20 ) - Subtracting: ( 2q = 36 ) ⇒ ( q = 18 ) - Both are integers and coprime.Great, so we found ( p = 20 ) and ( q = 18 ). Now, substituting back into the equation for ( n ):[n = frac{76 times 18^2}{20^2 - 18^2}]Calculating the denominator:[20^2 - 18^2 = 400 - 324 = 76]So,[n = frac{76 times 324}{76} = 324]Therefore, the value of ( n ) is 324.**Final Answer**The value of ( n ) is boxed{324}.

💡First, I recognize that the given expression is a fraction involving sine and cosine functions. Since I know the value of (tan alpha), which is (frac{sin alpha}{cos alpha}), I can simplify the expression by dividing both the numerator and the denominator by (cos alpha). This will allow me to express everything in terms of (tan alpha), making it easier to substitute the known value.Next, I divide each term in the numerator and the denominator by (cos alpha):[frac{2sin alpha - cos alpha}{sin alpha + 2cos alpha} = frac{frac{2sin alpha}{cos alpha} - frac{cos alpha}{cos alpha}}{frac{sin alpha}{cos alpha} + frac{2cos alpha}{cos alpha}} = frac{2tan alpha - 1}{tan alpha + 2}]Now, I substitute (tan alpha = 2) into the simplified expression:[frac{2(2) - 1}{2 + 2} = frac{4 - 1}{4} = frac{3}{4}]Therefore, the value of the given expression is (frac{3}{4}).

💡Okay, so I've got this problem with three parts, all related to the function ( f(x) = 4^x - 2^x ). Let me take it step by step.**Problem 1: Find the range of ( f(x) ) when its domain is ([-1, 1]).**Alright, so I need to figure out the minimum and maximum values of ( f(x) ) when ( x ) is between -1 and 1. First, I notice that ( 4^x ) can be written as ( (2^x)^2 ). So, let me set ( t = 2^x ). Then, ( f(x) ) becomes ( t^2 - t ). That simplifies things a bit because now I have a quadratic function in terms of ( t ).Next, I need to determine the range of ( t ) when ( x ) is in ([-1, 1]). Since ( 2^x ) is an exponential function, when ( x = -1 ), ( t = 2^{-1} = frac{1}{2} ), and when ( x = 1 ), ( t = 2^1 = 2 ). So, ( t ) ranges from ( frac{1}{2} ) to ( 2 ).Now, the function ( f(t) = t^2 - t ) is a quadratic. To find its range over the interval ( [frac{1}{2}, 2] ), I can analyze it. The quadratic ( t^2 - t ) has its vertex at ( t = frac{1}{2} ) because the vertex of ( at^2 + bt + c ) is at ( t = -frac{b}{2a} ). Here, ( a = 1 ) and ( b = -1 ), so ( t = frac{1}{2} ).Calculating ( f ) at the endpoints and the vertex:- At ( t = frac{1}{2} ): ( f(frac{1}{2}) = (frac{1}{2})^2 - frac{1}{2} = frac{1}{4} - frac{1}{2} = -frac{1}{4} ).- At ( t = 2 ): ( f(2) = 2^2 - 2 = 4 - 2 = 2 ).Since the quadratic opens upwards (because the coefficient of ( t^2 ) is positive), the minimum value is at the vertex, which is ( -frac{1}{4} ), and the maximum is at ( t = 2 ), which is 2.So, the range of ( f(x) ) is from ( -frac{1}{4} ) to 2.**Problem 2: Find the functional relationship ( b = g(a) ) and the domain of ( g(a) ).**Given that ( f(s) + f(t) = 0 ), and ( a = 2^s + 2^t ), ( b = 2^{s+t} ).First, let's write out ( f(s) + f(t) = 0 ):( 4^s - 2^s + 4^t - 2^t = 0 )I can rewrite ( 4^s ) as ( (2^s)^2 ) and similarly ( 4^t = (2^t)^2 ). Let me set ( u = 2^s ) and ( v = 2^t ). Then the equation becomes:( u^2 - u + v^2 - v = 0 )Combine like terms:( u^2 + v^2 - u - v = 0 )Now, notice that ( a = u + v ) and ( b = uv ) because ( 2^{s+t} = 2^s cdot 2^t = u cdot v ).So, let's express ( u^2 + v^2 ) in terms of ( a ) and ( b ). We know that ( u^2 + v^2 = (u + v)^2 - 2uv = a^2 - 2b ).Substituting back into the equation:( (a^2 - 2b) - (u + v) = 0 )But ( u + v = a ), so:( a^2 - 2b - a = 0 )Solving for ( b ):( a^2 - a = 2b )( b = frac{a^2 - a}{2} )So, ( b = g(a) = frac{a^2 - a}{2} ).Now, I need to find the domain of ( g(a) ). That is, the possible values of ( a ).Since ( a = 2^s + 2^t ), and ( s ) and ( t ) are real numbers, ( 2^s ) and ( 2^t ) are positive. So, ( a > 0 ).But we also have the condition from Problem 1 that ( f(s) + f(t) = 0 ). From Problem 1, we know that ( f(x) ) can take values from ( -frac{1}{4} ) to 2. So, ( f(s) = -f(t) ), meaning ( f(s) ) and ( f(t) ) are negatives of each other.Therefore, ( f(s) ) must be in the range ( [-frac{1}{4}, 2] ), so ( f(t) = -f(s) ) must be in ( [-2, frac{1}{4}] ). But since ( f(t) ) must also be within ( [-frac{1}{4}, 2] ), the overlap is ( [-frac{1}{4}, frac{1}{4}] ).Wait, that might complicate things. Maybe another approach is better.Since ( a = 2^s + 2^t ) and ( b = 2^{s+t} ), we can use the AM-GM inequality, which states that ( frac{u + v}{2} geq sqrt{uv} ). So, ( a geq 2sqrt{b} ).From the equation ( b = frac{a^2 - a}{2} ), substitute into the inequality:( a geq 2sqrt{frac{a^2 - a}{2}} )Simplify:( a geq 2 cdot sqrt{frac{a^2 - a}{2}} )Square both sides to eliminate the square root:( a^2 geq 4 cdot frac{a^2 - a}{2} )Simplify the right side:( a^2 geq 2(a^2 - a) )Expand:( a^2 geq 2a^2 - 2a )Bring all terms to one side:( -a^2 + 2a geq 0 )Multiply both sides by -1 (remember to reverse the inequality):( a^2 - 2a leq 0 )Factor:( a(a - 2) leq 0 )So, the solutions are ( 0 leq a leq 2 ).But since ( a = 2^s + 2^t > 0 ), the domain is ( 0 < a leq 2 ).However, we also have the condition from ( f(s) + f(t) = 0 ). Let's see if there are more restrictions.From ( f(s) + f(t) = 0 ), we have ( 4^s - 2^s + 4^t - 2^t = 0 ). As before, ( u^2 - u + v^2 - v = 0 ), which led us to ( b = frac{a^2 - a}{2} ).Since ( b = 2^{s+t} > 0 ), we have ( frac{a^2 - a}{2} > 0 ), which implies ( a^2 - a > 0 ). So, ( a(a - 1) > 0 ). This inequality holds when ( a > 1 ) or ( a < 0 ). But since ( a > 0 ), the domain is ( a > 1 ).Combining this with the earlier result ( a leq 2 ), the domain of ( g(a) ) is ( (1, 2] ).**Problem 3: Find the range of ( 8^s + 8^t ).**Again, let me express ( 8^s ) and ( 8^t ) in terms of ( u ) and ( v ). Since ( 8 = 2^3 ), we have ( 8^s = (2^s)^3 = u^3 ) and ( 8^t = v^3 ). So, ( 8^s + 8^t = u^3 + v^3 ).I can factor ( u^3 + v^3 ) as ( (u + v)(u^2 - uv + v^2) ). We already know ( u + v = a ) and ( uv = b ). Also, ( u^2 + v^2 = a^2 - 2b ).So, ( u^3 + v^3 = a(a^2 - 3b) ).From Problem 2, we have ( b = frac{a^2 - a}{2} ). Substitute this into the expression:( u^3 + v^3 = aleft(a^2 - 3 cdot frac{a^2 - a}{2}right) )Simplify inside the parentheses:( a^2 - frac{3a^2 - 3a}{2} = frac{2a^2 - 3a^2 + 3a}{2} = frac{-a^2 + 3a}{2} )So, ( u^3 + v^3 = a cdot frac{-a^2 + 3a}{2} = frac{-a^3 + 3a^2}{2} ).Thus, ( 8^s + 8^t = frac{-a^3 + 3a^2}{2} ).Let me denote this as ( h(a) = frac{-a^3 + 3a^2}{2} ). To find the range of ( h(a) ), I need to analyze it over the domain ( a in (1, 2] ).First, let's find the derivative of ( h(a) ) to check for extrema.( h(a) = frac{-a^3 + 3a^2}{2} )( h'(a) = frac{-3a^2 + 6a}{2} = frac{-3a(a - 2)}{2} )Set ( h'(a) = 0 ):( frac{-3a(a - 2)}{2} = 0 )Solutions are ( a = 0 ) and ( a = 2 ). But our domain is ( (1, 2] ), so the critical point within the domain is ( a = 2 ).Now, let's evaluate ( h(a) ) at the endpoints and critical points.- At ( a = 1 ): ( h(1) = frac{-1 + 3}{2} = frac{2}{2} = 1 ).- At ( a = 2 ): ( h(2) = frac{-8 + 12}{2} = frac{4}{2} = 2 ).Since ( h'(a) ) is positive in ( (1, 2) ) (because for ( 1 < a < 2 ), ( -3a(a - 2) ) is positive since ( a - 2 ) is negative and multiplied by negative gives positive), the function is increasing on ( (1, 2] ).Therefore, the range of ( h(a) ) is ( (1, 2] ).**Final Answer**1. The range of ( f(x) ) is (boxed{[-frac{1}{4}, 2]}).2. The functional relationship is ( b = frac{a^2 - a}{2} ) with the domain of ( g(a) ) being (boxed{(1, 2]}).3. The range of ( 8^s + 8^t ) is (boxed{(1, 2]}).

💡Okay, so I have this problem where I need to convert a point from cylindrical coordinates to rectangular coordinates. The point given is (7, π/3, -3). Hmm, I remember that cylindrical coordinates are represented as (r, θ, z), where r is the radius, θ is the angle, and z is the height. Rectangular coordinates, on the other hand, are (x, y, z). So, I need to find x and y in terms of r, θ, and z.Let me recall the conversion formulas. I think they involve trigonometric functions. From what I remember, x is equal to r multiplied by the cosine of θ, and y is equal to r multiplied by the sine of θ. The z-coordinate remains the same in both cylindrical and rectangular systems. So, the formulas should be:x = r * cos(θ)y = r * sin(θ)z = zAlright, so plugging in the values from the given point (7, π/3, -3):First, let's compute x. That would be 7 multiplied by the cosine of π/3. I remember that π/3 radians is 60 degrees. The cosine of 60 degrees is 0.5, right? So, 7 times 0.5 is 3.5. So, x is 3.5.Next, let's find y. That's 7 multiplied by the sine of π/3. The sine of 60 degrees is √3/2, which is approximately 0.866. So, 7 times √3/2 is (7√3)/2. I can leave it in exact form since it's more precise.And finally, z remains the same, which is -3.So, putting it all together, the rectangular coordinates should be (3.5, (7√3)/2, -3). Let me double-check my calculations to make sure I didn't make any mistakes.For x: 7 * cos(π/3) = 7 * 0.5 = 3.5. That seems correct.For y: 7 * sin(π/3) = 7 * (√3/2) = (7√3)/2. Yep, that looks right.And z is just -3, so no issues there.I think that's all. I don't see any mistakes in my reasoning. So, the rectangular coordinates corresponding to the cylindrical coordinates (7, π/3, -3) are indeed (3.5, (7√3)/2, -3).