answer：First, I need to determine how many beaded necklaces Kylie made on Monday. I'll start by calculating the total number of beads used for the jewelry made on Tuesday and Wednesday.On Tuesday, Kylie made 2 beaded necklaces. Each necklace requires 20 beads, so the total beads used for necklaces on Tuesday are:2 necklaces * 20 beads/necklace = 40 beadsOn Wednesday, Kylie made 5 beaded bracelets and 7 beaded earrings. The beads used for bracelets are:5 bracelets * 10 beads/bracelet = 50 beadsThe beads used for earrings are:7 earrings * 5 beads/earring = 35 beadsAdding up the beads used on Tuesday and Wednesday:40 beads (necklaces) + 50 beads (bracelets) + 35 beads (earrings) = 125 beadsKylie used a total of 325 beads. To find out how many beads were used on Monday:325 beads (total) - 125 beads (Tuesday and Wednesday) = 200 beadsSince each necklace requires 20 beads, the number of necklaces made on Monday is:200 beads / 20 beads/necklace = 10 necklacesTherefore, Kylie made 10 beaded necklaces on Monday.

answer：First, I need to determine the total number of students who tried out for the basketball team. There are 14 boys, and an unknown number of girls, which I'll denote as G.Next, I know that only 2 students were called back, and 21 students didn't make the cut. This means the total number of students who tried out is the sum of those who were called back and those who didn't make the cut.So, the total number of students is 2 (called back) + 21 (didn't make the cut) = 23 students.Now, I can set up the equation: G (number of girls) + 14 (number of boys) = 23 (total students).Solving for G, I subtract 14 from both sides: G = 23 - 14 = 9.Therefore, there were 9 girls who tried out for the team.

answer：Alright, so I have this problem about a sequence where each term satisfies a certain condition. The sequence is defined by the recurrence relation ( a_n a_{n+1} a_{n+2} = a_n + a_{n+1} + a_{n+2} ), with the initial terms ( a_1 = 1 ) and ( a_2 = 2 ). I need to find the value of ( A ) in the expression ( a_n = A sin(omega n + phi) + c ), where ( omega > 0 ) and ( |phi| < frac{pi}{2} ).First, I should probably try to compute the next few terms of the sequence to see if there's a pattern. Let me start by finding ( a_3 ).For ( n = 1 ):( a_1 a_2 a_3 = a_1 + a_2 + a_3 )Plugging in the known values:( 1 times 2 times a_3 = 1 + 2 + a_3 )Simplify:( 2a_3 = 3 + a_3 )Subtract ( a_3 ) from both sides:( a_3 = 3 )Okay, so ( a_3 = 3 ). Now, let's find ( a_4 ).For ( n = 2 ):( a_2 a_3 a_4 = a_2 + a_3 + a_4 )Plugging in the known values:( 2 times 3 times a_4 = 2 + 3 + a_4 )Simplify:( 6a_4 = 5 + a_4 )Subtract ( a_4 ) from both sides:( 5a_4 = 5 )Divide both sides by 5:( a_4 = 1 )Hmm, ( a_4 = 1 ). That's interesting, it's the same as ( a_1 ). Let's see if this pattern continues.For ( n = 3 ):( a_3 a_4 a_5 = a_3 + a_4 + a_5 )Plugging in the known values:( 3 times 1 times a_5 = 3 + 1 + a_5 )Simplify:( 3a_5 = 4 + a_5 )Subtract ( a_5 ) from both sides:( 2a_5 = 4 )Divide both sides by 2:( a_5 = 2 )Wait a second, ( a_5 = 2 ) which is the same as ( a_2 ). So, it seems like the sequence is repeating every three terms: 1, 2, 3, 1, 2, 3, and so on. So, it's periodic with period 3.Given that, the general term ( a_n ) can be expressed as a periodic function with period 3. The problem suggests that it can be written in the form ( A sin(omega n + phi) + c ). Since the period is 3, the angular frequency ( omega ) should satisfy ( frac{2pi}{omega} = 3 ), so ( omega = frac{2pi}{3} ).So, the expression becomes:( a_n = A sinleft(frac{2pi}{3}n + phiright) + c )Now, I need to find the constants ( A ), ( phi ), and ( c ). I have three initial terms: ( a_1 = 1 ), ( a_2 = 2 ), and ( a_3 = 3 ). Let's plug these into the equation.For ( n = 1 ):( 1 = A sinleft(frac{2pi}{3} times 1 + phiright) + c )Simplify:( 1 = A sinleft(frac{2pi}{3} + phiright) + c ) --- Equation (1)For ( n = 2 ):( 2 = A sinleft(frac{2pi}{3} times 2 + phiright) + c )Simplify:( 2 = A sinleft(frac{4pi}{3} + phiright) + c ) --- Equation (2)For ( n = 3 ):( 3 = A sinleft(frac{2pi}{3} times 3 + phiright) + c )Simplify:( 3 = A sinleft(2pi + phiright) + c )But ( sin(2pi + phi) = sin(phi) ), so:( 3 = A sin(phi) + c ) --- Equation (3)Now, I have three equations:1. ( 1 = A sinleft(frac{2pi}{3} + phiright) + c )2. ( 2 = A sinleft(frac{4pi}{3} + phiright) + c )3. ( 3 = A sin(phi) + c )I can use these to solve for ( A ), ( phi ), and ( c ). Let's subtract Equation (1) from Equation (3) to eliminate ( c ):( 3 - 1 = A sin(phi) - A sinleft(frac{2pi}{3} + phiright) )Simplify:( 2 = A left[ sin(phi) - sinleft(frac{2pi}{3} + phiright) right] )Similarly, subtract Equation (2) from Equation (3):( 3 - 2 = A sin(phi) - A sinleft(frac{4pi}{3} + phiright) )Simplify:( 1 = A left[ sin(phi) - sinleft(frac{4pi}{3} + phiright) right] )Now, I have two equations:1. ( 2 = A left[ sin(phi) - sinleft(frac{2pi}{3} + phiright) right] ) --- Equation (4)2. ( 1 = A left[ sin(phi) - sinleft(frac{4pi}{3} + phiright) right] ) --- Equation (5)I can use the sine subtraction formula or perhaps use trigonometric identities to simplify the terms inside the brackets.Recall that ( sin(A) - sin(B) = 2 cosleft(frac{A + B}{2}right) sinleft(frac{A - B}{2}right) ).Let's apply this identity to both equations.Starting with Equation (4):( sin(phi) - sinleft(frac{2pi}{3} + phiright) = 2 cosleft(frac{phi + frac{2pi}{3} + phi}{2}right) sinleft(frac{phi - left(frac{2pi}{3} + phiright)}{2}right) )Simplify the arguments:( = 2 cosleft(frac{2phi + frac{2pi}{3}}{2}right) sinleft(frac{-frac{2pi}{3}}{2}right) )( = 2 cosleft(phi + frac{pi}{3}right) sinleft(-frac{pi}{3}right) )Since ( sin(-x) = -sin(x) ):( = 2 cosleft(phi + frac{pi}{3}right) left(-frac{sqrt{3}}{2}right) )( = -sqrt{3} cosleft(phi + frac{pi}{3}right) )So, Equation (4) becomes:( 2 = A times left(-sqrt{3} cosleft(phi + frac{pi}{3}right)right) )( 2 = -A sqrt{3} cosleft(phi + frac{pi}{3}right) ) --- Equation (4a)Similarly, apply the identity to Equation (5):( sin(phi) - sinleft(frac{4pi}{3} + phiright) = 2 cosleft(frac{phi + frac{4pi}{3} + phi}{2}right) sinleft(frac{phi - left(frac{4pi}{3} + phiright)}{2}right) )Simplify the arguments:( = 2 cosleft(frac{2phi + frac{4pi}{3}}{2}right) sinleft(frac{-frac{4pi}{3}}{2}right) )( = 2 cosleft(phi + frac{2pi}{3}right) sinleft(-frac{2pi}{3}right) )Again, ( sin(-x) = -sin(x) ):( = 2 cosleft(phi + frac{2pi}{3}right) left(-frac{sqrt{3}}{2}right) )( = -sqrt{3} cosleft(phi + frac{2pi}{3}right) )So, Equation (5) becomes:( 1 = A times left(-sqrt{3} cosleft(phi + frac{2pi}{3}right)right) )( 1 = -A sqrt{3} cosleft(phi + frac{2pi}{3}right) ) --- Equation (5a)Now, I have Equations (4a) and (5a):1. ( 2 = -A sqrt{3} cosleft(phi + frac{pi}{3}right) )2. ( 1 = -A sqrt{3} cosleft(phi + frac{2pi}{3}right) )Let me denote ( theta = phi + frac{pi}{3} ). Then, ( phi + frac{2pi}{3} = theta + frac{pi}{3} ).So, Equation (4a) becomes:( 2 = -A sqrt{3} cos(theta) ) --- Equation (6)Equation (5a) becomes:( 1 = -A sqrt{3} cosleft(theta + frac{pi}{3}right) ) --- Equation (7)Now, I can write Equation (7) in terms of ( cos(theta) ) and ( sin(theta) ) using the cosine addition formula:( cosleft(theta + frac{pi}{3}right) = cos(theta)cosleft(frac{pi}{3}right) - sin(theta)sinleft(frac{pi}{3}right) )( = cos(theta) times frac{1}{2} - sin(theta) times frac{sqrt{3}}{2} )So, Equation (7) becomes:( 1 = -A sqrt{3} left( frac{1}{2} cos(theta) - frac{sqrt{3}}{2} sin(theta) right) )Simplify:( 1 = -A sqrt{3} times frac{1}{2} cos(theta) + A sqrt{3} times frac{sqrt{3}}{2} sin(theta) )( 1 = -frac{A sqrt{3}}{2} cos(theta) + frac{3A}{2} sin(theta) )From Equation (6), we have:( 2 = -A sqrt{3} cos(theta) )So, ( cos(theta) = -frac{2}{A sqrt{3}} )Let me plug this into Equation (7):( 1 = -frac{A sqrt{3}}{2} times left(-frac{2}{A sqrt{3}}right) + frac{3A}{2} sin(theta) )Simplify:( 1 = frac{A sqrt{3} times 2}{2 A sqrt{3}} + frac{3A}{2} sin(theta) )( 1 = 1 + frac{3A}{2} sin(theta) )Subtract 1 from both sides:( 0 = frac{3A}{2} sin(theta) )So, ( sin(theta) = 0 )But ( sin(theta) = 0 ) implies that ( theta = kpi ) for some integer ( k ). However, since ( |phi| < frac{pi}{2} ), let's see what this means for ( theta ).Recall that ( theta = phi + frac{pi}{3} ). Given ( |phi| < frac{pi}{2} ), then ( theta ) lies in the interval ( left(-frac{pi}{2} + frac{pi}{3}, frac{pi}{2} + frac{pi}{3}right) ), which is ( left(-frac{pi}{6}, frac{5pi}{6}right) ).So, ( theta ) is between ( -frac{pi}{6} ) and ( frac{5pi}{6} ). The solutions for ( sin(theta) = 0 ) in this interval are ( theta = 0 ) and ( theta = pi ). But ( pi ) is outside the upper bound of ( frac{5pi}{6} ), so the only solution is ( theta = 0 ).Therefore, ( theta = 0 ), which means ( phi + frac{pi}{3} = 0 ), so ( phi = -frac{pi}{3} ).Now, from Equation (6):( 2 = -A sqrt{3} cos(0) )Since ( cos(0) = 1 ):( 2 = -A sqrt{3} )So, ( A = -frac{2}{sqrt{3}} )Rationalizing the denominator:( A = -frac{2sqrt{3}}{3} )Let me check if this makes sense with Equation (3). From Equation (3):( 3 = A sin(phi) + c )We know ( A = -frac{2sqrt{3}}{3} ) and ( phi = -frac{pi}{3} ), so:( sin(-frac{pi}{3}) = -frac{sqrt{3}}{2} )Thus:( 3 = -frac{2sqrt{3}}{3} times left(-frac{sqrt{3}}{2}right) + c )Simplify:( 3 = frac{2sqrt{3} times sqrt{3}}{6} + c )( 3 = frac{6}{6} + c )( 3 = 1 + c )So, ( c = 2 )Now, let's verify with Equation (1):( 1 = A sinleft(frac{2pi}{3} + phiright) + c )Plug in ( A = -frac{2sqrt{3}}{3} ), ( phi = -frac{pi}{3} ), and ( c = 2 ):( 1 = -frac{2sqrt{3}}{3} sinleft(frac{2pi}{3} - frac{pi}{3}right) + 2 )Simplify the argument:( frac{2pi}{3} - frac{pi}{3} = frac{pi}{3} )So:( 1 = -frac{2sqrt{3}}{3} sinleft(frac{pi}{3}right) + 2 )( sinleft(frac{pi}{3}right) = frac{sqrt{3}}{2} )Thus:( 1 = -frac{2sqrt{3}}{3} times frac{sqrt{3}}{2} + 2 )Simplify:( 1 = -frac{6}{6} + 2 )( 1 = -1 + 2 )( 1 = 1 ) ✓Similarly, check Equation (2):( 2 = A sinleft(frac{4pi}{3} + phiright) + c )Plug in the values:( 2 = -frac{2sqrt{3}}{3} sinleft(frac{4pi}{3} - frac{pi}{3}right) + 2 )Simplify the argument:( frac{4pi}{3} - frac{pi}{3} = pi )( sin(pi) = 0 )So:( 2 = -frac{2sqrt{3}}{3} times 0 + 2 )( 2 = 0 + 2 )( 2 = 2 ) ✓Everything checks out. So, the value of ( A ) is ( -frac{2sqrt{3}}{3} ).**Final Answer**The value of the real number ( A ) is boxed{-dfrac{2sqrt{3}}{3}}.

answer：Okay, so I have this equation to solve: (2 cdot 2^x + sqrt[3]{8 cdot 8^x} = 32). Hmm, that looks a bit complicated, but maybe I can break it down step by step. Let me think about how to approach this.First, I notice that both terms involve exponents with base 2 and 8. I know that 8 is a power of 2, specifically (8 = 2^3). Maybe I can rewrite everything in terms of base 2 to make it easier to handle. That often helps when dealing with exponential equations.Starting with the first term, (2 cdot 2^x). Well, that's straightforward. It's just 2 multiplied by 2 raised to the power of x. I can leave that as it is for now.Now, looking at the second term, (sqrt[3]{8 cdot 8^x}). Since 8 is (2^3), I can rewrite 8 as (2^3). So, let's substitute that in:[sqrt[3]{8 cdot 8^x} = sqrt[3]{2^3 cdot (2^3)^x}]Wait, ((2^3)^x) is the same as (2^{3x}), right? So, the expression inside the cube root becomes:[2^3 cdot 2^{3x} = 2^{3 + 3x}]So now, the cube root of (2^{3 + 3x}) is:[sqrt[3]{2^{3 + 3x}} = 2^{(3 + 3x)/3} = 2^{1 + x}]Oh, that's simpler! So, the second term simplifies to (2^{x + 1}).Now, let's rewrite the original equation with these simplifications:[2 cdot 2^x + 2^{x + 1} = 32]Hmm, both terms on the left side have (2^x) in them. Maybe I can factor that out. Let's see:First term: (2 cdot 2^x = 2^{1} cdot 2^x = 2^{x + 1})Second term: (2^{x + 1})So, both terms are actually the same! That means:[2^{x + 1} + 2^{x + 1} = 32]Which simplifies to:[2 cdot 2^{x + 1} = 32]Wait, no, that's not quite right. Let me think again. If both terms are (2^{x + 1}), then adding them together is like:[2^{x + 1} + 2^{x + 1} = 2 cdot 2^{x + 1} = 2^{x + 2}]Oh, right! Because (2^{x + 1} + 2^{x + 1} = 2 cdot 2^{x + 1} = 2^{1} cdot 2^{x + 1} = 2^{x + 2}). So, the equation becomes:[2^{x + 2} = 32]Now, I need to solve for x. I know that 32 is a power of 2 as well. Specifically, (32 = 2^5). So, I can rewrite the equation as:[2^{x + 2} = 2^5]Since the bases are the same, the exponents must be equal. Therefore:[x + 2 = 5]Solving for x:[x = 5 - 2][x = 3]Let me double-check my steps to make sure I didn't make a mistake. I converted 8 to (2^3), simplified the cube root correctly, recognized that both terms were the same, combined them, and then solved for x. It all seems to add up. Plugging x = 3 back into the original equation:First term: (2 cdot 2^3 = 2 cdot 8 = 16)Second term: (sqrt[3]{8 cdot 8^3} = sqrt[3]{8 cdot 512} = sqrt[3]{4096} = 16)Adding them together: 16 + 16 = 32, which matches the right side of the equation. So, x = 3 is indeed the correct solution.

answer：Alright, so I have this problem about a rectangle PQRS with an area of 2016. There's also an ellipse involved that passes through points P and R, and it has its foci at Q and S. The area of the ellipse is given as 2016π. I need to find the perimeter of the rectangle. Hmm, okay, let me try to break this down step by step.First, let me recall some properties of rectangles and ellipses. A rectangle has opposite sides equal and all angles are right angles. The area of a rectangle is length times width, so if I let the sides be x and y, then the area is x*y = 2016. That's straightforward.Now, moving on to the ellipse. The area of an ellipse is π*a*b, where a is the semi-major axis and b is the semi-minor axis. Here, the area is given as 2016π, so that means π*a*b = 2016π. If I divide both sides by π, I get a*b = 2016. So, the product of the semi-major and semi-minor axes is 2016.Next, the ellipse passes through points P and R, which are opposite corners of the rectangle. The foci of the ellipse are at Q and S, the other two opposite corners. So, the foci are at the other two vertices of the rectangle. Interesting.I remember that for an ellipse, the distance between the foci is 2c, where c is the distance from the center to each focus. Also, for an ellipse, the relationship between a, b, and c is a^2 = b^2 + c^2. So, if I can find c, I can relate it to a and b.But how does this ellipse relate to the rectangle? The ellipse passes through P and R, which are diagonal points of the rectangle. The distance between P and R is the diagonal of the rectangle. Let me calculate that. In a rectangle with sides x and y, the diagonal length is sqrt(x^2 + y^2). So, the distance between P and R is sqrt(x^2 + y^2).Now, since the ellipse passes through P and R, these points must satisfy the ellipse's equation. But I'm not sure how to directly use that information. Maybe I need to think about the properties of the ellipse related to the rectangle.Wait, the foci are at Q and S, which are the other two corners of the rectangle. So, the distance between Q and S is also the diagonal of the rectangle, which is the same as the distance between P and R, right? So, the distance between the foci is sqrt(x^2 + y^2). But in an ellipse, the distance between the foci is 2c. Therefore, 2c = sqrt(x^2 + y^2). So, c = (sqrt(x^2 + y^2))/2.Earlier, I had that a*b = 2016, and a^2 = b^2 + c^2. So, if I can express c in terms of x and y, maybe I can find a relationship between a, b, x, and y.Let me write down what I have so far:1. x*y = 2016 (area of the rectangle)2. a*b = 2016 (area of the ellipse)3. c = sqrt(x^2 + y^2)/2 (distance from center to focus)4. a^2 = b^2 + c^2 (relationship in an ellipse)So, I have four equations here. Maybe I can solve for a and b in terms of x and y, and then find a relationship that allows me to solve for x and y.Let me substitute c into the fourth equation:a^2 = b^2 + (sqrt(x^2 + y^2)/2)^2a^2 = b^2 + (x^2 + y^2)/4But I also know that a*b = 2016. Hmm, so maybe I can express a in terms of b or vice versa.Let me solve for a from equation 2: a = 2016 / bSubstitute this into equation 4:(2016 / b)^2 = b^2 + (x^2 + y^2)/4Simplify:(2016^2) / b^2 = b^2 + (x^2 + y^2)/4Multiply both sides by b^2 to eliminate the denominator:2016^2 = b^4 + (x^2 + y^2)/4 * b^2Hmm, this is getting a bit complicated. Maybe I need another approach.I remember that in an ellipse, the sum of the distances from any point on the ellipse to the two foci is constant and equal to 2a. So, for points P and R, which lie on the ellipse, the sum of their distances to the foci Q and S should be equal to 2a.Let me calculate the distance from P to Q and from P to S. Wait, P is a vertex of the rectangle, and Q and S are the other two vertices. So, the distance from P to Q is one side of the rectangle, say x, and the distance from P to S is the other side, y. But wait, that's not correct because Q and S are opposite corners, not adjacent.Wait, no. Let me think again. If P is one corner, then Q and S are the other two corners adjacent to P? No, in a rectangle, each corner is connected to two adjacent corners. So, if P is connected to Q and S, then Q and S are adjacent to P, but in reality, in a rectangle, each corner is connected to two adjacent corners, and the opposite corner is connected via the diagonal.Wait, maybe I'm getting confused. Let me label the rectangle properly. Let's say rectangle PQRS has vertices P, Q, R, S in order, so that PQ, QR, RS, and SP are the sides. Then, the diagonals are PR and QS. So, points P and R are opposite, as are Q and S.So, the foci are at Q and S, which are opposite corners, and the ellipse passes through P and R, which are the other two opposite corners.So, the distance from P to Q is one side, say x, and the distance from P to S is the other side, y. Similarly, the distance from R to Q is y, and from R to S is x.But since the ellipse passes through P and R, the sum of the distances from P to Q and P to S should be equal to 2a, right? Because for any point on the ellipse, the sum of the distances to the two foci is 2a.So, for point P, the sum of distances to Q and S is PQ + PS = x + y. Similarly, for point R, the sum of distances to Q and S is RQ + RS = y + x. So, both give the same sum, which is x + y. Therefore, 2a = x + y. So, a = (x + y)/2.That's a useful relationship. So, a is half the sum of the sides of the rectangle.Now, going back to the area of the ellipse, which is π*a*b = 2016π, so a*b = 2016.We also have the relationship a^2 = b^2 + c^2, where c is the distance from the center to each focus.Earlier, I found that c = sqrt(x^2 + y^2)/2, since the distance between the foci is the diagonal of the rectangle.So, let me write down:a = (x + y)/2c = sqrt(x^2 + y^2)/2a^2 = b^2 + c^2a*b = 2016So, let's substitute a and c into the equation a^2 = b^2 + c^2:[(x + y)/2]^2 = b^2 + [sqrt(x^2 + y^2)/2]^2Let me compute each term:Left side: [(x + y)/2]^2 = (x^2 + 2xy + y^2)/4Right side: b^2 + [sqrt(x^2 + y^2)/2]^2 = b^2 + (x^2 + y^2)/4So, setting them equal:(x^2 + 2xy + y^2)/4 = b^2 + (x^2 + y^2)/4Subtract (x^2 + y^2)/4 from both sides:(2xy)/4 = b^2Simplify:xy/2 = b^2But we know from the area of the rectangle that x*y = 2016, so:2016/2 = b^2 => 1008 = b^2 => b = sqrt(1008)Okay, so b is sqrt(1008). Now, since a*b = 2016, we can find a:a = 2016 / b = 2016 / sqrt(1008)Simplify sqrt(1008). Let's see, 1008 factors into 16*63, so sqrt(16*63) = 4*sqrt(63). And sqrt(63) is 3*sqrt(7), so sqrt(1008) = 4*3*sqrt(7) = 12*sqrt(7). Wait, let me check that:Wait, 1008 divided by 16 is 63, yes. So sqrt(1008) = sqrt(16*63) = 4*sqrt(63). Then sqrt(63) is sqrt(9*7) = 3*sqrt(7). So, sqrt(1008) = 4*3*sqrt(7) = 12*sqrt(7). So, b = 12*sqrt(7).Therefore, a = 2016 / (12*sqrt(7)) = (2016 / 12) / sqrt(7) = 168 / sqrt(7). Rationalizing the denominator, multiply numerator and denominator by sqrt(7):a = (168*sqrt(7)) / 7 = 24*sqrt(7)So, a = 24*sqrt(7). Earlier, we had a = (x + y)/2, so:(x + y)/2 = 24*sqrt(7) => x + y = 48*sqrt(7)Now, the perimeter of the rectangle is 2*(x + y). So, perimeter = 2*(48*sqrt(7)) = 96*sqrt(7)Wait, but let me double-check my steps to make sure I didn't make a mistake.Starting from b^2 = 1008, so b = sqrt(1008) = 12*sqrt(7). Then a = 2016 / (12*sqrt(7)) = 168 / sqrt(7) = 24*sqrt(7). Then a = (x + y)/2, so x + y = 48*sqrt(7). Therefore, perimeter is 2*(x + y) = 96*sqrt(7). That seems consistent.But let me verify the relationship a^2 = b^2 + c^2:a = 24*sqrt(7), so a^2 = (24^2)*(7) = 576*7 = 4032b = 12*sqrt(7), so b^2 = 144*7 = 1008c = sqrt(x^2 + y^2)/2. We know that x*y = 2016, and x + y = 48*sqrt(7). Let me compute x^2 + y^2.We know that (x + y)^2 = x^2 + 2xy + y^2, so x^2 + y^2 = (x + y)^2 - 2xy = (48*sqrt(7))^2 - 2*2016Compute (48*sqrt(7))^2 = 48^2 * 7 = 2304 * 7 = 16128Then, 2*2016 = 4032So, x^2 + y^2 = 16128 - 4032 = 12096Therefore, c = sqrt(12096)/2Compute sqrt(12096). Let's factor 12096:12096 divided by 16 is 756.756 divided by 4 is 189.189 is 9*21, which is 9*3*7.So, 12096 = 16 * 4 * 9 * 3 * 7 = 16*4*9*21Wait, sqrt(12096) = sqrt(16*4*9*21) = 4*2*3*sqrt(21) = 24*sqrt(21)Therefore, c = 24*sqrt(21)/2 = 12*sqrt(21)Now, let's check a^2 = b^2 + c^2:a^2 = 4032b^2 + c^2 = 1008 + (12*sqrt(21))^2 = 1008 + 144*21 = 1008 + 3024 = 4032Yes, that checks out. So, all the relationships are consistent.Therefore, the perimeter of the rectangle is 96*sqrt(7). But let me see if that can be simplified further or if I made a miscalculation.Wait, 96*sqrt(7) is already simplified. So, that should be the answer.But just to recap:- Area of rectangle: x*y = 2016- Area of ellipse: π*a*b = 2016π => a*b = 2016- Sum of distances from any point on ellipse to foci is 2a, which for points P and R gives 2a = x + y => a = (x + y)/2- Relationship in ellipse: a^2 = b^2 + c^2, where c is distance from center to focus- Distance between foci is diagonal of rectangle: 2c = sqrt(x^2 + y^2) => c = sqrt(x^2 + y^2)/2- Solved for b^2 = 1008, so b = 12*sqrt(7)- Then a = 24*sqrt(7), so x + y = 48*sqrt(7)- Perimeter = 2*(x + y) = 96*sqrt(7)Everything seems to add up correctly. I think that's the answer.**Final Answer**The perimeter of the rectangle is boxed{96sqrt{7}}.

answer：Alright, so I've got this function here: ( f(x) = -sin^2 x - a sin x + b + 1 ). There are two parts to this problem. Let me tackle them one by one.Starting with part (1): If ( a = -frac{2}{3} ) and ( b = -1 ), I need to find the range of ( f(x) ) when ( 0 leq x leq frac{2pi}{3} ).First, let me substitute the given values of ( a ) and ( b ) into the function. So, substituting ( a = -frac{2}{3} ) and ( b = -1 ), the function becomes:( f(x) = -sin^2 x - (-frac{2}{3})sin x + (-1) + 1 )Simplifying that, the ( -1 ) and ( +1 ) cancel out, so we have:( f(x) = -sin^2 x + frac{2}{3}sin x )Hmm, okay. So, ( f(x) ) is a quadratic in terms of ( sin x ). Maybe I can make a substitution to make it easier. Let me let ( t = sin x ). Then, the function becomes:( f(t) = -t^2 + frac{2}{3}t )Now, I need to find the range of this quadratic function. But first, I should figure out the possible values of ( t ) given the domain of ( x ). Since ( x ) is between ( 0 ) and ( frac{2pi}{3} ), ( sin x ) will range from ( sin 0 = 0 ) to ( sin frac{2pi}{3} = frac{sqrt{3}}{2} ). So, ( t ) is in the interval ( [0, frac{sqrt{3}}{2}] ).Wait, hold on. The problem says ( 0 leq x leq frac{2pi}{3} ), so ( sin x ) actually goes from 0 up to ( sin frac{pi}{2} = 1 ) when ( x = frac{pi}{2} ), which is less than ( frac{2pi}{3} ). So, actually, ( t ) ranges from 0 to 1. Because ( sin frac{2pi}{3} ) is ( frac{sqrt{3}}{2} ), but since ( frac{pi}{2} ) is within the interval, the maximum value of ( sin x ) is 1. So, ( t in [0, 1] ).Okay, so now I have ( f(t) = -t^2 + frac{2}{3}t ), where ( t in [0, 1] ). This is a quadratic function, and since the coefficient of ( t^2 ) is negative, it opens downward, meaning it has a maximum at its vertex.To find the vertex, the formula for the vertex of a quadratic ( at^2 + bt + c ) is at ( t = -frac{b}{2a} ). Here, ( a = -1 ) and ( b = frac{2}{3} ), so:( t = -frac{frac{2}{3}}{2(-1)} = -frac{frac{2}{3}}{-2} = frac{1}{3} )So, the vertex is at ( t = frac{1}{3} ). Let me compute the value of ( f(t) ) at this point:( fleft(frac{1}{3}right) = -left(frac{1}{3}right)^2 + frac{2}{3} times frac{1}{3} = -frac{1}{9} + frac{2}{9} = frac{1}{9} )So, the maximum value of ( f(t) ) is ( frac{1}{9} ).Now, to find the minimum value, since the parabola opens downward, the minimums will occur at the endpoints of the interval ( [0, 1] ). Let me evaluate ( f(t) ) at ( t = 0 ) and ( t = 1 ):At ( t = 0 ):( f(0) = -0 + 0 = 0 )At ( t = 1 ):( f(1) = -1 + frac{2}{3} = -frac{1}{3} )So, the minimum value is ( -frac{1}{3} ).Therefore, the range of ( f(x) ) is from ( -frac{1}{3} ) to ( frac{1}{9} ).Wait, hold on. But in the original problem, when ( x ) is between ( 0 ) and ( frac{2pi}{3} ), ( sin x ) doesn't actually reach 1? Or does it? Because ( frac{pi}{2} ) is within ( 0 ) to ( frac{2pi}{3} ), so yes, ( sin x ) does reach 1. So, my initial thought was correct, ( t ) does go up to 1.But just to double-check, let me compute ( sin frac{2pi}{3} ). ( sin frac{2pi}{3} ) is ( sin (pi - frac{pi}{3}) = sin frac{pi}{3} = frac{sqrt{3}}{2} approx 0.866 ). So, actually, ( sin x ) reaches 1 at ( x = frac{pi}{2} ), which is less than ( frac{2pi}{3} ). So, yes, ( t ) does go up to 1.Therefore, the range is from ( -frac{1}{3} ) to ( frac{1}{9} ).Alright, that seems solid. So, part (1) is done.Moving on to part (2): Given ( a geq 0 ), if the maximum value of ( f(x) ) is 0 and the minimum value is -4, find the values of ( a ) and ( b ).So, the function is ( f(x) = -sin^2 x - a sin x + b + 1 ). Let me rewrite it:( f(x) = -sin^2 x - a sin x + (b + 1) )Again, this is a quadratic in terms of ( sin x ). Let me make the substitution ( t = sin x ), so ( t in [-1, 1] ). Then, the function becomes:( f(t) = -t^2 - a t + (b + 1) )But since ( a geq 0 ), I need to consider the behavior of this quadratic function over ( t in [-1, 1] ).First, let me write it in standard quadratic form:( f(t) = -t^2 - a t + (b + 1) )Alternatively, I can complete the square to find the vertex.Let me factor out the coefficient of ( t^2 ):( f(t) = -left(t^2 + a tright) + (b + 1) )Completing the square inside the parentheses:( t^2 + a t = left(t + frac{a}{2}right)^2 - frac{a^2}{4} )So, substituting back:( f(t) = -left(left(t + frac{a}{2}right)^2 - frac{a^2}{4}right) + (b + 1) )Simplify:( f(t) = -left(t + frac{a}{2}right)^2 + frac{a^2}{4} + b + 1 )So, the function is:( f(t) = -left(t + frac{a}{2}right)^2 + left(b + 1 + frac{a^2}{4}right) )This is a downward-opening parabola with vertex at ( t = -frac{a}{2} ) and the maximum value is ( b + 1 + frac{a^2}{4} ).But since ( a geq 0 ), the vertex is at ( t = -frac{a}{2} ), which is less than or equal to 0. So, the vertex is at the left side of the interval ( [-1, 1] ).Therefore, the maximum value of ( f(t) ) occurs at the vertex if the vertex is within the interval. But since ( t = -frac{a}{2} ) is less than or equal to 0, and our interval is from -1 to 1, the vertex is inside the interval only if ( -frac{a}{2} geq -1 ), which is always true since ( a geq 0 ).Wait, actually, if ( a ) is greater than 2, then ( -frac{a}{2} ) would be less than -1, which is outside the interval. So, let me think about that.Case 1: ( 0 leq a leq 2 ). Then, ( -frac{a}{2} geq -1 ), so the vertex is inside the interval ( [-1, 1] ). Therefore, the maximum value is at the vertex, which is ( b + 1 + frac{a^2}{4} ), and the minimum value occurs at one of the endpoints.Case 2: ( a > 2 ). Then, ( -frac{a}{2} < -1 ), so the vertex is outside the interval on the left. Therefore, the maximum value would occur at the left endpoint ( t = -1 ), and the minimum value would occur at the right endpoint ( t = 1 ).Wait, but the function is a downward-opening parabola, so if the vertex is outside the interval on the left, then on the interval ( [-1, 1] ), the function is increasing from ( t = -1 ) to ( t = 1 ). So, the maximum would be at ( t = 1 ) and the minimum at ( t = -1 ).Wait, actually, no. Since it's a downward-opening parabola, if the vertex is to the left of the interval, then the function is decreasing on the interval. So, the maximum would be at the left endpoint ( t = -1 ), and the minimum at the right endpoint ( t = 1 ).Wait, let me think again. For a downward-opening parabola, to the left of the vertex, the function is increasing, and to the right, it's decreasing. So, if the vertex is to the left of the interval ( [-1, 1] ), then on ( [-1, 1] ), the function is decreasing. So, maximum at ( t = -1 ), minimum at ( t = 1 ).Similarly, if the vertex is inside the interval, the maximum is at the vertex, and the minimum is at one of the endpoints.So, let's formalize this.Case 1: ( 0 leq a leq 2 ). Then, vertex at ( t = -frac{a}{2} in [-1, 0] ). So, maximum at vertex, minimum at either ( t = -1 ) or ( t = 1 ). Wait, but since the parabola is opening downward, the function is increasing from ( t = -1 ) to the vertex, then decreasing from the vertex to ( t = 1 ). So, the minimum could be at either ( t = -1 ) or ( t = 1 ), whichever gives a lower value.Case 2: ( a > 2 ). Vertex at ( t = -frac{a}{2} < -1 ). So, on ( [-1, 1] ), the function is decreasing, so maximum at ( t = -1 ), minimum at ( t = 1 ).Given that, let's compute the maximum and minimum for both cases.Given that the maximum value is 0 and the minimum is -4, we can set up equations accordingly.First, let's handle Case 1: ( 0 leq a leq 2 ).In this case, maximum at vertex: ( f(t)_{text{max}} = b + 1 + frac{a^2}{4} = 0 ).So, equation 1: ( b + 1 + frac{a^2}{4} = 0 ).Now, for the minimum, we need to evaluate ( f(t) ) at both endpoints ( t = -1 ) and ( t = 1 ), and see which one is smaller.Compute ( f(-1) ):( f(-1) = -(-1)^2 - a(-1) + (b + 1) = -1 + a + b + 1 = a + b )Compute ( f(1) ):( f(1) = -(1)^2 - a(1) + (b + 1) = -1 - a + b + 1 = -a + b )So, ( f(-1) = a + b ) and ( f(1) = -a + b ). Since ( a geq 0 ), ( f(-1) geq f(1) ) because ( a + b geq -a + b ) when ( a geq 0 ). Therefore, the minimum occurs at ( t = 1 ), which is ( f(1) = -a + b ).Given that the minimum is -4, so:Equation 2: ( -a + b = -4 )So, from equation 1: ( b + 1 + frac{a^2}{4} = 0 ) => ( b = -1 - frac{a^2}{4} )Substitute into equation 2:( -a + (-1 - frac{a^2}{4}) = -4 )Simplify:( -a -1 - frac{a^2}{4} = -4 )Multiply both sides by 4 to eliminate the fraction:( -4a -4 - a^2 = -16 )Bring all terms to one side:( -a^2 -4a -4 +16 = 0 )Simplify:( -a^2 -4a +12 = 0 )Multiply both sides by -1:( a^2 + 4a -12 = 0 )Now, solve for ( a ):Using quadratic formula:( a = frac{-4 pm sqrt{16 + 48}}{2} = frac{-4 pm sqrt{64}}{2} = frac{-4 pm 8}{2} )So, two solutions:1. ( a = frac{-4 + 8}{2} = frac{4}{2} = 2 )2. ( a = frac{-4 - 8}{2} = frac{-12}{2} = -6 )But since ( a geq 0 ), we discard ( a = -6 ). So, ( a = 2 ).Then, from equation 1: ( b = -1 - frac{(2)^2}{4} = -1 - 1 = -2 )So, in Case 1, we have ( a = 2 ) and ( b = -2 ). Now, we need to check if this satisfies the condition ( 0 leq a leq 2 ). Since ( a = 2 ), it's on the boundary. So, this is acceptable.Now, let's check Case 2: ( a > 2 ).In this case, the maximum occurs at ( t = -1 ), which is ( f(-1) = a + b ), and the minimum occurs at ( t = 1 ), which is ( f(1) = -a + b ).Given that the maximum is 0 and the minimum is -4, we have:Equation 1: ( a + b = 0 )Equation 2: ( -a + b = -4 )Let's solve these equations.From equation 1: ( b = -a )Substitute into equation 2:( -a + (-a) = -4 )( -2a = -4 )( a = 2 )But in this case, ( a > 2 ), so ( a = 2 ) is not in this case. Therefore, there is no solution in Case 2.Therefore, the only solution is ( a = 2 ) and ( b = -2 ).Let me double-check these values.Substitute ( a = 2 ) and ( b = -2 ) into the original function:( f(x) = -sin^2 x - 2 sin x + (-2) + 1 = -sin^2 x - 2 sin x -1 )Let me compute the maximum and minimum.Expressed as ( f(t) = -t^2 - 2t -1 ), where ( t = sin x ), ( t in [-1, 1] ).Complete the square:( f(t) = -(t^2 + 2t) -1 = -[(t + 1)^2 -1] -1 = -(t + 1)^2 +1 -1 = -(t +1)^2 )So, ( f(t) = -(t +1)^2 ). Therefore, the maximum value occurs when ( (t +1)^2 ) is minimized, which is when ( t = -1 ), giving ( f(t) = 0 ). The minimum occurs when ( (t +1)^2 ) is maximized, which is when ( t = 1 ), giving ( f(t) = -(1 +1)^2 = -4 ). So, yes, that checks out.Therefore, the solution is ( a = 2 ) and ( b = -2 ).**Final Answer**(1) The range of ( f(x) ) is boxed{left[ -dfrac{1}{3}, dfrac{1}{9} right]}.(2) The values of ( a ) and ( b ) are boxed{a = 2} and boxed{b = -2}.