💡Okay, so I have this problem here where a 10 by 20 rectangle is cut into two congruent hexagons, and then these hexagons are rearranged to form a square. I need to find the length y, which is a fourth of the square's side. The options are given, so that might help if I get stuck, but I should try to figure it out myself first.Alright, let's start by understanding the problem. We have a rectangle that's 10 units by 20 units. So, the area of the rectangle is 10 times 20, which is 200 square units. Since the rectangle is being cut into two congruent hexagons, each hexagon must have half the area of the rectangle, right? So, each hexagon has an area of 100 square units.Now, these two hexagons are rearranged to form a square. So, the total area of the square must also be 200 square units because we're just rearranging the pieces without changing their size. Therefore, the area of the square is 200. To find the side length of the square, I can take the square root of the area. So, the side length s is sqrt(200). Let me calculate that.sqrt(200) is the same as sqrt(100*2), which is 10*sqrt(2). So, the side length of the square is 10*sqrt(2). Now, the problem says that y is a fourth of the square's side. So, y is (10*sqrt(2))/4. Let me simplify that.Dividing 10 by 4 gives me 2.5, which is the same as 5/2. So, y is (5/2)*sqrt(2). That can also be written as (5*sqrt(2))/2. Let me check the answer options to see if that's one of them.Looking back, option A is (5*sqrt(2))/2, which matches what I got. So, that should be the answer. But wait, let me make sure I didn't skip any steps or make any mistakes.First, the area of the rectangle is definitely 10*20=200. Then, since it's split into two congruent hexagons, each has 100. When rearranged into a square, the total area remains 200, so the side length is sqrt(200)=10*sqrt(2). Then, y is a fourth of that, so y=(10*sqrt(2))/4= (5*sqrt(2))/2. Yep, that seems right.I think I'm confident with that answer. It's always good to double-check, but I don't see any errors in my reasoning. So, I'll go with option A.

💡First, I need to determine the total number of baseball cards Luke has. He has 3 new cards and 9 old cards.Adding these together gives:3 new cards + 9 old cards = 12 cardsNext, I need to figure out how many pages Luke will use if he places 3 cards on each page. To do this, I divide the total number of cards by the number of cards per page.12 cards ÷ 3 cards per page = 4 pagesTherefore, Luke will use 4 pages to organize all his baseball cards.

💡Okay, so I have this geometry problem here, and I need to figure out the measure of angle B in triangle ABC. Let me try to visualize the problem first.Alright, triangle ABC is given with AB equal to BC. Hmm, so that makes triangle ABC an isosceles triangle with AB = BC. Wait, hold on, in an isosceles triangle, the sides opposite equal angles are equal. So if AB = BC, then the angles opposite them should be equal. That would mean angle BAC is equal to angle BCA, right? Let me write that down:AB = BC ⇒ ∠BAC = ∠BCA.Okay, got that. Now, point D is on AC such that BD bisects angle BAC. So BD is the angle bisector of angle BAC. That means angle BAD is equal to angle CAD. Let me note that:BD bisects ∠BAC ⇒ ∠BAD = ∠CAD.Also, it's given that CD = BD. So triangle BCD has two sides equal: CD and BD. That makes triangle BCD isosceles with CD = BD. Therefore, the angles opposite these sides should be equal. So in triangle BCD, angle BCD is equal to angle BDC.So, ∠BCD = ∠BDC.Hmm, okay. Let me try to draw this triangle to get a better picture. So ABC is isosceles with AB = BC. Let me place point B at the top, A and C at the base. Then D is somewhere on AC. BD is the angle bisector of angle BAC, so it splits angle BAC into two equal parts. Also, CD equals BD, so triangle BCD is isosceles.Let me label the angles. Let me denote angle BAC as 2θ, so each of the angles BAD and CAD is θ. Since triangle ABC is isosceles with AB = BC, angles at A and C are equal. So angle BAC = angle BCA = 2θ. Then angle ABC would be 180° - 2*(2θ) = 180° - 4θ.Now, looking at triangle BCD. Since CD = BD, angles at C and D are equal. So angle BCD = angle BDC. Let me denote each of these angles as φ. So in triangle BCD, angles at C and D are φ, and angle at B is something else. The sum of angles in a triangle is 180°, so angle at B in triangle BCD is 180° - 2φ.But angle at B in triangle BCD is also part of angle ABC in triangle ABC. Wait, angle ABC in triangle ABC is 180° - 4θ, and in triangle BCD, angle at B is 180° - 2φ. So, is angle ABC equal to angle at B in triangle BCD? Hmm, not exactly. Because angle ABC is the angle at point B in triangle ABC, which is split into two angles by BD: angle ABD and angle CBD.Wait, BD is the angle bisector of angle BAC, which is 2θ, so angle BAD = angle CAD = θ. But BD is also a side in triangle BCD. Hmm, maybe I need to look at triangle ABD as well.In triangle ABD, we have BD as a side, and AB is equal to BC. Wait, AB = BC, but not necessarily equal to BD or CD. Hmm. Let me see.Since BD bisects angle BAC, which is 2θ, into two angles of θ each. So in triangle ABD, we have angle at A is θ, angle at B is something, and angle at D is something. Similarly, in triangle CBD, we have angle at C is φ, angle at D is φ, and angle at B is 180° - 2φ.Wait, maybe I can relate angles in triangle ABD and triangle CBD.Let me denote angle ABD as α. Then, since angle ABC is 180° - 4θ, angle ABD + angle CBD = 180° - 4θ. So angle ABD is α, and angle CBD is 180° - 4θ - α.But in triangle BCD, angle at B is 180° - 2φ, which is equal to angle CBD. So:180° - 2φ = 180° - 4θ - α.Simplifying, we get:-2φ = -4θ - α ⇒ 2φ = 4θ + α ⇒ φ = 2θ + α/2.Hmm, okay. Now, let's look at triangle ABD. In triangle ABD, we have angles at A, B, and D. Angle at A is θ, angle at B is α, so angle at D is 180° - θ - α.But in triangle BCD, angle at D is φ. So angle at D in triangle ABD is equal to angle at D in triangle BCD? Wait, no, because point D is common, but the angles at D in triangles ABD and BCD are different. Wait, actually, in triangle ABD, angle at D is adjacent to angle at D in triangle BCD. So the sum of angles at D in both triangles should be 180°, because they form a straight line along AC.Wait, no, AC is a straight line, so the angles at D on either side should add up to 180°. So angle ADB (in triangle ABD) + angle CDB (in triangle BCD) = 180°. But angle CDB is φ, and angle ADB is 180° - θ - α. So:(180° - θ - α) + φ = 180° ⇒ -θ - α + φ = 0 ⇒ φ = θ + α.But earlier, we had φ = 2θ + α/2. So:θ + α = 2θ + α/2.Let me solve for α:θ + α = 2θ + α/2 ⇒ α - α/2 = 2θ - θ ⇒ α/2 = θ ⇒ α = 2θ.So angle ABD is 2θ. Then, from angle ABC = 180° - 4θ, and angle ABC is also angle ABD + angle CBD = 2θ + angle CBD. So:2θ + angle CBD = 180° - 4θ ⇒ angle CBD = 180° - 4θ - 2θ = 180° - 6θ.But angle CBD is also equal to 180° - 2φ from triangle BCD. So:180° - 2φ = 180° - 6θ ⇒ -2φ = -6θ ⇒ φ = 3θ.But earlier, we had φ = θ + α, and α = 2θ, so φ = θ + 2θ = 3θ. That's consistent. So φ = 3θ.Now, let's look at triangle BCD. The sum of angles is 180°, so:φ + φ + (180° - 2φ) = 180°, which is just 180° = 180°, so that doesn't give new information.Wait, maybe I should look at triangle ABD. In triangle ABD, angles are θ at A, 2θ at B, and angle at D is 180° - θ - 2θ = 180° - 3θ.But angle at D in triangle ABD is also adjacent to angle at D in triangle BCD, which is φ = 3θ. So:(180° - 3θ) + 3θ = 180°, which is again consistent, but doesn't give new info.Hmm, maybe I need to use the fact that CD = BD. So in triangle BCD, sides CD = BD, so it's isosceles, and we have angles at C and D equal, which we denoted as φ.But also, in triangle ABC, sides AB = BC, so it's isosceles, and angles at A and C are equal, which we denoted as 2θ each.Let me try to relate the sides. Maybe using the Law of Sines or Law of Cosines.In triangle ABD, sides are AB, BD, and AD. In triangle CBD, sides are BC, BD, and CD. Since AB = BC, and BD is common, but CD = BD.Wait, CD = BD, so in triangle CBD, sides CD = BD, so it's isosceles with base BC. Wait, no, BC is a side of triangle ABC, not necessarily related directly.Wait, maybe I can apply the Law of Sines in triangle ABD and triangle CBD.In triangle ABD:AB / sin(angle ADB) = BD / sin(angle BAD) = AD / sin(angle ABD)Similarly, in triangle CBD:BC / sin(angle CDB) = BD / sin(angle BCD) = CD / sin(angle CBD)But AB = BC, BD = CD, angle BAD = θ, angle BCD = φ = 3θ, angle ABD = 2θ, angle CBD = 180° - 6θ, angle ADB = 180° - 3θ, angle CDB = φ = 3θ.So let's write the Law of Sines for triangle ABD:AB / sin(180° - 3θ) = BD / sin(θ) = AD / sin(2θ)Similarly, for triangle CBD:BC / sin(3θ) = BD / sin(3θ) = CD / sin(180° - 6θ)But AB = BC, and BD = CD, so let's equate the ratios.From triangle ABD:AB / sin(180° - 3θ) = BD / sin(θ)From triangle CBD:BC / sin(3θ) = BD / sin(3θ)But AB = BC, so let's set AB = BC = k, and BD = CD = m.So from triangle ABD:k / sin(3θ) = m / sin(θ) ⇒ k = m * sin(3θ) / sin(θ)From triangle CBD:k / sin(3θ) = m / sin(3θ) ⇒ k = mWait, that would imply that k = m, but from triangle ABD, k = m * sin(3θ) / sin(θ). So:m = m * sin(3θ) / sin(θ) ⇒ 1 = sin(3θ) / sin(θ) ⇒ sin(3θ) = sin(θ)Hmm, sin(3θ) = sin(θ). When does that happen? Well, sin(3θ) = sin(θ) implies that either 3θ = θ + 2πn or 3θ = π - θ + 2πn for some integer n. But since we're dealing with angles in a triangle, θ must be between 0 and 90°, so let's consider the principal solutions.Case 1: 3θ = θ + 2πn ⇒ 2θ = 2πn ⇒ θ = πn. But θ must be positive and less than 90°, so n=0 gives θ=0, which is invalid. So no solution here.Case 2: 3θ = π - θ + 2πn ⇒ 4θ = π + 2πn ⇒ θ = π/4 + πn/2. Again, θ must be less than 90°, so n=0 gives θ=π/4=45°, which is 45°, and n=1 gives θ=3π/4=135°, which is too big. So θ=45°.Wait, θ=45°, so angle BAC=2θ=90°, angle BCA=90°, and angle ABC=180°-2*90°=0°, which is impossible. That can't be right. So maybe I made a mistake somewhere.Wait, let's go back. From triangle ABD, we have:AB / sin(angle ADB) = BD / sin(angle BAD)Which is:AB / sin(180° - 3θ) = BD / sin(θ)But sin(180° - 3θ) = sin(3θ), so:AB / sin(3θ) = BD / sin(θ) ⇒ AB = BD * sin(3θ) / sin(θ)Similarly, from triangle CBD:BC / sin(angle CDB) = BD / sin(angle BCD)Which is:BC / sin(3θ) = BD / sin(3θ) ⇒ BC = BDBut AB = BC, so AB = BD. Therefore, AB = BD.So from triangle ABD, AB = BD, so triangle ABD is isosceles with AB = BD. Therefore, angles opposite these sides are equal. So angle at A (which is θ) equals angle at D (which is 180° - 3θ). Wait, that can't be right because θ is not equal to 180° - 3θ unless θ=45°, which we saw leads to a contradiction.Wait, maybe I messed up the Law of Sines application. Let me double-check.In triangle ABD, sides are AB, BD, AD. Angles opposite are angle ADB, angle BAD, angle ABD.So Law of Sines:AB / sin(angle ADB) = BD / sin(angle BAD) = AD / sin(angle ABD)So AB / sin(180° - 3θ) = BD / sin(θ)Which is AB / sin(3θ) = BD / sin(θ)Similarly, in triangle CBD:BC / sin(angle CDB) = BD / sin(angle BCD)Which is BC / sin(3θ) = BD / sin(3θ)So BC = BDBut AB = BC, so AB = BDTherefore, in triangle ABD, AB = BD, so it's isosceles with AB = BD. Therefore, angles opposite these sides are equal. So angle at A (θ) equals angle at D (180° - 3θ). So:θ = 180° - 3θ ⇒ 4θ = 180° ⇒ θ = 45°Again, θ=45°, which leads to angle BAC=90°, angle BCA=90°, and angle ABC=0°, which is impossible. So something's wrong here.Wait, maybe my assumption that angle ADB = 180° - 3θ is incorrect. Let me re-examine triangle ABD.In triangle ABD, angles are:At A: θAt B: 2θAt D: 180° - θ - 2θ = 180° - 3θSo angle ADB = 180° - 3θBut in triangle BCD, angle at D is φ = 3θBut angle ADB and angle CDB are supplementary because they form a straight line along AC. So:angle ADB + angle CDB = 180° ⇒ (180° - 3θ) + φ = 180° ⇒ φ = 3θWhich is consistent with earlier.But when I applied the Law of Sines, I got AB = BD, leading to θ=45°, which is impossible.Wait, maybe I need to consider the Law of Cosines instead.In triangle ABD, using Law of Cosines:AD² = AB² + BD² - 2*AB*BD*cos(angle ABD)Similarly, in triangle CBD, using Law of Cosines:CD² = BC² + BD² - 2*BC*BD*cos(angle CBD)But CD = BD, and AB = BC, so let's denote AB = BC = k, BD = CD = m.So in triangle ABD:AD² = k² + m² - 2*k*m*cos(2θ)In triangle CBD:m² = k² + m² - 2*k*m*cos(angle CBD)Simplify the second equation:m² = k² + m² - 2*k*m*cos(angle CBD) ⇒ 0 = k² - 2*k*m*cos(angle CBD) ⇒ k² = 2*k*m*cos(angle CBD) ⇒ k = 2*m*cos(angle CBD)But angle CBD = 180° - 6θ, so:k = 2*m*cos(180° - 6θ) = 2*m*(-cos(6θ)) = -2*m*cos(6θ)But k and m are lengths, so they are positive. Therefore, cos(6θ) must be negative, which implies that 6θ > 90°, so θ > 15°.But from earlier, θ=45°, which would make 6θ=270°, but cos(270°)=0, which would make k=0, which is impossible. So maybe this approach isn't working.Wait, maybe I need to consider the ratio of sides in triangle ABD and triangle CBD.Since AB = BC = k, BD = CD = m, and angle ABD = 2θ, angle CBD = 180° - 6θ.Using the Law of Sines in triangle ABD:AB / sin(angle ADB) = BD / sin(angle BAD) ⇒ k / sin(180° - 3θ) = m / sin(θ) ⇒ k / sin(3θ) = m / sin(θ)Similarly, in triangle CBD:BC / sin(angle CDB) = BD / sin(angle BCD) ⇒ k / sin(3θ) = m / sin(3θ) ⇒ k = mSo k = m, meaning AB = BD = BC = CD.So AB = BC = BD = CD.So triangle ABD has AB = BD, so it's isosceles with angles at A and D equal. But angle at A is θ, so angle at D is also θ. But earlier, angle at D in triangle ABD was 180° - 3θ. So:θ = 180° - 3θ ⇒ 4θ = 180° ⇒ θ = 45°Again, θ=45°, leading to angle BAC=90°, angle BCA=90°, and angle ABC=0°, which is impossible.This suggests that my approach is flawed. Maybe I need to consider a different method.Let me try to assign coordinates to the points to make it easier.Let me place point B at (0,0), point C at (c,0), and since AB = BC, point A must be somewhere such that distance from A to B equals distance from B to C, which is c.Let me assume point A is at (0,a), so AB = sqrt((0-0)^2 + (a-0)^2) = a, and BC = c, so a = c.So point A is at (0,c), point B at (0,0), point C at (c,0).Now, point D is on AC such that BD bisects angle BAC.Let me find coordinates of D.First, equation of AC: from (0,c) to (c,0). The parametric equations can be written as x = ct, y = c(1 - t), where t ranges from 0 to 1.So point D is (ct, c(1 - t)) for some t.Now, BD bisects angle BAC. So the angle between BA and BD is equal to the angle between BD and BC.Wait, actually, BD bisects angle BAC, which is the angle at A between BA and AC.Wait, no, angle BAC is at point A between BA and AC. So BD is the bisector of angle BAC, which is at point A.Wait, but BD is a line from B to D on AC. So it's the angle bisector from A to BD.Wait, maybe I need to use the Angle Bisector Theorem.The Angle Bisector Theorem states that the angle bisector divides the opposite side in the ratio of the adjacent sides.In triangle ABC, angle BAC is being bisected by BD, which meets AC at D.So according to the Angle Bisector Theorem:AB / BC = AD / DCBut AB = BC, so AB / BC = 1 = AD / DC ⇒ AD = DCSo D is the midpoint of AC.Wait, that's interesting. So D is the midpoint of AC.But it's also given that CD = BD.So CD = BD, and since D is the midpoint, CD = AD = AC/2.So BD = AC/2.Hmm, let's compute coordinates.Point A is (0,c), point C is (c,0), so midpoint D is ((0 + c)/2, (c + 0)/2) = (c/2, c/2).So D is at (c/2, c/2).Now, BD is the distance from B(0,0) to D(c/2, c/2), which is sqrt((c/2)^2 + (c/2)^2) = sqrt(c²/4 + c²/4) = sqrt(c²/2) = c/√2.But CD is the distance from C(c,0) to D(c/2, c/2), which is sqrt((c - c/2)^2 + (0 - c/2)^2) = sqrt((c/2)^2 + (c/2)^2) = same as BD, which is c/√2. So CD = BD, which satisfies the given condition.Now, let's find the measure of angle B.In triangle ABC, points are A(0,c), B(0,0), C(c,0). So triangle ABC is an isosceles right triangle with legs AB and BC, each of length c, and hypotenuse AC of length c√2.Wait, but in that case, angle at B is 90°, because triangle ABC is right-angled at B.But wait, in the problem, it's given that AB = BC, which in this coordinate system, AB = BC = c, and AC = c√2, so yes, triangle ABC is a right-angled isosceles triangle with angle B = 90°.But wait, earlier when I tried to solve it using angles, I ended up with a contradiction, suggesting angle B was 0°, which was wrong. But with coordinates, it makes sense that angle B is 90°.Wait, but in the problem, it's given that CD = BD, which in this case, CD = BD = c/√2, which is true. So everything checks out.Therefore, angle B is 90°.Wait, but in the initial problem statement, it's said that AB = BC, and D is on AC such that BD bisects angle BAC, and CD = BD. So according to this coordinate analysis, angle B is 90°, which seems correct.But earlier, when I tried to solve it using angles, I got θ=45°, leading to angle BAC=90°, which would make angle B=0°, which was wrong. So I must have made a mistake in that approach.Wait, maybe I confused the angles. Let me check again.In the coordinate system, angle BAC is the angle at A between BA and AC. BA is from A(0,c) to B(0,0), which is straight down along the y-axis. AC is from A(0,c) to C(c,0), which is a diagonal line. So angle BAC is 45°, because AC makes a 45° angle with the y-axis.Wait, no, angle BAC is the angle between BA and AC. BA is along the negative y-axis, and AC is going to the right and down at 45°. So the angle between them is 45°, so angle BAC=45°, not 90°.Wait, but in the coordinate system, AB = BC = c, and AC = c√2, so triangle ABC is right-angled at B, with angle B=90°, and angles at A and C each 45°, because it's an isosceles right triangle.So angle BAC=45°, which is bisected by BD into two angles of 22.5° each.Wait, but in the problem, it's given that BD bisects angle BAC, so angle BAD=angle CAD=22.5°, and CD=BD.In the coordinate system, D is the midpoint of AC, so AD=DC= c√2 / 2.But BD is c/√2, which is equal to CD, as we saw earlier.So everything fits.Therefore, angle B is 90°, which is the measure we're supposed to find.So the measure of angle B is 90 degrees.

💡Okay, so I have this functional equation to solve: ( f(xy) = f(x)f(y) - 2xy ) for all real numbers ( x ) and ( y ). Hmm, functional equations can be tricky, but let me try to break it down step by step.First, I remember that functional equations often require substituting specific values for ( x ) and ( y ) to find patterns or properties of the function ( f ). Maybe I can start by plugging in some simple values, like ( x = 1 ) or ( y = 1 ), to see if that helps.Let me try substituting ( x = 1 ) and ( y = 1 ). Then the equation becomes:( f(1 cdot 1) = f(1)f(1) - 2 cdot 1 cdot 1 )Simplifying that, I get:( f(1) = [f(1)]^2 - 2 )Hmm, that's a quadratic equation in terms of ( f(1) ). Let me write it as:( [f(1)]^2 - f(1) - 2 = 0 )To solve this quadratic, I can use the quadratic formula. The quadratic is ( a = 1 ), ( b = -1 ), and ( c = -2 ). Plugging into the formula:( f(1) = frac{-b pm sqrt{b^2 - 4ac}}{2a} = frac{1 pm sqrt{1 + 8}}{2} = frac{1 pm 3}{2} )So, ( f(1) ) can be either ( frac{1 + 3}{2} = 2 ) or ( frac{1 - 3}{2} = -1 ). Okay, so ( f(1) ) is either 2 or -1. That gives me two cases to consider.**Case 1: ( f(1) = 2 )**Let me substitute ( y = 1 ) into the original equation:( f(x cdot 1) = f(x)f(1) - 2x cdot 1 )Simplifying:( f(x) = f(x) cdot 2 - 2x )Hmm, let's solve for ( f(x) ):( f(x) = 2f(x) - 2x )Subtract ( 2f(x) ) from both sides:( -f(x) = -2x )Multiply both sides by -1:( f(x) = 2x )Okay, so in this case, the function is ( f(x) = 2x ). Let me check if this satisfies the original equation.Substitute ( f(x) = 2x ) into ( f(xy) = f(x)f(y) - 2xy ):Left side: ( f(xy) = 2(xy) = 2xy )Right side: ( f(x)f(y) - 2xy = (2x)(2y) - 2xy = 4xy - 2xy = 2xy )Yes, both sides are equal. So ( f(x) = 2x ) is a valid solution.**Case 2: ( f(1) = -1 )**Again, substitute ( y = 1 ) into the original equation:( f(x cdot 1) = f(x)f(1) - 2x cdot 1 )Simplifying:( f(x) = f(x) cdot (-1) - 2x )Let me solve for ( f(x) ):( f(x) = -f(x) - 2x )Add ( f(x) ) to both sides:( 2f(x) = -2x )Divide both sides by 2:( f(x) = -x )So, in this case, the function is ( f(x) = -x ). Let me verify this solution as well.Substitute ( f(x) = -x ) into ( f(xy) = f(x)f(y) - 2xy ):Left side: ( f(xy) = -xy )Right side: ( f(x)f(y) - 2xy = (-x)(-y) - 2xy = xy - 2xy = -xy )Both sides are equal again. So ( f(x) = -x ) is also a valid solution.Wait, so I have two possible solutions: ( f(x) = 2x ) and ( f(x) = -x ). Are there any other possibilities? Let me think.I considered both cases for ( f(1) ), which were 2 and -1. Each case led me to a unique function, and both functions satisfy the original equation. So, unless there's another approach that might reveal more solutions, I think these are the only ones.Just to be thorough, let me consider if there could be any other functions that satisfy the equation without assuming ( f(1) ) is 2 or -1. Maybe a function that isn't linear? Let's see.Suppose ( f ) is a quadratic function, say ( f(x) = ax^2 + bx + c ). Then, substituting into the equation:( f(xy) = a(xy)^2 + b(xy) + c )And ( f(x)f(y) - 2xy = (ax^2 + bx + c)(ay^2 + by + c) - 2xy )That seems complicated. Let me multiply it out:( (ax^2 + bx + c)(ay^2 + by + c) = a^2x^2y^2 + abx^2y + acx^2 + abxy^2 + b^2xy + bcx + acy^2 + bcy + c^2 )So, subtracting ( 2xy ):( a^2x^2y^2 + abx^2y + acx^2 + abxy^2 + b^2xy + bcx + acy^2 + bcy + c^2 - 2xy )Comparing this to ( f(xy) = a(xy)^2 + b(xy) + c ), which is ( a x^2 y^2 + b xy + c ).For these to be equal for all ( x ) and ( y ), the coefficients of corresponding terms must be equal.Looking at the highest degree term ( x^2 y^2 ):Coefficient in ( f(xy) ): ( a )Coefficient in ( f(x)f(y) - 2xy ): ( a^2 )So, ( a = a^2 ), which implies ( a = 0 ) or ( a = 1 ).If ( a = 0 ), then ( f(x) = bx + c ). Let's see if this can satisfy the original equation.If ( f(x) = bx + c ), then:( f(xy) = b(xy) + c )And ( f(x)f(y) - 2xy = (bx + c)(by + c) - 2xy = b^2xy + bcx + bcy + c^2 - 2xy )Set equal:( bxy + c = b^2xy + bcx + bcy + c^2 - 2xy )Comparing coefficients:For ( xy ): ( b = b^2 - 2 )For ( x ): ( 0 = bc )For ( y ): ( 0 = bc )Constant term: ( c = c^2 )So, from ( 0 = bc ), either ( b = 0 ) or ( c = 0 ).If ( b = 0 ), then from ( b = b^2 - 2 ), we get ( 0 = 0 - 2 ), which is false. So ( b neq 0 ), hence ( c = 0 ).Then, from ( c = c^2 ), since ( c = 0 ), that's satisfied.From ( b = b^2 - 2 ):( b^2 - b - 2 = 0 )Solving quadratic:( b = frac{1 pm sqrt{1 + 8}}{2} = frac{1 pm 3}{2} )So, ( b = 2 ) or ( b = -1 ).Therefore, if ( a = 0 ), ( f(x) = 2x ) or ( f(x) = -x ), which are the same solutions as before.If ( a = 1 ), then from the coefficient of ( x^2 y^2 ), ( a = 1 ). Let's see if this leads to a solution.From the previous expansion, with ( a = 1 ):( f(xy) = x^2 y^2 + bxy + c )And ( f(x)f(y) - 2xy = x^2 y^2 + bxy + c + terms... ) Wait, but when ( a = 1 ), the expansion is more complicated.Wait, actually, if ( a = 1 ), then ( f(x) = x^2 + bx + c ). Let's substitute into the original equation:( f(xy) = (xy)^2 + b(xy) + c )And ( f(x)f(y) - 2xy = (x^2 + bx + c)(y^2 + by + c) - 2xy )Expanding ( f(x)f(y) ):( x^2 y^2 + bx y^2 + c x^2 + b x^2 y + b^2 xy + bc x + c y^2 + bc y + c^2 )Subtracting ( 2xy ):( x^2 y^2 + bx y^2 + c x^2 + b x^2 y + b^2 xy + bc x + c y^2 + bc y + c^2 - 2xy )Set equal to ( f(xy) = x^2 y^2 + bxy + c ):So, equate coefficients:- ( x^2 y^2 ): 1 = 1, okay.- ( x^2 y ): 0 = b- ( x y^2 ): 0 = b- ( x^2 ): 0 = c- ( y^2 ): 0 = c- ( xy ): b = b^2 - 2- ( x ): 0 = bc- ( y ): 0 = bc- Constant term: c = c^2From ( x^2 y ) and ( x y^2 ), we get ( b = 0 ).From ( x^2 ) and ( y^2 ), we get ( c = 0 ).From ( xy ): ( b = b^2 - 2 ). But since ( b = 0 ), this becomes ( 0 = 0 - 2 ), which is false.So, no solution in this case.Therefore, the only solutions are the linear functions ( f(x) = 2x ) and ( f(x) = -x ).I think that's thorough enough. I considered both cases for ( f(1) ), checked if quadratic functions could work, and found that only linear functions satisfy the equation. So, the solutions are ( f(x) = 2x ) and ( f(x) = -x ).

💡Okay, so I have this problem here: I need to find the minimum value of ( n ) such that the expansion of the binomial ( (3x^2 - frac{2}{3x})^n ) contains a constant term. The options are A: 4, B: 5, C: 6, D: 7. Hmm, let me think about how to approach this.First, I remember that when expanding a binomial like ( (a + b)^n ), each term in the expansion is given by the binomial theorem, which is ( T_{r+1} = C(n, r) cdot a^{n-r} cdot b^r ), where ( C(n, r) ) is the combination of ( n ) things taken ( r ) at a time. So, in this case, ( a = 3x^2 ) and ( b = -frac{2}{3x} ).Let me write that out. The general term ( T_{r+1} ) in the expansion would be:[T_{r+1} = C(n, r) cdot (3x^2)^{n - r} cdot left(-frac{2}{3x}right)^r]Simplifying this, I can break it down into the coefficients and the variables separately.First, let's handle the coefficients:- ( (3x^2)^{n - r} = 3^{n - r} cdot x^{2(n - r)} )- ( left(-frac{2}{3x}right)^r = (-1)^r cdot left(frac{2}{3}right)^r cdot x^{-r} )Multiplying these together, the coefficient part becomes:[3^{n - r} cdot (-1)^r cdot left(frac{2}{3}right)^r]And the variable part is:[x^{2(n - r)} cdot x^{-r} = x^{2n - 2r - r} = x^{2n - 3r}]So, combining these, the general term is:[T_{r+1} = C(n, r) cdot 3^{n - r} cdot (-1)^r cdot left(frac{2}{3}right)^r cdot x^{2n - 3r}]Simplifying the coefficients further:[3^{n - r} cdot left(frac{2}{3}right)^r = 3^{n - r} cdot 2^r cdot 3^{-r} = 3^{n - 2r} cdot 2^r]So, the general term simplifies to:[T_{r+1} = C(n, r) cdot (-1)^r cdot 3^{n - 2r} cdot 2^r cdot x^{2n - 3r}]Now, we're looking for the constant term in the expansion. A constant term means that the exponent of ( x ) is zero. So, we set the exponent equal to zero:[2n - 3r = 0]Solving for ( n ):[2n = 3r implies n = frac{3r}{2}]Hmm, so ( n ) must be equal to ( frac{3r}{2} ). But ( n ) is a positive integer, and ( r ) is also an integer because it's the term number in the expansion. Therefore, ( frac{3r}{2} ) must be an integer. This implies that ( r ) must be an even number because 3 is odd, and for the fraction to simplify to an integer, the denominator 2 must divide into the numerator. So, ( r ) must be even.Let me denote ( r = 2k ) where ( k ) is an integer. Substituting back into the equation for ( n ):[n = frac{3(2k)}{2} = 3k]So, ( n ) must be a multiple of 3. The smallest positive integer value for ( k ) is 1, which gives ( n = 3 ). But wait, let me check the options given: A:4, B:5, C:6, D:7. 3 isn't an option. Hmm, maybe I made a mistake.Wait, let's go back. The exponent was ( 2n - 3r = 0 ), so ( 2n = 3r ). So, ( n = frac{3r}{2} ). Therefore, ( r ) must be even, as I thought, so let me set ( r = 2k ), then ( n = 3k ). So, the smallest ( n ) is 3, but 3 isn't an option. Maybe I need to consider that ( r ) must be less than or equal to ( n ) because in the binomial expansion, ( r ) ranges from 0 to ( n ).So, if ( n = 3k ), then ( r = 2k ) must be less than or equal to ( n ). So, ( 2k leq 3k ), which is always true for ( k geq 1 ). So, the smallest ( n ) is 3, but since 3 isn't an option, perhaps I need to consider the next multiple.Wait, maybe I made a mistake in simplifying the exponent. Let me double-check.The term is ( (3x^2)^{n - r} cdot left(-frac{2}{3x}right)^r ). So, the exponent of ( x ) is ( 2(n - r) + (-1)r = 2n - 2r - r = 2n - 3r ). That seems correct.So, ( 2n - 3r = 0 implies 2n = 3r implies n = frac{3r}{2} ). So, ( r ) must be even. Let me try ( r = 2 ), then ( n = 3 ). But 3 isn't an option. Next, ( r = 4 ), then ( n = 6 ). 6 is option C. So, is 6 the minimum value of ( n )?Wait, but if ( n = 6 ), then ( r = 4 ), which is less than ( n ), so that's valid. But is there a smaller ( n ) that satisfies the condition? Let me check ( n = 4 ). If ( n = 4 ), then ( 2n = 8 = 3r implies r = 8/3 ), which is not an integer. So, ( r ) must be an integer, so ( n = 4 ) doesn't work.Similarly, ( n = 5 ): ( 2*5 = 10 = 3r implies r = 10/3 ), which is not an integer. So, ( n = 5 ) doesn't work either. Next, ( n = 6 ): ( 2*6 = 12 = 3r implies r = 4 ), which is an integer. So, ( n = 6 ) works.Wait, but earlier I thought ( n = 3 ) works, but 3 isn't an option. So, perhaps the answer is 6, which is option C. But wait, let me check the original problem again.The problem says ( n in mathbb{N}^* ), which means positive integers. So, 3 is a positive integer, but it's not an option. So, the next possible is 6, which is an option. So, the minimum value of ( n ) is 6, which is option C.Wait, but in the initial thought process, I thought the answer was 7, but that might have been a mistake. Let me double-check.If ( n = 6 ), then ( r = 4 ). Let me compute the term to see if it's indeed a constant term.Compute ( T_{5} = C(6, 4) cdot (3x^2)^{2} cdot left(-frac{2}{3x}right)^4 ).Compute each part:- ( C(6, 4) = 15 )- ( (3x^2)^2 = 9x^4 )- ( left(-frac{2}{3x}right)^4 = left(frac{16}{81}right) x^{-4} )Multiply them together:- Coefficients: ( 15 cdot 9 cdot frac{16}{81} )- Variables: ( x^4 cdot x^{-4} = x^0 = 1 )Compute coefficients:- ( 15 cdot 9 = 135 )- ( 135 cdot frac{16}{81} = frac{135 cdot 16}{81} = frac{2160}{81} = 26.666... ) Wait, that's not an integer. Hmm, that's strange.Wait, maybe I made a mistake in the calculation. Let me compute it again.( 15 cdot 9 = 135 )( 135 cdot frac{16}{81} = frac{135}{81} cdot 16 = frac{5}{3} cdot 16 = frac{80}{3} ). Hmm, that's still not an integer. But the term should be a constant term, but the coefficient is a fraction. Is that possible?Wait, but the problem doesn't specify that the constant term has to be an integer, just that it's a constant term. So, even if the coefficient is a fraction, it's still a constant term. So, ( n = 6 ) does give a constant term, even though the coefficient is ( frac{80}{3} ).But wait, let me check ( n = 3 ). If ( n = 3 ), then ( r = 2 ). Let's compute the term.( T_{3} = C(3, 2) cdot (3x^2)^{1} cdot left(-frac{2}{3x}right)^2 )Compute each part:- ( C(3, 2) = 3 )- ( (3x^2)^1 = 3x^2 )- ( left(-frac{2}{3x}right)^2 = frac{4}{9x^2} )Multiply them together:- Coefficients: ( 3 cdot 3 cdot frac{4}{9} = 3 cdot 3 = 9; 9 cdot frac{4}{9} = 4 )- Variables: ( x^2 cdot x^{-2} = x^0 = 1 )So, the term is 4, which is a constant term. So, ( n = 3 ) does give a constant term. But 3 isn't an option. The options start at 4. So, maybe the problem is that ( n ) must be such that the term is present, but perhaps ( r ) must be less than or equal to ( n ). For ( n = 3 ), ( r = 2 ) is valid. So, why isn't 3 an option?Wait, maybe I made a mistake in the initial setup. Let me check the exponent again.Wait, in the term ( (3x^2 - frac{2}{3x})^n ), the exponents are ( x^2 ) and ( x^{-1} ). So, when we raise them to powers, the exponents add up. So, for the general term, the exponent of ( x ) is ( 2(n - r) - r = 2n - 3r ). So, setting that equal to zero gives ( 2n = 3r ), so ( n = frac{3r}{2} ). So, ( r ) must be even, as I thought.But for ( n = 3 ), ( r = 2 ), which is valid, but 3 isn't an option. So, perhaps the problem is that the term must be present in the expansion, but for ( n = 3 ), the term is present, but maybe the problem is considering ( n ) starting from 1, but 3 is still smaller than the options given.Wait, the options are 4,5,6,7. So, maybe the problem is that ( r ) must be less than or equal to ( n ), but for ( n = 3 ), ( r = 2 ) is okay, but perhaps the problem expects ( n ) to be such that ( r ) is an integer, but maybe I need to consider that ( r ) must be an integer, so ( n ) must be a multiple of 3/2, but since ( n ) is an integer, ( r ) must be even.Wait, perhaps I need to find the smallest ( n ) such that ( r ) is an integer. So, ( n = frac{3r}{2} ), so ( r = frac{2n}{3} ). So, ( r ) must be an integer, so ( 2n ) must be divisible by 3, meaning ( n ) must be a multiple of 3. So, the smallest ( n ) is 3, but since 3 isn't an option, the next multiple is 6, which is option C.But wait, earlier when I calculated ( n = 6 ), the coefficient was ( frac{80}{3} ), which is a fraction, but still a constant term. So, maybe the answer is 6.But wait, let me check ( n = 4 ). If ( n = 4 ), then ( 2n = 8 = 3r implies r = 8/3 ), which is not an integer. So, no constant term.( n = 5 ): ( 2*5 = 10 = 3r implies r = 10/3 ), not an integer.( n = 6 ): ( r = 4 ), which is integer, so constant term exists.( n = 7 ): ( 2*7 = 14 = 3r implies r = 14/3 ), not integer.So, the next possible ( n ) after 6 is 9, but that's not an option. So, the minimum ( n ) is 6, which is option C.Wait, but earlier I thought the answer was 7, but that might have been a mistake. Let me check again.Wait, in the initial problem, the options are A:4, B:5, C:6, D:7. So, 6 is an option, and it's the smallest ( n ) that works. So, the answer should be C:6.But wait, in the initial thought process, I thought the answer was 7, but that was a mistake. So, I need to correct that.So, to summarize: The exponent of ( x ) in the general term is ( 2n - 3r ). Setting this equal to zero gives ( 2n = 3r implies n = frac{3r}{2} ). For ( n ) to be an integer, ( r ) must be even. The smallest ( r ) is 2, giving ( n = 3 ), but since 3 isn't an option, the next possible is ( r = 4 ), giving ( n = 6 ), which is option C.Therefore, the minimum value of ( n ) is 6.

💡First, I need to find the square root of 500 to determine the least integer greater than it.I know that 22 squared is 484 and 23 squared is 529. Since 500 is between 484 and 529, the square root of 500 must be between 22 and 23.To find a more precise value, I can use the linear approximation method. I'll start with an initial guess of 22.4 and calculate its square.22.4 squared equals 501.76, which is slightly higher than 500. To adjust, I'll subtract a small value from 22.4 and recalculate.After a few iterations, I find that 22.36 squared is approximately 500. This means that the square root of 500 is approximately 22.36.Since 22.36 is not an integer, the least integer greater than the square root of 500 is 23.