💡Okay, so I have this equation: x² + y² + 4x - 2y - 4 = 0, and I need to find the maximum value of x² + y². Hmm, let me think about how to approach this.First, I remember that x² + y² is the equation of a circle, but it's not in the standard form. Maybe I should try to rewrite it in the standard form by completing the square for both x and y terms. That usually helps in identifying the center and radius of the circle.Let me start with the x terms: x² + 4x. To complete the square, I take half of the coefficient of x, which is 4, so half of that is 2, and then square it, getting 4. So, I can write x² + 4x as (x + 2)² - 4.Now, for the y terms: y² - 2y. Similarly, half of -2 is -1, and squaring that gives 1. So, y² - 2y can be written as (y - 1)² - 1.Putting it all back into the original equation:(x + 2)² - 4 + (y - 1)² - 1 - 4 = 0Simplifying the constants: -4 -1 -4 = -9So, the equation becomes:(x + 2)² + (y - 1)² - 9 = 0Which means:(x + 2)² + (y - 1)² = 9Wait, that's a circle with center at (-2, 1) and radius 3, since 9 is 3 squared.Okay, so the given equation represents a circle centered at (-2, 1) with radius 3.Now, I need to find the maximum value of x² + y². Hmm, x² + y² is the square of the distance from any point (x, y) to the origin (0, 0). So, essentially, I need to find the point on the circle that is farthest from the origin and then compute the square of that distance.To find the maximum distance from the origin to a point on the circle, I can use the distance from the origin to the center of the circle and then add the radius. Because the farthest point will be in the direction away from the origin.First, let's find the distance from the origin (0, 0) to the center of the circle (-2, 1). Using the distance formula:Distance = √[(x2 - x1)² + (y2 - y1)²]= √[(-2 - 0)² + (1 - 0)²]= √[4 + 1]= √5So, the distance from the origin to the center is √5. The radius of the circle is 3. Therefore, the maximum distance from the origin to any point on the circle is √5 + 3.But wait, is that correct? Let me visualize it. If the center is at (-2, 1), which is in the second quadrant, and the origin is at (0,0). The distance between them is √5, which is approximately 2.236. The radius is 3, which is larger than √5, so the origin is inside the circle? Wait, no, because the radius is 3, and the distance from the center to the origin is √5 ≈ 2.236, which is less than 3. So, the origin is inside the circle.Wait, hold on. If the origin is inside the circle, then the maximum distance from the origin to a point on the circle would be the distance from the origin to the center plus the radius. But since the origin is inside, the maximum distance is indeed the distance from the origin to the center plus the radius.Wait, no, actually, if the origin is inside the circle, then the maximum distance would be the distance from the origin to the center plus the radius. Because the farthest point would be in the direction away from the origin.But let me confirm: if the origin is inside the circle, the maximum distance from the origin to a point on the circle is equal to the distance from the origin to the center plus the radius. Yes, that makes sense.So, the maximum distance is √5 + 3.Therefore, the maximum value of x² + y² is (√5 + 3)².Let me compute that:(√5 + 3)² = (√5)² + 2*√5*3 + 3² = 5 + 6√5 + 9 = 14 + 6√5.So, the maximum value of x² + y² is 14 + 6√5.Wait, but let me think again. Is the origin inside the circle? The distance from the center to the origin is √5 ≈ 2.236, and the radius is 3. Since 2.236 < 3, yes, the origin is inside the circle. Therefore, the maximum distance is indeed √5 + 3, and squaring that gives 14 + 6√5.Alternatively, another way to approach this is to parametrize the circle and then maximize x² + y².Let me try that method as well to confirm.Parametrizing the circle: since it's centered at (-2, 1) with radius 3, any point on the circle can be written as:x = -2 + 3 cosθy = 1 + 3 sinθThen, x² + y² = (-2 + 3 cosθ)² + (1 + 3 sinθ)²Let me expand that:= (4 - 12 cosθ + 9 cos²θ) + (1 + 6 sinθ + 9 sin²θ)= 4 + 1 - 12 cosθ + 6 sinθ + 9 cos²θ + 9 sin²θ= 5 - 12 cosθ + 6 sinθ + 9 (cos²θ + sin²θ)Since cos²θ + sin²θ = 1,= 5 - 12 cosθ + 6 sinθ + 9*1= 14 - 12 cosθ + 6 sinθSo, x² + y² = 14 - 12 cosθ + 6 sinθNow, to find the maximum value of this expression, we can consider the function f(θ) = 14 - 12 cosθ + 6 sinθTo find its maximum, we can find the amplitude of the sinusoidal part.The expression -12 cosθ + 6 sinθ can be written as R cos(θ - φ), where R is the amplitude.R = √[(-12)² + 6²] = √[144 + 36] = √180 = 6√5So, f(θ) = 14 + 6√5 cos(θ - φ)The maximum value of cos(θ - φ) is 1, so the maximum value of f(θ) is 14 + 6√5*1 = 14 + 6√5.So, that confirms the earlier result.Therefore, the maximum value of x² + y² is 14 + 6√5.I think that's solid. Both methods lead to the same answer, so I'm confident it's correct.

💡To determine how many different terms are produced when the expression ((a + b + c + d + e + f + g + h + i)^2) is expanded and simplified, we can analyze the process of expanding the product.1. **Expression Rewriting:** [ (a + b + c + d + e + f + g + h + i)^2 = (a + b + c + d + e + f + g + h + i)(a + b + c + d + e + f + g + h + i) ]2. **Calculating the Product:** When expanding the product of the two sums, we need to multiply every term in the first sum by every term in the second sum.3. **Initial Term Multiplication:** - Multiply the (a) in the first bracket by each term in the second bracket: [ a cdot a, a cdot b, a cdot c, a cdot d, a cdot e, a cdot f, a cdot g, a cdot h, a cdot i ] This produces 9 different terms for the (a) term alone.4. **Continuing the Process with the (b):** - Multiply (b) in the first bracket by the terms from (a) to (i) in the second bracket: [ b cdot a, b cdot b, b cdot c, b cdot d, b cdot e, b cdot f, b cdot g, b cdot h, b cdot i ] However, since (b cdot a) is already counted in the previous step, this produces 8 new terms.5. **Generalizing the Steps:** - Continuing this process for each subsequent letter, the count of new terms for each letter is reduced by one until the last letter (i) is multiplied by itself: [ text{New Terms:} quad 9, 8, 7, 6, 5, 4, 3, 2, 1 ]6. **Summing the Number of Different Terms:** Calculate the total number of different terms by adding the terms: [ 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = sum_{k=1}^{9} k ] The sum of the first (n) natural numbers is given by: [ frac{n(n+1)}{2} ] Substitute (n = 9) in the formula: [ frac{9(9+1)}{2} = frac{9 times 10}{2} = 45 ]### Conclusion:Therefore, the total number of different terms produced in the expansion and simplification of ((a + b + c + d + e + f + g + h + i)^2) is (boxed{45}).

💡Alright, so I've got this problem here: "If 3a + 4b = 0 and a is three less than twice b, what is 5b?" Hmm, okay, let's break this down step by step. I'm a bit nervous because I'm still getting the hang of these kinds of algebra problems, but I think I can figure it out.First, let's look at what we're given. There are two equations here. The first one is straightforward: 3a plus 4b equals zero. So, that's like a linear equation with two variables, a and b. The second part says that a is three less than twice b. Okay, so that's another equation, but it's expressed in words instead of symbols. I need to translate that into an equation I can work with.Alright, so let's start by writing down both equations clearly. The first equation is:3a + 4b = 0And the second statement is: a is three less than twice b. Translating that into math, "twice b" would be 2b, and "three less than that" would be 2b minus 3. So, the second equation is:a = 2b - 3Now, I have a system of two equations with two variables. This means I can solve for one variable and then substitute back to find the other. Since the second equation already expresses a in terms of b, it might be easiest to substitute this expression for a into the first equation.So, let's do that. I'll take the expression for a from the second equation and plug it into the first equation wherever I see a. That gives me:3(2b - 3) + 4b = 0Okay, now I need to simplify this equation to solve for b. Let's start by distributing the 3 into the parentheses:3 * 2b = 6b3 * (-3) = -9So, after distributing, the equation becomes:6b - 9 + 4b = 0Now, I can combine like terms. The terms with b are 6b and 4b, and the constant term is -9. Adding 6b and 4b gives me 10b. So, the equation simplifies to:10b - 9 = 0Now, I need to solve for b. Let's isolate b by moving the constant term to the other side of the equation. I'll add 9 to both sides:10b - 9 + 9 = 0 + 910b = 9Now, to solve for b, I'll divide both sides by 10:b = 9/10Okay, so b is 9/10. But the question asks for 5b, not just b. So, I need to multiply b by 5 to find the value of 5b.Let's do that:5b = 5 * (9/10)Multiplying 5 by 9 gives me 45, and the denominator remains 10. So, 5b is 45/10. I can simplify this fraction by dividing both the numerator and the denominator by 5:45 ÷ 5 = 910 ÷ 5 = 2So, 45/10 simplifies to 9/2. Therefore, 5b equals 9/2.Wait a second, let me double-check my steps to make sure I didn't make any mistakes. Starting from the beginning:1. I translated "a is three less than twice b" correctly into a = 2b - 3.2. Substituted a into the first equation: 3(2b - 3) + 4b = 0.3. Distributed the 3: 6b - 9 + 4b = 0.4. Combined like terms: 10b - 9 = 0.5. Solved for b: 10b = 9 → b = 9/10.6. Calculated 5b: 5*(9/10) = 45/10 = 9/2.Everything seems to check out. I don't see any errors in my calculations. So, I'm confident that 5b is indeed 9/2.

💡Okay, so I have this problem where I need to prove that for any triangle with sides a, b, and c, the sum of the squares of the sides is less than twice the sum of the products of the sides taken two at a time. In other words, I need to show that (a^2 + b^2 + c^2 < 2(ab + bc + ca)).Hmm, let me start by recalling some basic properties of triangles. I know that the triangle inequality theorem states that the sum of any two sides of a triangle must be greater than the third side. So, that gives me three inequalities:1. (a + b > c)2. (b + c > a)3. (c + a > b)Maybe I can use these inequalities somehow to derive the desired result. Let me think about how to relate these to the expression I need to prove.I also remember that sometimes squaring both sides of an inequality can help, but I have to be careful because squaring can sometimes introduce extraneous solutions or reverse the inequality if negative numbers are involved. However, since a, b, and c are lengths of sides, they are positive, so squaring should maintain the inequality direction.Let me try squaring each of the triangle inequalities:1. ((a + b)^2 > c^2)2. ((b + c)^2 > a^2)3. ((c + a)^2 > b^2)Expanding each of these squared terms:1. (a^2 + 2ab + b^2 > c^2)2. (b^2 + 2bc + c^2 > a^2)3. (c^2 + 2ca + a^2 > b^2)Now, if I add all three inequalities together, what do I get?Adding the left sides: (a^2 + 2ab + b^2 + b^2 + 2bc + c^2 + c^2 + 2ca + a^2)Adding the right sides: (c^2 + a^2 + b^2)So, simplifying the left side:(2a^2 + 2b^2 + 2c^2 + 2ab + 2bc + 2ca)And the right side is:(a^2 + b^2 + c^2)Putting it all together:(2a^2 + 2b^2 + 2c^2 + 2ab + 2bc + 2ca > a^2 + b^2 + c^2)Subtracting (a^2 + b^2 + c^2) from both sides:(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca > 0)Wait, that doesn't seem helpful. Maybe I took a wrong turn. Let me try another approach.I recall that in any triangle, the Law of Cosines relates the sides and the cosine of an angle. It states that (c^2 = a^2 + b^2 - 2abcos(C)), where C is the angle opposite side c. Maybe I can use this to express each side squared in terms of the other sides and the cosine of the included angle.Let me write down the Law of Cosines for each angle:1. (a^2 = b^2 + c^2 - 2bccos(A))2. (b^2 = a^2 + c^2 - 2accos(B))3. (c^2 = a^2 + b^2 - 2abcos(C))Now, if I add all three equations together:(a^2 + b^2 + c^2 = (b^2 + c^2 - 2bccos(A)) + (a^2 + c^2 - 2accos(B)) + (a^2 + b^2 - 2abcos(C)))Simplifying the right side:(a^2 + b^2 + c^2 = 2a^2 + 2b^2 + 2c^2 - 2bccos(A) - 2accos(B) - 2abcos(C))Subtracting (a^2 + b^2 + c^2) from both sides:(0 = a^2 + b^2 + c^2 - 2bccos(A) - 2accos(B) - 2abcos(C))Hmm, not sure if that helps directly. Maybe I need to think differently.Another thought: perhaps I can rearrange the original inequality I need to prove. Let's write it as:(a^2 + b^2 + c^2 - 2ab - 2bc - 2ca < 0)Which can be rewritten as:(-(ab + bc + ca) + (a^2 + b^2 + c^2 - ab - bc - ca) < 0)Wait, that doesn't seem particularly useful. Maybe I can factor the left side somehow.Alternatively, I know that (2(ab + bc + ca) - (a^2 + b^2 + c^2)) is a known expression. Let me compute that:(2(ab + bc + ca) - (a^2 + b^2 + c^2) = (2ab - a^2 - b^2) + (2bc - b^2 - c^2) + (2ca - c^2 - a^2))Each of these terms can be written as (- (a - b)^2), (- (b - c)^2), and (- (c - a)^2) respectively.So,(2(ab + bc + ca) - (a^2 + b^2 + c^2) = -[(a - b)^2 + (b - c)^2 + (c - a)^2])Since squares are always non-negative, the sum ((a - b)^2 + (b - c)^2 + (c - a)^2) is non-negative. Therefore, the entire expression is non-positive:(2(ab + bc + ca) - (a^2 + b^2 + c^2) leq 0)Which implies:(2(ab + bc + ca) leq a^2 + b^2 + c^2)Wait, but that's the opposite of what I need to prove. I need to show that (a^2 + b^2 + c^2 < 2(ab + bc + ca)), but according to this, it's the other way around.Hmm, maybe I made a mistake in the signs. Let me double-check.Starting from:(2(ab + bc + ca) - (a^2 + b^2 + c^2) = (2ab - a^2 - b^2) + (2bc - b^2 - c^2) + (2ca - c^2 - a^2))Each term:(2ab - a^2 - b^2 = - (a^2 - 2ab + b^2) = - (a - b)^2)Similarly for the others:(2bc - b^2 - c^2 = - (b - c)^2)(2ca - c^2 - a^2 = - (c - a)^2)So, indeed,(2(ab + bc + ca) - (a^2 + b^2 + c^2) = -[(a - b)^2 + (b - c)^2 + (c - a)^2])Which is always less than or equal to zero because the sum of squares is non-negative. Therefore,(2(ab + bc + ca) leq a^2 + b^2 + c^2)But this contradicts what I need to prove. So, where is the mistake?Wait a second, in the problem statement, it says "an arbitrary triangle". Maybe in degenerate triangles, the inequality becomes equality, but for non-degenerate triangles, it's strict. Let me think.In a degenerate triangle where one side equals the sum of the other two, the triangle becomes a straight line, and the area is zero. In such a case, the inequality becomes equality because the sum of the squares equals twice the sum of the products.But for a non-degenerate triangle, where the sides satisfy the strict triangle inequality, the expression ((a - b)^2 + (b - c)^2 + (c - a)^2) is positive, so:(2(ab + bc + ca) - (a^2 + b^2 + c^2) = -[(a - b)^2 + (b - c)^2 + (c - a)^2] < 0)Which implies:(2(ab + bc + ca) < a^2 + b^2 + c^2)Wait, that's still the opposite of what I need. I must have messed up the direction somewhere.Let me go back to the original problem. It says to prove (a^2 + b^2 + c^2 < 2(ab + bc + ca)). But according to my derivation, it's the other way around. So, either the problem is incorrect, or my reasoning is flawed.Wait, maybe I misapplied the Law of Cosines. Let me double-check that part.Law of Cosines: (c^2 = a^2 + b^2 - 2abcos(C)). So, if I add all three, I get:(a^2 + b^2 + c^2 = 2a^2 + 2b^2 + 2c^2 - 2abcos(C) - 2bccos(A) - 2cacos(B))Wait, that seems correct. Then, moving terms:(0 = a^2 + b^2 + c^2 - 2abcos(C) - 2bccos(A) - 2cacos(B))But I don't see how that helps with the original inequality.Maybe another approach: consider that in any triangle, the angles are less than 180 degrees, and the cosines of the angles are less than 1. So, perhaps I can use that to bound the terms.From the Law of Cosines:(c^2 = a^2 + b^2 - 2abcos(C))Since (cos(C) < 1) (because angle C is less than 180 degrees in a non-degenerate triangle), then:(c^2 > a^2 + b^2 - 2ab)Similarly,(a^2 > b^2 + c^2 - 2bc)(b^2 > a^2 + c^2 - 2ac)Adding these three inequalities:(a^2 + b^2 + c^2 > (a^2 + b^2 - 2ab) + (b^2 + c^2 - 2bc) + (a^2 + c^2 - 2ac))Simplifying the right side:(a^2 + b^2 + c^2 > 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac)Subtracting (a^2 + b^2 + c^2) from both sides:(0 > a^2 + b^2 + c^2 - 2ab - 2bc - 2ac)Which can be rewritten as:(2(ab + bc + ac) > a^2 + b^2 + c^2)Ah, there we go! So, by using the fact that the cosine of each angle is less than 1, we derived the desired inequality.So, to summarize:1. Start with the Law of Cosines for each angle.2. Use the fact that (cos(C) < 1) for each angle in a non-degenerate triangle.3. Derive inequalities for each side squared.4. Add these inequalities together.5. Simplify to get the desired result.Therefore, the inequality (a^2 + b^2 + c^2 < 2(ab + bc + ca)) holds for any non-degenerate triangle.

💡First, I need to convert the repeating decimals (0.overline{15}) and (2.overline{403}) into fractions.For (0.overline{15}), I'll let (x = 0.overline{15}). Multiplying both sides by 100 gives (100x = 15.overline{15}). Subtracting the original equation from this, I get (99x = 15), so (x = frac{15}{99}), which simplifies to (frac{5}{33}).Next, for (2.overline{403}), I'll let (y = 2.overline{403}). Multiplying both sides by 1000 gives (1000y = 2403.overline{403}). Subtracting the original equation from this, I get (999y = 2401), so (y = frac{2401}{999}).Now, I'll form the fraction (frac{frac{5}{33}}{frac{2401}{999}}), which simplifies to (frac{5 times 999}{33 times 2401}). Calculating this gives (frac{4995}{79233}).Finally, I'll check if the fraction can be simplified further. Since 4995 and 79233 have no common factors other than 1, the fraction is already in its simplest form.

💡Okay, so I'm trying to figure out whether it's possible for a zoo to have hippos but no giraffes and no rhinos, based on the given conditions. Let me break down the problem step by step.First, let's list out the conditions:1. All zoos that have hippos and rhinos have no giraffes.2. All zoos that have rhinos and no giraffes have hippos.3. All zoos that have hippos and giraffes have rhinos.I need to see if there's a way for a zoo to have hippos without having giraffes or rhinos. Let's try to understand each condition and see how they interact.Starting with the first condition: If a zoo has both hippos and rhinos, then it cannot have giraffes. So, if H (hippos) and R (rhinos) are both present, G (giraffes) must be absent. Symbolically, this can be written as:H ∧ R → ¬GNext, the second condition says that if a zoo has rhinos and no giraffes, then it must have hippos. So, if R is present and G is absent, then H must be present. Symbolically:R ∧ ¬G → HThe third condition states that if a zoo has hippos and giraffes, then it must also have rhinos. So, if H and G are both present, then R must be present. Symbolically:H ∧ G → RNow, I need to see if it's possible for a zoo to have H (hippos) but not G (giraffes) and not R (rhinos). In other words, is H ∧ ¬G ∧ ¬R possible?Let's analyze this. If a zoo has hippos but no giraffes and no rhinos, does this violate any of the given conditions?1. The first condition says that if a zoo has both hippos and rhinos, then it cannot have giraffes. In this case, the zoo doesn't have rhinos, so this condition doesn't apply here.2. The second condition says that if a zoo has rhinos and no giraffes, then it must have hippos. Again, since our zoo doesn't have rhinos, this condition doesn't impose any restrictions on our scenario.3. The third condition says that if a zoo has hippos and giraffes, then it must have rhinos. Our zoo doesn't have giraffes, so this condition also doesn't apply.So, none of the given conditions directly prevent a zoo from having hippos without giraffes and rhinos. Therefore, it seems possible for such a zoo to exist.But wait, let me double-check to make sure I'm not missing something. Could there be an indirect implication that I'm overlooking?Suppose there's a zoo with hippos but no giraffes and no rhinos. Does this create any contradictions with the given conditions?- The first condition only applies when both hippos and rhinos are present, which isn't the case here.- The second condition applies when rhinos and no giraffes are present, which again isn't the case here.- The third condition applies when hippos and giraffes are present, which isn't the case here either.Since none of the conditions are violated in this scenario, it seems that such a zoo is indeed possible.Another way to think about it is to consider the contrapositive of the given conditions. For example, the contrapositive of the first condition is: If a zoo has giraffes, then it does not have both hippos and rhinos. Similarly, the contrapositive of the second condition is: If a zoo does not have hippos, then it does not have rhinos and no giraffes. And the contrapositive of the third condition is: If a zoo does not have rhinos, then it does not have both hippos and giraffes.None of these contrapositives prevent a zoo from having hippos without giraffes and rhinos.Therefore, after carefully analyzing each condition and considering their implications, I conclude that it is possible for a zoo to have hippos but no giraffes and no rhinos.