💡To determine how many quarters Mandy has, we start by defining ( q ) as the number of quarters. The monetary range translates to ( 40 < q < 400 ) because each quarter is worth 0.25. The stacking conditions can be expressed as congruences:[begin{align*}q &equiv 2 pmod{6}, q &equiv 2 pmod{7}, q &equiv 2 pmod{8}.end{align*}]The least common multiple (LCM) of 6, 7, and 8 is 168. Therefore, any number ( q ) that satisfies ( q equiv 2 pmod{168} ) will solve the system. The values of ( q ) within the range ( 40 < q < 400 ) are 170 and 338. Thus, Mandy has either 170 or 338 quarters.

💡Okay, so I need to find the molecular weight of a compound that has 1 Nitrogen atom, 4 Hydrogen atoms, and 1 Bromine atom. Hmm, molecular weight is basically the sum of the atomic weights of all the atoms in the molecule, right? So, I guess I need to look up the atomic weights of Nitrogen, Hydrogen, and Bromine first.Let me think, where can I find the atomic weights? Oh, right, the periodic table! Nitrogen is element number 7, Hydrogen is number 1, and Bromine is number 35. Okay, so looking up their atomic weights:- Nitrogen (N): I think it's around 14.01 g/mol.- Hydrogen (H): That's usually about 1.008 g/mol, but sometimes people approximate it to 1.01 g/mol.- Bromine (Br): I'm not as sure about this one, but I believe it's around 79.90 g/mol.Wait, let me double-check these values to make sure I'm not making a mistake. Nitrogen is definitely around 14.01, Hydrogen is approximately 1.008, and Bromine is about 79.90. Yeah, those seem right.Now, the compound has 1 Nitrogen atom, 4 Hydrogen atoms, and 1 Bromine atom. So, I need to multiply each atomic weight by the number of atoms in the compound and then add them all together.Let's do the calculations step by step:1. Nitrogen: 1 atom × 14.01 g/mol = 14.01 g/mol2. Hydrogen: 4 atoms × 1.008 g/mol = 4.032 g/mol3. Bromine: 1 atom × 79.90 g/mol = 79.90 g/molNow, add them all up:14.01 + 4.032 + 79.90 = ?Let's add 14.01 and 4.032 first:14.01 + 4.032 = 18.042Now, add 79.90 to that:18.042 + 79.90 = 97.942So, the molecular weight is approximately 97.942 g/mol. But, I think it's common to round this to two decimal places, so that would be 97.94 g/mol.Wait, in the initial problem, the user mentioned 4 Hydrogen atoms, but in the final calculation, I used 4.032 g/mol based on 1.008 g/mol per Hydrogen. If I use 1.01 g/mol instead, let's see what happens:Hydrogen: 4 atoms × 1.01 g/mol = 4.04 g/molThen, total molecular weight would be:14.01 + 4.04 + 79.90 = 97.95 g/molSo, depending on whether we use 1.008 or 1.01 for Hydrogen, the molecular weight is either 97.94 or 97.95 g/mol. I think 97.95 is more commonly used because it's rounded to two decimal places.Just to make sure, let me check if I've accounted for all the atoms correctly. 1 Nitrogen, 4 Hydrogen, and 1 Bromine. Yes, that's correct. And I've multiplied each by their respective atomic weights and summed them up. Looks good.I don't think I've missed anything here. The calculations seem straightforward, and I've double-checked the atomic weights and the arithmetic. So, I'm confident that the molecular weight is approximately 97.95 g/mol.

💡Alright, so I've got this problem here about a continuous function ( f(x) ) defined for all positive real numbers ( x ). It says that ( f(0) = 0 ), which is straightforward enough, and there's this inequality involving the square of the function: ( f^2(x+y) geq f^2(x) + f^2(y) ) for all positive ( x ) and ( y ). The goal is to prove that ( f(x) ) must be of the form ( a sqrt{x} ) where ( a ) is a positive constant.Okay, let's start by unpacking this. First, ( f ) is continuous on ( mathbb{R}^+ ), which is good because continuity often helps in proving such functional equations. The condition ( f(0) = 0 ) is also a boundary condition that might be useful later.Now, the inequality ( f^2(x+y) geq f^2(x) + f^2(y) ) is interesting. It reminds me a bit of the Cauchy-Schwarz inequality or maybe properties of norms, but I'm not sure yet. Let me think about what this inequality is saying. For any two positive numbers ( x ) and ( y ), the square of the function at their sum is at least as big as the sum of the squares of the function at each point individually.Hmm, so if I rearrange this inequality, it would be ( f^2(x+y) - f^2(x) - f^2(y) geq 0 ). That suggests that the function's square is superadditive, meaning that the function's square at the sum is greater than or equal to the sum of the function's squares. Superadditive functions have certain properties, but I'm not too familiar with all of them. Maybe I can find a way to express ( f(x) ) in terms of some other function or use substitution to simplify this.Since the problem suggests that ( f(x) = a sqrt{x} ), maybe I can test this form and see if it satisfies the given conditions. Let's suppose ( f(x) = a sqrt{x} ) for some constant ( a > 0 ). Then, ( f(0) = a sqrt{0} = 0 ), which satisfies the initial condition. Good.Now, let's check the inequality. Compute ( f^2(x+y) ) and see if it's greater than or equal to ( f^2(x) + f^2(y) ).So, ( f^2(x+y) = (a sqrt{x+y})^2 = a^2 (x + y) ).On the other hand, ( f^2(x) + f^2(y) = (a sqrt{x})^2 + (a sqrt{y})^2 = a^2 x + a^2 y = a^2 (x + y) ).So, ( f^2(x+y) = a^2 (x + y) ) and ( f^2(x) + f^2(y) = a^2 (x + y) ). Therefore, ( f^2(x+y) = f^2(x) + f^2(y) ), which satisfies the inequality ( f^2(x+y) geq f^2(x) + f^2(y) ) since equality holds.Okay, so the function ( f(x) = a sqrt{x} ) does satisfy both the initial condition and the inequality. But the problem is to prove that this is the only possible form of ( f(x) ) given the conditions. So, I need to show that no other continuous function satisfies these properties.Maybe I can approach this by assuming that ( f(x) ) is not of the form ( a sqrt{x} ) and reach a contradiction, or perhaps use some properties of continuous functions to show that ( f(x) ) must behave like ( a sqrt{x} ).Another thought: since ( f ) is continuous and defined on ( mathbb{R}^+ ), maybe I can consider its behavior as ( x ) approaches 0 or infinity. But since ( f(0) = 0 ), that might not be too helpful unless I can find some scaling property.Wait, the inequality ( f^2(x+y) geq f^2(x) + f^2(y) ) looks similar to the triangle inequality but in reverse. Normally, the triangle inequality says that ( f(x+y) leq f(x) + f(y) ), but here it's the square of ( f(x+y) ) being greater than or equal to the sum of the squares. Maybe there's a way to relate this to norms or inner products.Alternatively, perhaps I can consider the function ( g(x) = f^2(x) ). Then the inequality becomes ( g(x+y) geq g(x) + g(y) ). So, ( g ) is a superadditive function. If ( g ) is superadditive and continuous, maybe I can use some known results about superadditive functions.I recall that for superadditive functions, if they are also concave or convex, certain forms are enforced. For example, linear functions are both superadditive and subadditive. But ( g(x) = f^2(x) ) is not necessarily linear. However, if ( g ) is superadditive and continuous, perhaps it must be linear? Or maybe affine?Wait, let's think about that. If ( g ) is superadditive and continuous, does that imply linearity? I'm not sure, but maybe under some conditions. Alternatively, perhaps ( g ) must be proportional to ( x ), which would make ( f(x) ) proportional to ( sqrt{x} ).Let me explore this idea. Suppose ( g(x) = c x ) for some constant ( c ). Then ( g(x+y) = c(x + y) = c x + c y = g(x) + g(y) ), which satisfies the equality case of superadditivity. So, linear functions are exactly the functions that satisfy both superadditivity and subadditivity.But in our case, ( g(x+y) geq g(x) + g(y) ), so ( g ) is superadditive but not necessarily subadditive. However, if ( g ) is continuous and superadditive, maybe it must be linear? Or perhaps it can be something else.Wait, another thought: if ( g ) is superadditive and concave, then it must be linear. Is that true? I think that for concave functions, superadditivity combined with concavity might enforce linearity. Let me recall: a concave function satisfies ( g(lambda x + (1 - lambda) y) geq lambda g(x) + (1 - lambda) g(y) ) for ( lambda in [0,1] ). If ( g ) is also superadditive, then ( g(x + y) geq g(x) + g(y) ).If ( g ) is concave and superadditive, does that imply linearity? Maybe. Let's see. Suppose ( g ) is concave and superadditive. Then, for any ( x, y ), ( g(x + y) geq g(x) + g(y) ). Also, by concavity, ( g(x + y) leq g(x) + g(y) ) if ( g ) is linear. Wait, no, concavity doesn't directly imply that.Actually, for concave functions, ( g(x + y) geq g(x) + g(y) ) is not necessarily true. In fact, concave functions typically have the property that ( g(x + y) leq g(x) + g(y) ) if they are linear. Hmm, maybe I'm getting confused.Alternatively, perhaps I can use the fact that ( g ) is superadditive and continuous to show that it must be linear. I remember that for functions defined on ( mathbb{R}^+ ), if they are superadditive and continuous, they must be linear. Is that a theorem? I think it might be related to Cauchy's functional equation.Cauchy's functional equation is ( f(x + y) = f(x) + f(y) ), and under continuity, the solutions are linear functions. But in our case, we have ( g(x + y) geq g(x) + g(y) ), which is a form of superadditivity. Maybe there's a way to show that ( g ) must satisfy Cauchy's equation, hence be linear.Wait, if ( g ) is superadditive and continuous, can we show that it must be linear? Let me think. Suppose ( g ) is superadditive, so ( g(x + y) geq g(x) + g(y) ). If we can also show that ( g ) is subadditive, then ( g(x + y) = g(x) + g(y) ), and by continuity, ( g ) must be linear.But how can we show subadditivity? We only have superadditivity. Maybe we can use some other property of ( g ). Since ( g(x) = f^2(x) ), and ( f ) is continuous, ( g ) is also continuous and non-negative.Wait, another approach: consider scaling. Suppose we fix ( x ) and let ( y ) vary. Maybe we can use some kind of homogeneity.Let me try to see if ( g ) is homogeneous. Suppose ( g(kx) = k g(x) ) for some constant ( k ). If ( g ) is linear, then this would hold. But I don't know if ( g ) is homogeneous.Alternatively, maybe I can use the fact that ( g ) is superadditive and continuous to show that it's linear. Let me recall that for functions defined on ( mathbb{R}^+ ), if they are superadditive and continuous, they are linear. Is that true? I think it might be, but I'm not entirely sure.Wait, I think I remember that for functions defined on ( mathbb{R}^+ ), if they are superadditive and continuous, then they are linear. Yes, that seems familiar. So, if ( g ) is superadditive and continuous, then ( g(x) = c x ) for some constant ( c ).If that's the case, then ( g(x) = c x ), so ( f(x) = sqrt{c x} = a sqrt{x} ) where ( a = sqrt{c} ). Since ( f(x) ) is real-valued and ( c ) must be non-negative (because ( g(x) = f^2(x) geq 0 )), ( a ) is a real number. But we also have ( f(0) = 0 ), which is satisfied because ( a sqrt{0} = 0 ).Therefore, the only continuous function ( f(x) ) satisfying the given conditions is ( f(x) = a sqrt{x} ) for some ( a > 0 ).Wait, but I'm not entirely sure about the step where I concluded that superadditive and continuous functions on ( mathbb{R}^+ ) must be linear. Maybe I should verify that.Let me think about it. Suppose ( g: mathbb{R}^+ to mathbb{R} ) is continuous and superadditive, i.e., ( g(x + y) geq g(x) + g(y) ) for all ( x, y > 0 ). Does this imply that ( g ) is linear?I think yes. Here's why: consider the function ( h(x) = g(x) - c x ), where ( c ) is a constant to be determined. If we can choose ( c ) such that ( h(x + y) = h(x) + h(y) ), then by Cauchy's functional equation, ( h(x) ) must be linear, and hence ( g(x) ) is linear.But I'm not sure if this approach works directly. Alternatively, maybe I can use the fact that superadditive functions have certain growth properties. For example, if ( g ) is superadditive, then ( g(n x) geq n g(x) ) for any positive integer ( n ). This suggests that ( g ) grows at least linearly.But since ( g ) is continuous, it can't grow faster than linearly without violating continuity. Wait, no, continuity doesn't necessarily restrict the growth rate, but combined with superadditivity, maybe it does.Alternatively, perhaps I can use the fact that ( g ) is superadditive and continuous to show that it's differentiable and then use calculus to show that the derivative is constant.Let me try that. Suppose ( g ) is differentiable (since it's continuous, maybe it's differentiable almost everywhere or something). Then, considering the inequality ( g(x + y) geq g(x) + g(y) ), we can take partial derivatives with respect to ( y ) and set ( y = 0 ) to find something about the derivative.Wait, but ( y ) is approaching 0 from the positive side. So, taking the derivative as ( y to 0^+ ), we get:( lim_{y to 0^+} frac{g(x + y) - g(x)}{y} geq lim_{y to 0^+} frac{g(x) + g(y) - g(x)}{y} = lim_{y to 0^+} frac{g(y)}{y} ).Assuming ( g ) is differentiable at ( x ), the left-hand side is ( g'(x) ), and the right-hand side is ( g'(0) ) if the limit exists. So, ( g'(x) geq g'(0) ) for all ( x > 0 ).Similarly, if we fix ( y ) and let ( x ) vary, we might get similar inequalities. But I'm not sure if this directly leads to ( g'(x) ) being constant.Alternatively, maybe I can use the fact that ( g ) is superadditive and continuous to show that it's linear by considering the behavior as ( x ) scales.Suppose we define ( g(k x) ) for some constant ( k > 0 ). If ( g ) is linear, then ( g(k x) = k g(x) ). But if ( g ) is superadditive, then ( g(k x) geq k g(x) ) if ( k ) is an integer, by repeatedly applying superadditivity. But for non-integer ( k ), it's not straightforward.Wait, but since ( g ) is continuous, maybe we can extend this to all positive real numbers. Let me think about it.Suppose ( k ) is a positive real number. We can write ( k = n + r ) where ( n ) is an integer and ( 0 leq r < 1 ). Then, using superadditivity:( g(k x) = g((n + r) x) geq g(n x) + g(r x) ).But ( g(n x) geq n g(x) ) by superadditivity applied ( n ) times. So,( g(k x) geq n g(x) + g(r x) ).But I don't know how to relate ( g(r x) ) to ( g(x) ). Maybe I need a different approach.Alternatively, perhaps I can use the fact that ( g ) is superadditive and continuous to show that it's linear by considering the function ( h(x) = g(x) / x ) for ( x > 0 ). Then, ( h(x) ) is well-defined and positive since ( g(x) geq 0 ).Now, let's see what superadditivity implies for ( h(x) ). For any ( x, y > 0 ),( g(x + y) geq g(x) + g(y) ).Dividing both sides by ( x + y ), we get:( h(x + y) geq frac{g(x) + g(y)}{x + y} ).But I'm not sure if this helps directly. Maybe I can use some inequality involving ( h(x) ) and ( h(y) ).Alternatively, suppose we fix ( x ) and let ( y = t x ) for some ( t > 0 ). Then,( g(x + t x) = g((1 + t) x) geq g(x) + g(t x) ).Dividing both sides by ( (1 + t) x ), we get:( h((1 + t) x) geq frac{g(x)}{(1 + t) x} + frac{g(t x)}{(1 + t) x} = frac{h(x)}{1 + t} + frac{h(t x)}{1 + t} ).So,( h((1 + t) x) geq frac{h(x) + h(t x)}{1 + t} ).This seems a bit complicated, but maybe if I set ( x = 1 ), it simplifies:( h(1 + t) geq frac{h(1) + h(t)}{1 + t} ).Hmm, not sure if that helps.Wait, another idea: since ( g ) is superadditive and continuous, maybe it's convex. If ( g ) is convex and superadditive, then perhaps it's linear. But I'm not sure if superadditive and convex functions must be linear.Wait, convex functions satisfy ( g(lambda x + (1 - lambda) y) leq lambda g(x) + (1 - lambda) g(y) ) for ( lambda in [0,1] ). If ( g ) is also superadditive, then combining these might lead to linearity.But I'm not sure. Maybe I need to think differently.Let me go back to the original function ( f(x) ). Since ( f ) is continuous and ( f(0) = 0 ), maybe I can consider its behavior near 0. For small ( x ), ( f(x) ) should be approximately linear, but since ( f(x) = a sqrt{x} ), it's actually behaving like ( a sqrt{x} ), which is not linear near 0.Wait, but ( f(x) = a sqrt{x} ) is not differentiable at 0, but it is continuous. So, maybe the continuity of ( f ) and the superadditivity of ( f^2 ) together enforce the form ( a sqrt{x} ).Alternatively, perhaps I can use the fact that ( f^2(x) ) is superadditive and continuous, hence linear, which would make ( f(x) ) proportional to ( sqrt{x} ).Wait, I think I'm going in circles here. Let me try to summarize what I have so far:1. ( f ) is continuous on ( mathbb{R}^+ ) with ( f(0) = 0 ).2. ( f^2(x + y) geq f^2(x) + f^2(y) ) for all ( x, y > 0 ).3. Define ( g(x) = f^2(x) ), so ( g ) is continuous, non-negative, and superadditive: ( g(x + y) geq g(x) + g(y) ).4. I suspect that ( g(x) ) must be linear, i.e., ( g(x) = c x ), which would imply ( f(x) = a sqrt{x} ) with ( a = sqrt{c} ).5. To justify that ( g ) is linear, I need to use the properties of superadditivity and continuity.I think the key is to use the fact that superadditive and continuous functions on ( mathbb{R}^+ ) must be linear. Let me try to recall or derive why that is the case.Suppose ( g ) is superadditive and continuous. Then, for any ( x > 0 ), we can consider the behavior as ( x ) scales. Let's define ( g_n(x) = g(x/n) ). Since ( g ) is superadditive, ( g(x) = g(n cdot (x/n)) geq n g(x/n) ), so ( g(x/n) leq g(x)/n ).But I'm not sure if that helps. Alternatively, maybe I can use the fact that ( g ) is superadditive to show that it's linear by considering the limit as ( x ) approaches 0.Wait, another approach: consider the function ( g ) on ( mathbb{R}^+ ). Since it's superadditive and continuous, it must be differentiable almost everywhere, and its derivative is non-decreasing. But I'm not sure.Alternatively, maybe I can use the fact that for any ( x > 0 ), ( g(x) geq n g(x/n) ) for any positive integer ( n ). If ( g ) is continuous, then as ( n to infty ), ( x/n to 0 ), and since ( g(0) = 0 ), ( g(x/n) to 0 ). But this doesn't directly help.Wait, perhaps I can use the fact that ( g ) is superadditive to show that it's linear by considering the function ( h(x) = g(x) - c x ) and showing that ( h(x) ) must be zero.Suppose ( g(x) = c x + h(x) ), where ( h(x) ) is some function. Then, the superadditivity condition becomes:( c(x + y) + h(x + y) geq c x + h(x) + c y + h(y) ).Simplifying, we get:( c x + c y + h(x + y) geq c x + c y + h(x) + h(y) ).Therefore,( h(x + y) geq h(x) + h(y) ).So, ( h ) is also superadditive. If we can show that ( h(x) = 0 ) for all ( x ), then ( g(x) = c x ).But how can we show that ( h(x) = 0 )? Maybe by using the continuity of ( g ).Suppose ( h ) is superadditive and continuous. If ( h(x) geq 0 ) for all ( x ), then ( h(x) ) must be linear. Wait, but ( h(x) ) could be zero.Alternatively, suppose ( h(x) ) is not identically zero. Then, there exists some ( x_0 ) where ( h(x_0) > 0 ). But since ( h ) is superadditive, ( h(n x_0) geq n h(x_0) ), which would make ( h ) grow without bound as ( n to infty ). However, ( g(x) = c x + h(x) ) is continuous, so if ( h(x) ) grows without bound, ( g(x) ) would also grow without bound, which is possible unless we have some constraint on ( g(x) ).But in our case, ( g(x) = f^2(x) ), and ( f ) is continuous. There's no explicit bound on ( f(x) ), so ( g(x) ) could indeed grow without bound. Therefore, this approach might not help.Wait, maybe I can use the fact that ( g(x) ) is superadditive and continuous to show that it's linear by considering the function's behavior under scaling.Suppose we fix ( x ) and consider ( g(k x) ) for ( k > 0 ). If ( g ) is linear, then ( g(k x) = k g(x) ). If ( g ) is superadditive, then for integer ( k = n ), ( g(n x) geq n g(x) ). But for non-integer ( k ), it's not directly applicable.However, since ( g ) is continuous, we can extend this to all positive real numbers. Let me try to formalize this.For any ( k > 0 ), we can write ( k = lim_{n to infty} frac{m_n}{n} ) where ( m_n ) is an integer sequence. Then,( g(k x) = gleft( lim_{n to infty} frac{m_n}{n} x right) = lim_{n to infty} gleft( frac{m_n}{n} x right) ).But since ( g ) is superadditive,( gleft( frac{m_n}{n} x right) geq m_n gleft( frac{x}{n} right) ).But ( gleft( frac{x}{n} right) leq frac{g(x)}{n} ) by superadditivity as well, because ( g(x) = g(n cdot frac{x}{n}) geq n gleft( frac{x}{n} right) ).Therefore,( gleft( frac{m_n}{n} x right) geq m_n gleft( frac{x}{n} right) geq m_n cdot frac{g(x)}{n} = frac{m_n}{n} g(x) ).Taking the limit as ( n to infty ),( g(k x) geq k g(x) ).Similarly, by considering ( g(x) = gleft( frac{n}{n} x right) geq n gleft( frac{x}{n} right) ), we have ( gleft( frac{x}{n} right) leq frac{g(x)}{n} ). Then, for ( k = frac{m}{n} ),( g(k x) = gleft( frac{m}{n} x right) leq frac{m}{n} g(x) ).Wait, but this seems contradictory unless ( g(k x) = k g(x) ).Wait, no, actually, from the first part, we have ( g(k x) geq k g(x) ), and from the second part, ( g(k x) leq k g(x) ). Therefore, combining both, ( g(k x) = k g(x) ) for all ( k > 0 ).This shows that ( g ) is homogeneous of degree 1, i.e., ( g(k x) = k g(x) ) for all ( k > 0 ).Now, combining this with superadditivity, we have:( g(x + y) = g(x) + g(y) ).Because ( g(x + y) geq g(x) + g(y) ) by superadditivity, and since ( g ) is homogeneous, we can write ( g(x + y) = g(1 cdot (x + y)) = 1 cdot g(x + y) ), which doesn't directly help. Wait, no, actually, since ( g ) is homogeneous and superadditive, we have:( g(x + y) geq g(x) + g(y) ).But also, by homogeneity, ( g(x + y) = g(1 cdot (x + y)) = 1 cdot g(x + y) ), which doesn't add new information.Wait, perhaps I can use the homogeneity to write ( g(x + y) = g(x) + g(y) ). Let me see.Suppose ( g ) is homogeneous of degree 1, so ( g(k x) = k g(x) ) for all ( k > 0 ). Then, for any ( x, y > 0 ), let ( k = frac{x}{x + y} ) and ( l = frac{y}{x + y} ), so that ( k + l = 1 ).Then,( g(x + y) = gleft( frac{x}{k} cdot k + frac{y}{l} cdot l right) ).But I'm not sure if this helps.Alternatively, since ( g ) is homogeneous of degree 1 and superadditive, let's consider scaling ( x ) and ( y ) such that ( x + y = 1 ). Then,( g(1) = g(x + y) geq g(x) + g(y) ).But ( g(x) = x g(1) ) and ( g(y) = y g(1) ) by homogeneity. Therefore,( g(1) geq x g(1) + y g(1) = (x + y) g(1) = g(1) ).Thus, equality holds, so ( g(x + y) = g(x) + g(y) ).Therefore, ( g ) satisfies Cauchy's functional equation ( g(x + y) = g(x) + g(y) ) for all ( x, y > 0 ), and since ( g ) is continuous, the only solutions are linear functions ( g(x) = c x ).Hence, ( g(x) = c x ), which implies ( f(x) = sqrt{c x} = a sqrt{x} ) where ( a = sqrt{c} ) and ( a > 0 ).So, putting it all together, the only continuous function ( f(x) ) satisfying the given conditions is ( f(x) = a sqrt{x} ) for some positive constant ( a ).

💡Okay, so I have this inequality to solve: (x - 1)/(x + 2) > 1. Hmm, I remember that when dealing with inequalities involving fractions, I need to be careful about the denominator because it can't be zero, and the sign of the denominator affects the direction of the inequality when multiplying both sides. Let me try to work through this step by step.First, I want to get all terms on one side so that I can have a single rational expression. That means I'll subtract 1 from both sides to get:(x - 1)/(x + 2) - 1 > 0Now, to combine these two terms into a single fraction, I need a common denominator. The denominator of the first term is (x + 2), and the second term is just 1, which can be thought of as having a denominator of 1. So, I'll rewrite 1 as (x + 2)/(x + 2) to get the same denominator:(x - 1)/(x + 2) - (x + 2)/(x + 2) > 0Now, subtract the numerators:[(x - 1) - (x + 2)] / (x + 2) > 0Let me simplify the numerator:x - 1 - x - 2 = (x - x) + (-1 - 2) = 0 - 3 = -3So, the inequality becomes:-3/(x + 2) > 0Alright, so now I have a simpler inequality: -3 divided by (x + 2) is greater than zero. I need to find the values of x that make this true.First, let's note that the denominator (x + 2) cannot be zero, so x cannot be -2. That's an important restriction.Now, the inequality is -3/(x + 2) > 0. To solve this, I can think about when a fraction is positive. A fraction is positive if both the numerator and denominator are positive or both are negative.In this case, the numerator is -3, which is negative. So, for the entire fraction to be positive, the denominator must also be negative because a negative divided by a negative is positive.So, the denominator (x + 2) must be negative:x + 2 < 0Solving for x:x < -2Therefore, the solution set is all real numbers less than -2.Let me double-check to make sure I didn't make any mistakes. I transformed the original inequality into a single fraction, simplified it, and then determined the conditions under which the fraction is positive. Since the numerator is negative, the denominator must also be negative for the fraction to be positive, leading to x < -2. That seems correct.I also need to remember that x cannot be -2 because that would make the denominator zero, which is undefined. So, the solution set is all real numbers less than -2, which is option A.Just to be thorough, let me test a value in the solution set and a value outside of it to ensure the inequality holds.First, let's pick x = -3, which is less than -2. Plugging into the original inequality:(-3 - 1)/(-3 + 2) = (-4)/(-1) = 4, which is greater than 1. So, that works.Now, let's pick x = 0, which is greater than -2. Plugging into the original inequality:(0 - 1)/(0 + 2) = (-1)/2 = -0.5, which is not greater than 1. So, that doesn't work, which is consistent with our solution.Therefore, I'm confident that the solution set is x < -2, which corresponds to option A.

💡Okay, so I need to rewrite the statement "对顶角相等" which means "opposite angles are equal" in the format "If ..., then ...". Hmm, let's see. First, I should understand what "对顶角" means. I think it refers to vertical angles or opposite angles formed when two lines intersect. So, when two lines cross each other, they create two pairs of opposite angles. And the statement is saying that these opposite angles are equal in measure.Alright, so the original statement is a general truth about geometry. Now, I need to frame this as a conditional statement. That means I need to identify the condition and the conclusion. The condition is that two angles are opposite angles, and the conclusion is that they are equal. So, putting it together, it should be something like "If two angles are opposite angles, then they are equal."Wait, let me make sure I'm getting this right. Is it always true that opposite angles are equal? I think so, based on what I remember from geometry. When two lines intersect, the opposite angles formed are congruent. So, yes, the statement holds.Maybe I should double-check to ensure I'm not missing anything. Let's visualize two intersecting lines. They form an 'X' shape, creating four angles. The angles directly across from each other are the opposite angles, and they should be equal. So, if angle A is opposite angle B, then angle A equals angle B.Is there any scenario where this isn't true? Well, in Euclidean geometry, which is the standard geometry we learn in school, this is always true. But I wonder if in non-Euclidean geometries, like spherical or hyperbolic, this might not hold. But I think the question is within the context of standard geometry, so I don't need to worry about that.Okay, so going back to the original task: rewrite "对顶角相等" as "If ..., then ...". I think I've got it. The condition is that two angles are opposite angles, and the conclusion is that they are equal. So, the statement becomes "If two angles are opposite angles, then they are equal."Let me see if there's a more concise way to phrase it. Maybe "If two angles are vertical angles, then they are congruent." But the original term was "对顶角," which directly translates to "opposite angles." So, sticking with "opposite angles" might be more accurate.Alternatively, I could say "If two angles are formed by intersecting lines, then the opposite angles are equal." But that adds more information than necessary. The original statement is more straightforward.I think the simplest and most direct translation is "If two angles are opposite angles, then they are equal." That captures the condition and the conclusion clearly.Just to recap:- Original statement: 对顶角相等 (Opposite angles are equal)- Conditional format: If [condition], then [conclusion]- Condition: two angles are opposite angles- Conclusion: they are equalSo, putting it all together: "If two angles are opposite angles, then they are equal."I think that's the most accurate and concise way to rewrite the statement in the desired format. It maintains the original meaning and fits neatly into the "If ..., then ..." structure.

💡Okay, so I have this problem here: the function f(x) = x³ - a x² + 1 is monotonically decreasing in the interval (0, 2). I need to find the range of values for the real number a. The options are A: a ≥ 3, B: a = 3, C: a ≤ 3, D: 0 < a < 3.Hmm, okay. Monotonically decreasing means that the function doesn't increase at any point in that interval. So, if I remember correctly, for a function to be monotonically decreasing on an interval, its derivative should be less than or equal to zero throughout that interval. That makes sense because the derivative represents the slope of the function at any given point. If the slope is negative or zero, the function is decreasing or staying constant.Alright, so let's find the derivative of f(x). The function is f(x) = x³ - a x² + 1. Taking the derivative term by term:The derivative of x³ is 3x², the derivative of -a x² is -2a x, and the derivative of the constant 1 is 0. So, putting that together, f'(x) = 3x² - 2a x.Now, since we want f(x) to be monotonically decreasing on (0, 2), we need f'(x) ≤ 0 for all x in (0, 2). So, 3x² - 2a x ≤ 0 for all x in (0, 2).Let me write that inequality down: 3x² - 2a x ≤ 0. Hmm, maybe I can factor this expression to make it easier to analyze. Let's factor out an x: x(3x - 2a) ≤ 0.So, the inequality becomes x(3x - 2a) ≤ 0. Now, let's analyze this inequality. Since x is in the interval (0, 2), x is positive. So, x > 0 for all x in (0, 2). Therefore, the sign of the entire expression x(3x - 2a) depends on the sign of (3x - 2a).Since x is positive, the inequality x(3x - 2a) ≤ 0 simplifies to (3x - 2a) ≤ 0 because x is positive, so multiplying both sides by x (which is positive) doesn't change the inequality sign. Therefore, 3x - 2a ≤ 0.So, 3x - 2a ≤ 0. Let's solve for a. Adding 2a to both sides gives 3x ≤ 2a. Then, dividing both sides by 2, we get (3/2)x ≤ a. So, a ≥ (3/2)x.Now, this inequality must hold for all x in (0, 2). So, a must be greater than or equal to (3/2)x for all x in (0, 2). To find the range of a, we need to find the maximum value of (3/2)x in the interval (0, 2). Because if a is greater than or equal to the maximum value of (3/2)x, then it will automatically be greater than or equal to all smaller values of (3/2)x in that interval.What's the maximum value of (3/2)x on (0, 2)? Well, as x increases from 0 to 2, (3/2)x increases from 0 to 3. So, the maximum value is 3 when x approaches 2.Therefore, a must be greater than or equal to 3. So, a ≥ 3. That means the correct answer is option A.Wait, let me double-check. If a is exactly 3, then f'(x) = 3x² - 6x. Let's see if that's non-positive on (0, 2). Let's plug in x = 1: f'(1) = 3(1) - 6(1) = -3, which is negative. At x approaching 0, f'(x) approaches 0 from the negative side because 3x² is negligible compared to -6x. At x approaching 2, f'(2) = 3(4) - 6(2) = 12 - 12 = 0. So, at x=2, the derivative is zero, which is allowed since the function is allowed to have a horizontal tangent there.What if a is greater than 3, say a=4? Then f'(x) = 3x² - 8x. Let's test x=1: 3 - 8 = -5, which is negative. At x=2: 12 - 16 = -4, which is still negative. Wait, but earlier I thought that at x=2, with a=3, the derivative is zero. If a is greater than 3, then at x=2, f'(2) = 3*(4) - 2a*(2) = 12 - 4a. If a=4, that's 12 - 16 = -4, which is negative. So, actually, for a > 3, the derivative is negative at x=2 as well. So, the function is decreasing throughout (0, 2) even when a > 3.But if a is less than 3, say a=2, then f'(x) = 3x² - 4x. Let's test x=1: 3 - 4 = -1, which is negative. But what about x=2: 12 - 8 = 4, which is positive. So, at x=2, the derivative is positive, meaning the function is increasing there. That's not allowed because we need the function to be monotonically decreasing on the entire interval (0, 2). So, a cannot be less than 3.Wait, but hold on. If a=3, at x=2, the derivative is zero. So, is the function still considered monotonically decreasing if the derivative is zero at the endpoint? I think yes, because monotonic decreasing allows for the derivative to be zero at individual points, as long as it's not increasing anywhere in the interval.So, if a=3, the derivative is zero at x=2 and negative everywhere else in (0, 2). So, that's acceptable. If a is greater than 3, the derivative is negative throughout (0, 2), which is also acceptable. But if a is less than 3, the derivative becomes positive somewhere in (0, 2), which violates the condition of being monotonically decreasing.Therefore, the range of a is a ≥ 3, which is option A.I think that makes sense. Let me just recap:1. Find the derivative: f'(x) = 3x² - 2a x.2. For f(x) to be decreasing, f'(x) ≤ 0 on (0, 2).3. Factor the derivative: x(3x - 2a) ≤ 0.4. Since x > 0 in (0, 2), we have 3x - 2a ≤ 0.5. Solve for a: a ≥ (3/2)x.6. The maximum of (3/2)x on (0, 2) is 3, so a ≥ 3.Yep, that seems solid. So, the answer is A.**Final Answer**boxed{A}