💡Okay, so I have this geometry problem here where a square and a regular octagon are coplanar and share a common side AD. I need to find the degree measure of the exterior angle BAC. Hmm, let me try to visualize this. First, I know that in a square, all sides are equal and all angles are right angles, which are 90 degrees each. On the other hand, a regular octagon has all sides equal and all interior angles equal. I remember that the formula for the interior angle of a regular polygon is ((n-2) times 180^circ / n), where (n) is the number of sides. So for an octagon, that would be ((8-2) times 180^circ / 8 = 135^circ). Now, since the square and the octagon share a common side AD, points A and D are vertices of both the square and the octagon. Point B is a vertex of the octagon, and it's touching side AB. So, I'm assuming that point B is adjacent to point A in the octagon. I need to find the measure of the exterior angle BAC. Let me think about what that means. An exterior angle is formed by one side of a polygon and the extension of an adjacent side. In this case, angle BAC is formed by side AB and the extension of side AC. Wait, but where is point C? Is point C another vertex? Maybe it's part of the square or the octagon? Wait, the problem mentions that the diagram is analogous to the given heptagon problem. In that problem, points A and D are on the common side, and point B is a vertex of the octagon touching side AB. So, perhaps point C is another vertex of the square or the octagon? Let me try to sketch this mentally. We have a square with vertices A, D, and two others. The octagon shares side AD with the square, so it must be attached along that side. Point B is a vertex of the octagon adjacent to A, so from A, the octagon goes to B, and then continues around. If I consider point A, it's a corner of both the square and the octagon. The square has a right angle at A, which is 90 degrees. The octagon has an interior angle at A of 135 degrees. So, if I look at the angle inside the octagon at A, it's 135 degrees, and the angle inside the square at A is 90 degrees. But angle BAC is an exterior angle. Exterior angles are typically supplementary to the interior angles. Wait, no, that's for regular polygons. But in this case, angle BAC is formed by the side AB of the octagon and the extension of another side. Let me think. If I extend side AC beyond point A, forming an angle with side AB, that's the exterior angle BAC. So, to find this exterior angle, I might need to consider the angles around point A. At point A, we have two angles: one from the square (90 degrees) and one from the octagon (135 degrees). But since they are sharing a common side AD, the total angles around point A should add up to 360 degrees. Wait, no. Actually, around a point, the sum of all angles is 360 degrees. So, if we have the square contributing 90 degrees and the octagon contributing 135 degrees, the remaining angle would be the exterior angle BAC. So, let me calculate that. If the square has 90 degrees and the octagon has 135 degrees at point A, then the remaining angle around point A would be 360 - 90 - 135 = 135 degrees. Therefore, the exterior angle BAC is 135 degrees. Wait, let me double-check. If the interior angle of the octagon is 135 degrees, then the exterior angle would typically be 180 - 135 = 45 degrees. But in this case, since the square is also attached, maybe it's different. Hmm, perhaps I need to consider the angle between the square and the octagon. The square has a right angle, and the octagon has an interior angle of 135 degrees. So, the angle between them would be the difference between 180 degrees and the sum of these two angles? Wait, no, that doesn't make sense. Let me think again. At point A, the square contributes a 90-degree angle, and the octagon contributes a 135-degree angle. These two angles are adjacent and form part of the 360 degrees around point A. So, the remaining angle, which is the exterior angle BAC, would be 360 - 90 - 135 = 135 degrees. Yes, that seems correct. So, the exterior angle BAC is 135 degrees.

💡Alright, so I've got this problem about a function ( f(x) ) defined on all real numbers. It has some properties: for any ( x, y in mathbb{R} ), it satisfies ( f(x) + f(y) = f(x + y) + 1 ), and when ( x > 0 ), ( f(x) > 1 ). There are three parts to this problem, and I need to tackle them one by one.Starting with part (I): Let ( h(x) = f(x) - 1 ). I need to prove that ( h(x) ) is an odd function. Hmm, okay. An odd function satisfies ( h(-x) = -h(x) ) for all ( x ). So, I need to show that ( h(-x) = -h(x) ).First, let's see what we know. The functional equation is ( f(x) + f(y) = f(x + y) + 1 ). Maybe I can plug in specific values for ( x ) and ( y ) to find more information about ( f ). Let's try ( x = 0 ) and ( y = 0 ). Then, the equation becomes ( f(0) + f(0) = f(0 + 0) + 1 ), which simplifies to ( 2f(0) = f(0) + 1 ). Subtracting ( f(0) ) from both sides gives ( f(0) = 1 ). Okay, so ( f(0) = 1 ).Now, let's try ( y = -x ). Plugging into the functional equation: ( f(x) + f(-x) = f(x + (-x)) + 1 ), which simplifies to ( f(x) + f(-x) = f(0) + 1 ). Since we know ( f(0) = 1 ), this becomes ( f(x) + f(-x) = 2 ). Subtracting 1 from both sides gives ( f(x) - 1 + f(-x) - 1 = 0 ), which is ( h(x) + h(-x) = 0 ). Therefore, ( h(-x) = -h(x) ), which means ( h(x) ) is indeed an odd function. That wasn't too bad.Moving on to part (II): Prove that ( f(x) ) is increasing on ( mathbb{R} ). So, I need to show that if ( x_1 > x_2 ), then ( f(x_1) > f(x_2) ).Let me take two arbitrary real numbers ( x_1 ) and ( x_2 ) such that ( x_1 > x_2 ). I need to compare ( f(x_1) ) and ( f(x_2) ). Maybe I can express ( f(x_1) - f(x_2) ) in terms of the functional equation.Let's set ( x = x_1 ) and ( y = -x_2 ). Then, the functional equation becomes ( f(x_1) + f(-x_2) = f(x_1 - x_2) + 1 ). Rearranging, we get ( f(x_1) - f(x_2) = f(x_1 - x_2) - 1 ).Since ( x_1 > x_2 ), ( x_1 - x_2 > 0 ). From the given condition, when ( x > 0 ), ( f(x) > 1 ). Therefore, ( f(x_1 - x_2) > 1 ), which implies ( f(x_1 - x_2) - 1 > 0 ). Thus, ( f(x_1) - f(x_2) > 0 ), meaning ( f(x_1) > f(x_2) ). So, ( f(x) ) is increasing on ( mathbb{R} ).Alright, that makes sense. Now, part (III): Solve the inequality ( f(x^2) - f(3tx) + f(2t^2 + 2t - x) < 1 ) for ( x ), where ( t in mathbb{R} ).This looks a bit more complicated. Let me try to manipulate the inequality using the functional equation. The given inequality is:( f(x^2) - f(3tx) + f(2t^2 + 2t - x) < 1 )Let me rearrange this:( f(x^2) + f(2t^2 + 2t - x) < f(3tx) + 1 )Hmm, the functional equation is ( f(a) + f(b) = f(a + b) + 1 ). So, if I set ( a = x^2 ) and ( b = 2t^2 + 2t - x ), then:( f(x^2) + f(2t^2 + 2t - x) = f(x^2 + 2t^2 + 2t - x) + 1 )So, substituting back into the inequality:( f(x^2 + 2t^2 + 2t - x) + 1 < f(3tx) + 1 )Subtracting 1 from both sides:( f(x^2 + 2t^2 + 2t - x) < f(3tx) )Since ( f ) is increasing, as we proved in part (II), this inequality implies:( x^2 + 2t^2 + 2t - x < 3tx )Let me rewrite this:( x^2 + 2t^2 + 2t - x - 3tx < 0 )Combine like terms:( x^2 - (3t + 1)x + 2t^2 + 2t < 0 )This is a quadratic inequality in terms of ( x ). Let me write it as:( x^2 - (3t + 1)x + 2t^2 + 2t < 0 )To solve this inequality, I can factor the quadratic expression. Let me try to factor it:Looking for two numbers that multiply to ( 2t^2 + 2t ) and add up to ( -(3t + 1) ).Wait, let me see:( x^2 - (3t + 1)x + 2t^2 + 2t )Let me try to factor:Looking for factors of the form ( (x - a)(x - b) ), where ( a + b = 3t + 1 ) and ( ab = 2t^2 + 2t ).Let me solve for ( a ) and ( b ):We have:( a + b = 3t + 1 )( ab = 2t^2 + 2t )Let me try to factor ( 2t^2 + 2t ). It factors as ( 2t(t + 1) ). Hmm, so maybe ( a = 2t ) and ( b = t + 1 ). Let's check:( a + b = 2t + t + 1 = 3t + 1 ), which matches.So, the quadratic factors as:( (x - 2t)(x - (t + 1)) < 0 )So, the inequality becomes:( (x - 2t)(x - t - 1) < 0 )Now, to solve this inequality, I need to find the intervals where the product is negative. The critical points are ( x = 2t ) and ( x = t + 1 ). The sign of the expression depends on the order of these roots.Case 1: ( 2t < t + 1 )Solving ( 2t < t + 1 ) gives ( t < 1 ).In this case, the critical points are ordered as ( 2t < t + 1 ). The quadratic opens upwards (since the coefficient of ( x^2 ) is positive), so the expression is negative between the roots.Thus, the solution is ( 2t < x < t + 1 ).Case 2: ( 2t > t + 1 )Solving ( 2t > t + 1 ) gives ( t > 1 ).Here, the critical points are ordered as ( t + 1 < 2t ). Again, the quadratic opens upwards, so the expression is negative between the roots.Thus, the solution is ( t + 1 < x < 2t ).Case 3: ( 2t = t + 1 )This happens when ( t = 1 ). In this case, both roots coincide at ( x = 2 ). The quadratic becomes ( (x - 2)^2 < 0 ), which has no solution since a square is always non-negative.Therefore, summarizing:- If ( t < 1 ), the solution is ( 2t < x < t + 1 ).- If ( t > 1 ), the solution is ( t + 1 < x < 2t ).- If ( t = 1 ), there is no solution.So, that's the solution to the inequality.**Final Answer**(I) boxed{h(x) text{ is an odd function}}(II) boxed{f(x) text{ is increasing on } mathbb{R}}(III) The solution is (boxed{{x mid 2t < x < t + 1}}) when ( t < 1 ), (boxed{{x mid t + 1 < x < 2t}}) when ( t > 1 ), and no solution when ( t = 1 ).

💡Okay, so I have this problem about a factory's output over three years. Let me try to understand it step by step. The problem says that the output in the third year increased by 20% compared to the first year. They mention that the average annual growth rate is the same, denoted as x. I need to figure out whether x is 10%, less than 10%, more than 10%, or if it depends on the first year's output.Alright, let's break it down. First, let's denote the output in the first year as, say, O. Then, if the growth rate is x each year, the output in the second year would be O multiplied by (1 + x). Similarly, the output in the third year would be the second year's output multiplied by (1 + x) again. So, the third year's output would be O*(1 + x)^2.The problem states that the third year's output is 20% higher than the first year's. So, that means the third year's output is O + 20% of O, which is 1.2*O. So, putting that together, we have:O*(1 + x)^2 = 1.2*OHmm, okay, so if I divide both sides by O, assuming O is not zero, I get:(1 + x)^2 = 1.2Now, to solve for x, I need to take the square root of both sides. So, taking the square root of 1.2. Let me calculate that. The square root of 1.2 is approximately 1.0954. So, 1 + x ≈ 1.0954, which means x ≈ 0.0954, or 9.54%.So, x is approximately 9.54%, which is less than 10%. Therefore, the average annual growth rate is less than 10%.Wait, but let me double-check my steps to make sure I didn't make a mistake. Starting from the beginning: output in year 1 is O, year 2 is O*(1 + x), year 3 is O*(1 + x)^2. The output in year 3 is 20% more than year 1, so O*(1 + x)^2 = 1.2*O. Dividing both sides by O gives (1 + x)^2 = 1.2. Taking the square root, 1 + x = sqrt(1.2) ≈ 1.0954, so x ≈ 0.0954 or 9.54%. Yep, that seems correct.So, since 9.54% is less than 10%, the correct conclusion is that x is less than 10%. Therefore, the answer should be option B: x < 10%.I don't think the magnitude of x depends on the output of the first year because we assumed O is the output in the first year, and it cancels out in the equation. So, whether O is large or small doesn't affect the growth rate x. It only depends on the percentage increase from year 1 to year 3.Let me think if there's another way to interpret the problem. Maybe if the growth rate is compounded annually, but I think that's exactly what I did. So, if the growth rate is x each year, compounded over two years, it leads to a 20% increase over two years. So, the calculation seems right.Alternatively, if someone thought that the growth rate is 10% each year, then after two years, the output would be O*(1.1)^2 = 1.21*O, which is a 21% increase, which is more than 20%. So, to get exactly a 20% increase over two years, the growth rate needs to be slightly less than 10%, which is consistent with our calculation of approximately 9.54%.Therefore, I feel confident that the correct answer is B: x < 10%.

💡Okay, so I have this function f(x) that's defined for all real numbers, and it's an odd function. That means f(-x) = -f(x) for any x. Got that. It also satisfies the condition f(x) = f(1 - x) for any real number x. Hmm, interesting. And we're told that f(x) = 2x² when x is in the interval (0, 1/2]. So, I need to find f(3) + f(-5/2). Alright, let me break this down. Since f is odd, I know that f(-x) = -f(x). So, if I can find f(5/2), then f(-5/2) would just be -f(5/2). That might be helpful later.Now, the function also satisfies f(x) = f(1 - x). So, for any x, the value of the function at x is the same as its value at 1 - x. That seems like a symmetry around x = 1/2. Maybe this can help me find values of f(x) outside the given interval (0, 1/2].Let me see. If I plug in x = 3 into the equation f(x) = f(1 - x), I get f(3) = f(1 - 3) = f(-2). But since f is odd, f(-2) = -f(2). So, f(3) = -f(2). Hmm, but I don't know what f(2) is yet. Maybe I can find f(2) using the same property.Let's try f(2) = f(1 - 2) = f(-1). Again, since f is odd, f(-1) = -f(1). So, f(2) = -f(1). Therefore, f(3) = -f(2) = -(-f(1)) = f(1). So, f(3) is equal to f(1). But what is f(1)? Let's see. Since 1 is not in the interval (0, 1/2], I can't directly use the given formula f(x) = 2x². Maybe I can use the property f(x) = f(1 - x) again. Let's set x = 1, so f(1) = f(1 - 1) = f(0). But f(0) for an odd function must be 0 because f(-0) = -f(0) implies f(0) = -f(0), which means f(0) = 0. So, f(1) = 0, which means f(3) = 0.Alright, so f(3) is 0. Now, let's find f(-5/2). Since f is odd, f(-5/2) = -f(5/2). So, I need to find f(5/2). Again, using the property f(x) = f(1 - x), let's plug in x = 5/2. So, f(5/2) = f(1 - 5/2) = f(-3/2). But f(-3/2) = -f(3/2) because f is odd. So, f(5/2) = -f(3/2).Now, let's find f(3/2). Using f(x) = f(1 - x), set x = 3/2, so f(3/2) = f(1 - 3/2) = f(-1/2). Since f is odd, f(-1/2) = -f(1/2). So, f(3/2) = -f(1/2). But f(1/2) is given because 1/2 is in the interval (0, 1/2]. Wait, actually, 1/2 is the endpoint. The interval is (0, 1/2], so does 1/2 include? Yes, it does. So, f(1/2) = 2*(1/2)² = 2*(1/4) = 1/2. Therefore, f(3/2) = -f(1/2) = -1/2. So, going back, f(5/2) = -f(3/2) = -(-1/2) = 1/2. But wait, f(5/2) = 1/2, so f(-5/2) = -f(5/2) = -1/2. So, putting it all together, f(3) + f(-5/2) = 0 + (-1/2) = -1/2.Wait, let me double-check that. Starting from f(5/2):f(5/2) = f(1 - 5/2) = f(-3/2) = -f(3/2). f(3/2) = f(1 - 3/2) = f(-1/2) = -f(1/2) = -1/2. So, f(5/2) = -(-1/2) = 1/2. Therefore, f(-5/2) = -1/2. And f(3) = f(1) = 0. So, adding them together, 0 + (-1/2) = -1/2. I think that's correct. So, the final answer should be -1/2.

💡Okay, so I've got this problem here about three trucks carrying sand to a construction yard. Each truck starts with a certain amount of sand and loses some along the way. I need to figure out how much sand they have in total when they arrive. Hmm, let's break this down step by step.First, let's look at the first truck. It carries 4.1 pounds of sand and loses 2.4 pounds. So, to find out how much sand it has left when it arrives, I need to subtract the lost sand from the initial amount. That would be 4.1 minus 2.4. Let me do that calculation: 4.1 - 2.4 equals... hmm, 4.1 minus 2 is 2.1, and then minus another 0.4 would be 1.7. So, the first truck has 1.7 pounds of sand left.Next, the second truck carries 5.7 pounds and loses 3.6 pounds. Again, I'll subtract the lost sand from the initial amount. So, 5.7 minus 3.6. Let's see, 5 minus 3 is 2, and 0.7 minus 0.6 is 0.1, so altogether that's 2.1 pounds. So, the second truck has 2.1 pounds of sand left.Now, the third truck carries 8.2 pounds and loses 1.9 pounds. Subtracting the lost sand from the initial amount gives me 8.2 minus 1.9. Let's calculate that: 8 minus 1 is 7, and 0.2 minus 0.9... wait, that's a bit tricky because 0.2 is less than 0.9. Maybe I should think of it as 8.2 minus 2 plus 0.1, which would be 6.2 plus 0.1 equals 6.3. So, the third truck has 6.3 pounds of sand left.Okay, now I have the remaining sand for each truck: 1.7 pounds, 2.1 pounds, and 6.3 pounds. To find the total amount of sand when they arrive, I need to add these three amounts together. Let's add them up: 1.7 plus 2.1 is 3.8, and then adding 6.3 to that gives me 10.1 pounds.Wait, let me double-check to make sure I didn't make any mistakes. For the first truck, 4.1 minus 2.4 is indeed 1.7. For the second truck, 5.7 minus 3.6 is 2.1. And for the third truck, 8.2 minus 1.9 is 6.3. Adding those together: 1.7 plus 2.1 is 3.8, and 3.8 plus 6.3 is 10.1. Yep, that seems right.So, the total amount of sand the three trucks have when they arrive at the construction yard is 10.1 pounds.

💡Okay, so I have this problem where I'm given two vectors, a and b. Vector a is (3,2) and vector b is (2,-1). The question says that if I take lambda times vector a and add vector b, that resulting vector should be parallel to vector a plus lambda times vector b. I need to find the value of lambda that makes this true.Alright, let me start by writing down what lambda*a + b and a + lambda*b actually are. So, if I multiply vector a by lambda, that would be (3*lambda, 2*lambda). Then, adding vector b to that would give me (3*lambda + 2, 2*lambda - 1). Similarly, vector a is (3,2), and lambda times vector b is (2*lambda, -lambda). Adding those together gives me (3 + 2*lambda, 2 - lambda).Now, the key thing here is that these two resulting vectors need to be parallel. I remember that for two vectors to be parallel, one must be a scalar multiple of the other. That means there exists some scalar k such that (3*lambda + 2, 2*lambda - 1) = k*(3 + 2*lambda, 2 - lambda). So, breaking this down into components, I can set up two equations:1. 3*lambda + 2 = k*(3 + 2*lambda)2. 2*lambda - 1 = k*(2 - lambda)Now, I have two equations with two unknowns: lambda and k. I need to solve for lambda. Since both equations equal k times something, I can set the ratios equal to each other. That is, (3*lambda + 2)/(3 + 2*lambda) should equal (2*lambda - 1)/(2 - lambda). So, writing that out:(3*lambda + 2)/(3 + 2*lambda) = (2*lambda - 1)/(2 - lambda)To solve this equation, I'll cross-multiply to eliminate the denominators:(3*lambda + 2)*(2 - lambda) = (2*lambda - 1)*(3 + 2*lambda)Now, I'll expand both sides:Left side: (3*lambda)*(2) + (3*lambda)*(-lambda) + 2*(2) + 2*(-lambda)= 6*lambda - 3*lambda^2 + 4 - 2*lambdaRight side: (2*lambda)*(3) + (2*lambda)*(2*lambda) + (-1)*(3) + (-1)*(2*lambda)= 6*lambda + 4*lambda^2 - 3 - 2*lambdaSimplify both sides:Left side: (6*lambda - 2*lambda) + (-3*lambda^2) + 4= 4*lambda - 3*lambda^2 + 4Right side: (6*lambda - 2*lambda) + 4*lambda^2 - 3= 4*lambda + 4*lambda^2 - 3Now, set the left side equal to the right side:4*lambda - 3*lambda^2 + 4 = 4*lambda + 4*lambda^2 - 3Let me subtract 4*lambda from both sides to eliminate that term:-3*lambda^2 + 4 = 4*lambda^2 - 3Now, I'll bring all terms to one side to set the equation to zero:-3*lambda^2 - 4*lambda^2 + 4 + 3 = 0Combine like terms:-7*lambda^2 + 7 = 0Now, I can factor out a -7:-7*(lambda^2 - 1) = 0Divide both sides by -7:lambda^2 - 1 = 0This simplifies to:lambda^2 = 1Taking the square root of both sides gives:lambda = ±1So, lambda can be either 1 or -1.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the cross-multiplication:(3*lambda + 2)(2 - lambda) = (2*lambda - 1)(3 + 2*lambda)Expanding:Left side: 6*lambda - 3*lambda^2 + 4 - 2*lambda = 4*lambda - 3*lambda^2 + 4Right side: 6*lambda + 4*lambda^2 - 3 - 2*lambda = 4*lambda + 4*lambda^2 - 3Setting them equal:4*lambda - 3*lambda^2 + 4 = 4*lambda + 4*lambda^2 - 3Subtract 4*lambda:-3*lambda^2 + 4 = 4*lambda^2 - 3Bring all terms to left:-3*lambda^2 -4*lambda^2 +4 +3 = 0-7*lambda^2 +7 = 0Divide by -7:lambda^2 -1 =0So, lambda^2=1, lambda=±1.Yes, that seems correct. So, lambda can be 1 or -1.But wait, let me plug these back into the original vectors to make sure they are indeed parallel.First, lambda=1:lambda*a + b = 1*(3,2) + (2,-1) = (3+2, 2-1)=(5,1)a + lambda*b = (3,2) +1*(2,-1)=(3+2,2-1)=(5,1)So, both vectors are (5,1), which are obviously parallel.Now, lambda=-1:lambda*a + b = -1*(3,2) + (2,-1)= (-3+2, -2-1)=(-1,-3)a + lambda*b = (3,2) + (-1)*(2,-1)=(3-2,2+1)=(1,3)Now, (-1,-3) and (1,3). Are these parallel? Let's see if one is a scalar multiple of the other.If I multiply (1,3) by -1, I get (-1,-3), which is exactly the other vector. So yes, they are parallel.Therefore, both lambda=1 and lambda=-1 satisfy the condition.Wait, but in the initial problem statement, it just says "lambda equals to ______." It doesn't specify if there are multiple solutions. So, I should probably list both.But let me check if there's any restriction on lambda. The problem doesn't specify, so both solutions are valid.So, the values of lambda are 1 and -1.**Final Answer**The values of lambda are boxed{1} and boxed{-1}.