💡Okay, so I have this inequality: (lg left(20 - 5x^2right) > lg (a - x) + 1). I need to find the range of the real number (a) such that this inequality has a unique integer solution of 1. Hmm, let me try to break this down step by step.First, I remember that logarithmic inequalities can sometimes be tricky, but I think I can manipulate this to make it easier. The inequality is (lg left(20 - 5x^2right) > lg (a - x) + 1). I know that (lg) is the logarithm base 10, so maybe I can use properties of logarithms to simplify this.I recall that (lg A > lg B + C) can be rewritten by converting the constant term into a logarithmic term. Since (1 = lg 10), I can rewrite the inequality as:[lg left(20 - 5x^2right) > lg (a - x) + lg 10]Because (lg A + lg B = lg (AB)), this becomes:[lg left(20 - 5x^2right) > lg left(10(a - x)right)]Now, since the logarithm function is increasing, if (lg X > lg Y), then (X > Y). So, applying this property, I get:[20 - 5x^2 > 10(a - x)]Let me simplify this inequality. First, I'll expand the right side:[20 - 5x^2 > 10a - 10x]Now, I'll bring all terms to one side to form a quadratic inequality:[-5x^2 + 10x + 20 - 10a > 0]Hmm, I can factor out a -5 to make it simpler:[-5(x^2 - 2x - 4 + 2a) > 0]Dividing both sides by -5 (and remembering to reverse the inequality sign because I'm dividing by a negative number):[x^2 - 2x - 4 + 2a < 0]So, the inequality simplifies to:[x^2 - 2x + (2a - 4) < 0]This is a quadratic in terms of (x). For this quadratic to be less than zero, the quadratic must open upwards (since the coefficient of (x^2) is positive) and have real roots such that the quadratic is negative between the roots.But wait, the problem states that the inequality should have a unique integer solution of 1. That means (x = 1) is the only integer that satisfies this inequality. So, I need to ensure that (x = 1) is the only integer in the interval where the quadratic is negative.Let me denote the quadratic as (f(x) = x^2 - 2x + (2a - 4)). For (f(x) < 0), the roots of the quadratic must be such that the interval between them includes only (x = 1) as an integer.First, let's find the roots of the quadratic equation (x^2 - 2x + (2a - 4) = 0). Using the quadratic formula:[x = frac{2 pm sqrt{4 - 4(1)(2a - 4)}}{2}]Simplifying the discriminant:[sqrt{4 - 8a + 16} = sqrt{20 - 8a}]So, the roots are:[x = frac{2 pm sqrt{20 - 8a}}{2} = 1 pm frac{sqrt{20 - 8a}}{2}]Let me denote the roots as (x_1 = 1 - frac{sqrt{20 - 8a}}{2}) and (x_2 = 1 + frac{sqrt{20 - 8a}}{2}).For the quadratic to have real roots, the discriminant must be non-negative:[20 - 8a geq 0 implies 8a leq 20 implies a leq frac{20}{8} = frac{5}{2}]So, (a) must be less than or equal to (frac{5}{2}).Now, since the quadratic opens upwards, (f(x) < 0) between its roots (x_1) and (x_2). We need this interval ((x_1, x_2)) to contain only the integer (x = 1). That means that the interval should not include any other integers besides 1.To ensure that, the interval ((x_1, x_2)) must lie entirely between 0 and 2, but not including 0 or 2. Wait, actually, since (x = 1) is the only integer solution, the interval should not include 0 or 2 because those are integers adjacent to 1. So, we need:[0 < x_1 < 1 < x_2 < 2]Wait, no, actually, if (x_1 < 1 < x_2), but we need the interval ((x_1, x_2)) to include 1 but not include 0 or 2. So, perhaps:[x_1 < 1 < x_2]But we also need to ensure that the interval does not include any other integers. So, the interval should be such that (x_1) is greater than 0 and less than 1, and (x_2) is greater than 1 but less than 2. That way, only (x = 1) is inside the interval.So, let's write down the conditions:1. (x_1 < 1)2. (x_2 > 1)3. (x_1 > 0)4. (x_2 < 2)Let me express these conditions in terms of (a).First, (x_1 = 1 - frac{sqrt{20 - 8a}}{2}). We need (x_1 > 0):[1 - frac{sqrt{20 - 8a}}{2} > 0 implies frac{sqrt{20 - 8a}}{2} < 1 implies sqrt{20 - 8a} < 2]Squaring both sides:[20 - 8a < 4 implies -8a < -16 implies a > 2]So, (a > 2).Next, (x_2 = 1 + frac{sqrt{20 - 8a}}{2}). We need (x_2 < 2):[1 + frac{sqrt{20 - 8a}}{2} < 2 implies frac{sqrt{20 - 8a}}{2} < 1 implies sqrt{20 - 8a} < 2]Which is the same condition as above, leading to (a > 2).Additionally, we need (x_1 < 1) and (x_2 > 1), but since (x_1 = 1 - frac{sqrt{20 - 8a}}{2}) and (x_2 = 1 + frac{sqrt{20 - 8a}}{2}), these are automatically satisfied as long as (sqrt{20 - 8a} > 0), which is true as long as (a < frac{5}{2}), which we already have from the discriminant condition.Wait, but earlier we found that (a leq frac{5}{2}), but from the condition (a > 2), so combining these, we have:[2 < a leq frac{5}{2}]But wait, the problem states that the inequality should have a unique integer solution of 1. So, if (a = frac{5}{2}), what happens?Let me check when (a = frac{5}{2}). Then, the discriminant becomes:[20 - 8a = 20 - 8 times frac{5}{2} = 20 - 20 = 0]So, the quadratic has a repeated root at (x = 1). Therefore, the inequality (f(x) < 0) becomes (f(x) < 0), but since the quadratic touches the x-axis at (x = 1), the inequality is not satisfied at (x = 1). So, actually, there is no solution when (a = frac{5}{2}), because the quadratic is zero at (x = 1) and positive elsewhere.Therefore, (a) must be strictly less than (frac{5}{2}).So, combining the conditions:[2 < a < frac{5}{2}]But wait, let me verify this. If (a = 2), then the discriminant is:[20 - 8a = 20 - 16 = 4]So, the roots are:[x = 1 pm frac{sqrt{4}}{2} = 1 pm 1]Thus, (x_1 = 0) and (x_2 = 2). So, the inequality (f(x) < 0) holds for (0 < x < 2). But in this case, the integer solutions are (x = 1), which is the only integer in that interval. So, when (a = 2), the inequality still has a unique integer solution of 1.Wait, so maybe (a) can be equal to 2. Let me check.When (a = 2), the inequality becomes:[lg(20 - 5x^2) > lg(2 - x) + 1]But we need to ensure that the arguments of the logarithms are positive.For (lg(20 - 5x^2)), we need (20 - 5x^2 > 0 implies x^2 < 4 implies -2 < x < 2).For (lg(2 - x)), we need (2 - x > 0 implies x < 2).So, combining these, the domain is (x < 2). But since (x) must be an integer, the possible integer values are (x = 1, 0, -1, -2, ldots). However, the inequality (f(x) < 0) when (a = 2) is (0 < x < 2), so the only integer solution is (x = 1). Therefore, (a = 2) is acceptable.But earlier, when I considered (a = 2), the quadratic inequality becomes (x^2 - 2x < 0), which factors to (x(x - 2) < 0), so the solution is (0 < x < 2). Thus, the integer solution is only (x = 1). So, (a = 2) is valid.Therefore, the range of (a) should include 2, but exclude (frac{5}{2}), because at (a = frac{5}{2}), there is no solution.So, putting it all together, the range of (a) is:[2 leq a < frac{5}{2}]Let me double-check this. If (a = 2), as we saw, the solution is (x = 1). If (a) is slightly less than (frac{5}{2}), say (a = 2.4), then the roots would be:[x = 1 pm frac{sqrt{20 - 8 times 2.4}}{2} = 1 pm frac{sqrt{20 - 19.2}}{2} = 1 pm frac{sqrt{0.8}}{2} approx 1 pm 0.447]So, (x_1 approx 0.553) and (x_2 approx 1.447). Thus, the interval is approximately (0.553 < x < 1.447), which includes only (x = 1) as an integer solution.If (a) is just above 2, say (a = 2.1), then:[x = 1 pm frac{sqrt{20 - 8 times 2.1}}{2} = 1 pm frac{sqrt{20 - 16.8}}{2} = 1 pm frac{sqrt{3.2}}{2} approx 1 pm 0.894]So, (x_1 approx 0.106) and (x_2 approx 1.894). Thus, the interval is approximately (0.106 < x < 1.894), which still includes only (x = 1) as an integer solution.If (a) is exactly 2, as we saw, the interval is (0 < x < 2), which includes (x = 1) as the only integer solution.If (a) is less than 2, say (a = 1.9), then:[x = 1 pm frac{sqrt{20 - 8 times 1.9}}{2} = 1 pm frac{sqrt{20 - 15.2}}{2} = 1 pm frac{sqrt{4.8}}{2} approx 1 pm 1.095]So, (x_1 approx -0.095) and (x_2 approx 2.095). Thus, the interval is approximately (-0.095 < x < 2.095). But since (x) must be an integer, the possible integer solutions are (x = 0, 1, 2). However, we need to check if these satisfy the original inequality.Wait, but the domain of the original inequality is (20 - 5x^2 > 0) and (a - x > 0). For (a = 1.9), (a - x > 0) implies (x < 1.9). So, the integer solutions must satisfy (x < 1.9), which are (x = 1, 0, -1, ldots). But the quadratic inequality (f(x) < 0) is (-0.095 < x < 2.095), so the integer solutions are (x = 0, 1). Thus, there are two integer solutions, which violates the condition of having a unique integer solution of 1. Therefore, (a) cannot be less than 2.Hence, the range of (a) is indeed (2 leq a < frac{5}{2}).

💡Okay, so I have this problem where I need to find the unit vector(s) collinear with vector AB, given points A(1, 3) and B(4, -1). Hmm, let me try to remember what collinear vectors are. I think they are vectors that lie on the same line, meaning they point in the same or exactly opposite directions. So, unit vectors collinear with AB would just be vectors in the same or opposite direction as AB but with a magnitude of 1.First, I need to find vector AB. To do that, I subtract the coordinates of point A from point B. So, the x-component is 4 - 1, which is 3, and the y-component is -1 - 3, which is -4. So, vector AB is (3, -4). Now, I need to find the magnitude of vector AB because to get a unit vector, I have to divide each component by the magnitude. The formula for the magnitude of a vector (x, y) is sqrt(x² + y²). Plugging in the values, that would be sqrt(3² + (-4)²) = sqrt(9 + 16) = sqrt(25) = 5. Okay, so the magnitude is 5.To get the unit vector in the same direction as AB, I divide each component of AB by its magnitude. So, that would be (3/5, -4/5). But since collinear vectors can also point in the opposite direction, the unit vector in the opposite direction would be (-3/5, 4/5). Looking at the options, option C is (3/5, -4/5) or (-3/5, 4/5), which matches what I just calculated. Option A only gives one of the unit vectors, and option B is different. Option D has different components, so it's not correct. Wait, let me double-check my calculations. Vector AB is definitely (3, -4). The magnitude is sqrt(9 + 16) = 5. Dividing each component by 5 gives (3/5, -4/5). The opposite direction would flip the signs, so (-3/5, 4/5). Yep, that seems right. I think I got it. So, the correct answer should be option C.

💡First, calculate the total number of windows in Lucas' house. Since each of the 3 floors has 3 windows, the total number of windows is 3 multiplied by 3, which equals 9 windows.Next, determine the penalty for taking 6 days to complete the job. The penalty is 1 for every 3 days. Therefore, for 6 days, the penalty is 6 divided by 3, which equals 2.Lucas' father will pay him 16 after the penalty is deducted. To find the original amount before the penalty, add the penalty back to the final payment: 16 plus 2 equals 18.Finally, to find out how much Lucas' father pays him per window, divide the original amount by the total number of windows: 18 divided by 9 windows equals 2 per window.

💡First, I need to determine the area of the square region where the point ( Q ) is selected. The square has vertices at ( (pm 3, pm 3) ), which means each side of the square spans from (-3) to (3) on both the x-axis and y-axis. Calculating the side length of the square:[text{Side length} = 3 - (-3) = 6 text{ units}]Now, finding the area of the square:[text{Area of the square} = text{side length}^2 = 6^2 = 36 text{ square units}]Next, I consider the circle centered at the origin with a radius of 2 units. The area of a circle is given by the formula:[text{Area of the circle} = pi r^2]Plugging in the radius ( r = 2 ):[text{Area of the circle} = pi (2)^2 = 4pi text{ square units}]To find the probability that point ( Q ) lies within two units of the origin, I take the ratio of the area of the circle to the area of the square:[text{Probability} = frac{text{Area of the circle}}{text{Area of the square}} = frac{4pi}{36} = frac{pi}{9}]Therefore, the probability that ( Q ) is within two units of the origin is ( frac{pi}{9} ).

💡Okay, so I have this problem here: I need to find the equation of the axis of symmetry for the graph of the function ( y = cos(2x + frac{pi}{3}) ). The options given are A: ( x = -frac{pi}{6} ), B: ( x = -frac{pi}{12} ), C: ( x = frac{pi}{6} ), and D: ( x = frac{pi}{12} ). Hmm, I remember that the axis of symmetry for a cosine function is related to its maximum or minimum points, right? First, let me recall the general form of a cosine function. It's usually written as ( y = cos(Bx + C) ). In this case, our function is ( y = cos(2x + frac{pi}{3}) ), so ( B = 2 ) and ( C = frac{pi}{3} ). I think the axis of symmetry occurs at the points where the cosine function reaches its maximum or minimum values. Wait, actually, the axis of symmetry for a cosine graph is a vertical line that passes through the maximum or minimum points. For the standard cosine function ( y = cos(x) ), the maximum occurs at ( x = 0 ), ( x = 2pi ), etc., and the minimum occurs at ( x = pi ), ( x = 3pi ), and so on. So, the axis of symmetry would be at these points. But in our case, the function is transformed. There's a horizontal compression and a horizontal shift. The coefficient ( B = 2 ) affects the period of the cosine function. The period of ( cos(Bx + C) ) is ( frac{2pi}{B} ), so here it would be ( frac{2pi}{2} = pi ). That means the function completes a full cycle every ( pi ) units instead of ( 2pi ). Now, the phase shift is given by ( -frac{C}{B} ). Plugging in the values, that's ( -frac{pi/3}{2} = -frac{pi}{6} ). So, the graph is shifted to the left by ( frac{pi}{6} ). Wait, so does that mean the maximum point, which was originally at ( x = 0 ) for ( cos(x) ), is now shifted to ( x = -frac{pi}{6} )? Because the phase shift is negative, it's moving to the left. So, the first maximum after shifting would be at ( x = -frac{pi}{6} ). If that's the case, then the axis of symmetry should be at ( x = -frac{pi}{6} ), which is option A. But let me double-check to make sure I'm not making a mistake here. Alternatively, maybe I should find the critical points by taking the derivative and setting it to zero. The derivative of ( y = cos(2x + frac{pi}{3}) ) is ( y' = -2sin(2x + frac{pi}{3}) ). Setting this equal to zero gives ( -2sin(2x + frac{pi}{3}) = 0 ), which simplifies to ( sin(2x + frac{pi}{3}) = 0 ). Solving for ( x ), we have ( 2x + frac{pi}{3} = kpi ) where ( k ) is any integer. Then, ( 2x = kpi - frac{pi}{3} ), so ( x = frac{kpi}{2} - frac{pi}{6} ). Let me plug in some integer values for ( k ) to find the critical points. For ( k = 0 ): ( x = 0 - frac{pi}{6} = -frac{pi}{6} ). For ( k = 1 ): ( x = frac{pi}{2} - frac{pi}{6} = frac{pi}{3} ). For ( k = 2 ): ( x = pi - frac{pi}{6} = frac{5pi}{6} ). And so on. So, the critical points are at ( x = -frac{pi}{6} + frac{kpi}{2} ). Now, these critical points correspond to the maxima and minima of the cosine function. Since the cosine function alternates between maxima and minima every half-period, which in this case is ( frac{pi}{2} ). So, the maximum points occur at ( x = -frac{pi}{6} + kpi ) and the minimum points at ( x = frac{pi}{3} + kpi ), or something like that. Wait, maybe I should think differently. Since the period is ( pi ), the distance between consecutive maxima is ( pi ). So, if one maximum is at ( x = -frac{pi}{6} ), the next one would be at ( x = -frac{pi}{6} + pi = frac{5pi}{6} ), and so on. Similarly, the minima would be halfway between the maxima, so at ( x = -frac{pi}{6} + frac{pi}{2} = frac{pi}{3} ), which matches what I found earlier. Therefore, the axis of symmetry should pass through these maximum points. So, the first maximum is at ( x = -frac{pi}{6} ), which is option A. But wait, let me think again. The axis of symmetry is a vertical line that divides the graph into two mirror images. For a cosine function, which is symmetric about its maxima and minima, the axis of symmetry should pass through these points. So, if the maximum is at ( x = -frac{pi}{6} ), then the axis of symmetry is ( x = -frac{pi}{6} ). That makes sense. Alternatively, if I consider the graph of ( y = cos(2x + frac{pi}{3}) ), it's a horizontally compressed and shifted cosine graph. The compression factor is 2, so it's squished horizontally, making the period ( pi ). The shift is ( -frac{pi}{6} ), so it's moved to the left by ( frac{pi}{6} ). Therefore, the first peak (maximum) is at ( x = -frac{pi}{6} ), and the graph is symmetric about this vertical line. So, the axis of symmetry is indeed ( x = -frac{pi}{6} ). I think that's solid. I don't see any mistakes in my reasoning. So, the answer should be option A.

💡First, I need to determine the time spent on each activity related to the cat.1. **Petting the cat** takes 12 minutes.2. **Combing the cat** is one-third of the petting time, which is 1/3 * 12 = 4 minutes.3. **Brushing the cat's teeth** is one-fourth of the combing time, so 1/4 * 4 = 1 minute.4. **Playing with the cat** is half of the petting time, which is 1/2 * 12 = 6 minutes.5. **Feeding the cat** takes 5 minutes.6. **Cleaning the food bowl** is two-fifths of the feeding time, so 2/5 * 5 = 2 minutes.Next, I'll add up all these times to find the total time spent with the cat.Total time = Petting + Combing + Brushing teeth + Playing + Feeding + Cleaning bowlTotal time = 12 minutes + 4 minutes + 1 minute + 6 minutes + 5 minutes + 2 minutesTotal time = 30 minutesTherefore, Marcy spends a total of 30 minutes with her cat.