💡First, I need to determine how many balloons each of Dante's 5 friends initially received. To do this, I'll divide the total number of balloons by the number of friends.Next, Dante asks each friend to give him 11 balloons. This means each friend will have to return 11 balloons to Dante. I'll subtract 11 from the initial number of balloons each friend had to find out how many balloons they have now.Finally, I'll present the final number of balloons each friend has after giving back 11 balloons to Dante.

💡Let's solve the problem step by step.1. **Define Variables:** - Let ( x ) be Clea's walking speed in units per second. - Let ( y ) be the total distance of the escalator in units. - Let ( k ) be the speed of the escalator in units per second.2. **Set Up Equations Based on Given Information:** - When the escalator is not moving, Clea takes 75 seconds to walk down: [ 75x = y ] - When the escalator is moving, Clea takes 30 seconds to walk down: [ 30(x + k) = y ]3. **Equate the Two Expressions for ( y ):** [ 75x = 30(x + k) ]4. **Solve for ( k ):** [ 75x = 30x + 30k ] [ 45x = 30k ] [ k = frac{45x}{30} = 1.5x ]5. **Calculate the Time to Ride the Escalator Without Walking:** - The time ( t ) is given by: [ t = frac{y}{k} ] - Substitute ( y = 75x ) and ( k = 1.5x ): [ t = frac{75x}{1.5x} = 50 text{ seconds} ]**Conclusion:**The time it takes for Clea to ride down the escalator without walking is (boxed{textbf{(B)} 50}) seconds.

💡Okay, so I need to find the distance between the foci of the hyperbola given by the equation (x^2 - 4x - 12y^2 + 24y = -36). Hmm, I remember that hyperbolas have standard forms, and to find the foci, I probably need to rewrite the given equation into one of those standard forms. First, I think I need to complete the square for both the x terms and the y terms. Completing the square is a method to convert a quadratic equation into a perfect square trinomial, which should help me get the equation into the standard hyperbola form. Let me start with the x terms: (x^2 - 4x). To complete the square, I take the coefficient of x, which is -4, divide it by 2 to get -2, and then square it to get 4. So, I can write (x^2 - 4x) as ((x - 2)^2 - 4). That makes sense because expanding ((x - 2)^2) gives (x^2 - 4x + 4), so subtracting 4 brings it back to the original expression.Now, moving on to the y terms: (-12y^2 + 24y). Hmm, I notice that there's a coefficient in front of (y^2), which is -12. I think I should factor that out first to make completing the square easier. So, factoring out -12, I get (-12(y^2 - 2y)). Now, inside the parentheses, I have (y^2 - 2y). Completing the square here: take the coefficient of y, which is -2, divide by 2 to get -1, square it to get 1. So, (y^2 - 2y) becomes ((y - 1)^2 - 1). Therefore, the entire y part is (-12[(y - 1)^2 - 1]), which simplifies to (-12(y - 1)^2 + 12). Alright, so now I can rewrite the original equation with these completed squares. Let me substitute them back in:Original equation: (x^2 - 4x - 12y^2 + 24y = -36)After completing the square:((x - 2)^2 - 4 - 12(y - 1)^2 + 12 = -36)Wait, let me double-check that. The x part is ((x - 2)^2 - 4) and the y part is (-12(y - 1)^2 + 12). So combining them, it should be:((x - 2)^2 - 4 - 12(y - 1)^2 + 12 = -36)Simplify the constants: -4 + 12 is 8. So:((x - 2)^2 - 12(y - 1)^2 + 8 = -36)Wait, that doesn't seem right. Let me check my substitution again. The original equation after substitution is:((x - 2)^2 - 4 - 12(y - 1)^2 + 12 = -36)So, combining the constants: -4 + 12 is indeed 8. So:((x - 2)^2 - 12(y - 1)^2 + 8 = -36)But 8 is on the left side, so I need to move it to the right side by subtracting 8:((x - 2)^2 - 12(y - 1)^2 = -36 - 8)Which is:((x - 2)^2 - 12(y - 1)^2 = -44)Hmm, wait a second. Hyperbola standard forms usually have positive terms on the left. Here, the right side is negative, which might mean I need to rearrange the equation. Let me see.If I move the terms around, I can write:(-12(y - 1)^2 + (x - 2)^2 = -44)But that still doesn't look right. Maybe I should factor out the negative sign from the y term to make it positive. Let me try that.Starting from:((x - 2)^2 - 12(y - 1)^2 = -44)Multiply both sides by -1 to make the right side positive:(- (x - 2)^2 + 12(y - 1)^2 = 44)Or, rearranged:(12(y - 1)^2 - (x - 2)^2 = 44)Now, this looks more like a hyperbola equation, but I think I need to write it in the standard form where the right side is 1. So, I'll divide every term by 44 to normalize it.Dividing both sides by 44:(frac{12(y - 1)^2}{44} - frac{(x - 2)^2}{44} = 1)Simplify the fractions:(frac{(y - 1)^2}{frac{44}{12}} - frac{(x - 2)^2}{44} = 1)Simplify (frac{44}{12}) to (frac{11}{3}):(frac{(y - 1)^2}{frac{11}{3}} - frac{(x - 2)^2}{44} = 1)Wait, so this is a hyperbola that opens up and down because the positive term is with the y. In the standard form of a hyperbola, it's (frac{(y - k)^2}{a^2} - frac{(x - h)^2}{b^2} = 1), where (h, k) is the center. So, in this case, the center is at (2, 1). For hyperbolas, the distance to the foci from the center is given by c, where (c^2 = a^2 + b^2). From the equation, (a^2 = frac{11}{3}) and (b^2 = 44). So, let's compute c.First, calculate (a^2 + b^2):(a^2 + b^2 = frac{11}{3} + 44)Convert 44 to thirds: 44 = (frac{132}{3})So, (a^2 + b^2 = frac{11}{3} + frac{132}{3} = frac{143}{3})Therefore, (c = sqrt{frac{143}{3}})Simplify that: (sqrt{frac{143}{3}} = frac{sqrt{143}}{sqrt{3}} = frac{sqrt{143} cdot sqrt{3}}{3} = frac{sqrt{429}}{3})Wait, but hold on. The standard form I used was for a hyperbola that opens up and down, so the foci are located at (h, k ± c). So, the distance between the two foci is 2c.So, 2c = 2 * (frac{sqrt{429}}{3}) = (frac{2sqrt{429}}{3})But wait, let me double-check my calculations because I'm not sure if I did everything correctly.Wait, in the equation, after dividing by 44, I had:(frac{(y - 1)^2}{frac{11}{3}} - frac{(x - 2)^2}{44} = 1)So, (a^2 = frac{11}{3}) and (b^2 = 44). Therefore, (c^2 = a^2 + b^2 = frac{11}{3} + 44). Let me compute that again:(frac{11}{3} + 44 = frac{11}{3} + frac{132}{3} = frac{143}{3}). So, (c = sqrt{frac{143}{3}}).Simplify (sqrt{frac{143}{3}}): it's equal to (frac{sqrt{143}}{sqrt{3}}). Rationalizing the denominator, multiply numerator and denominator by (sqrt{3}):(frac{sqrt{143} cdot sqrt{3}}{3} = frac{sqrt{429}}{3})So, c is (frac{sqrt{429}}{3}), and the distance between the foci is 2c, which is (frac{2sqrt{429}}{3}).But wait, in my initial steps, I think I might have made a mistake when completing the square. Let me go back and check.Original equation: (x^2 - 4x - 12y^2 + 24y = -36)Completing the square for x:(x^2 - 4x = (x - 2)^2 - 4). That seems correct.For y:(-12y^2 + 24y = -12(y^2 - 2y)). Then, completing the square inside the parentheses:(y^2 - 2y = (y - 1)^2 - 1). So, substituting back:(-12[(y - 1)^2 - 1] = -12(y - 1)^2 + 12). That also seems correct.So, substituting back into the equation:((x - 2)^2 - 4 - 12(y - 1)^2 + 12 = -36)Simplify constants: -4 + 12 = 8. So:((x - 2)^2 - 12(y - 1)^2 + 8 = -36)Subtract 8 from both sides:((x - 2)^2 - 12(y - 1)^2 = -44)Then, I multiplied both sides by -1 to get:(- (x - 2)^2 + 12(y - 1)^2 = 44)Which is the same as:(12(y - 1)^2 - (x - 2)^2 = 44)Then, dividing by 44:(frac{12(y - 1)^2}{44} - frac{(x - 2)^2}{44} = 1)Simplify fractions:(frac{(y - 1)^2}{frac{44}{12}} - frac{(x - 2)^2}{44} = 1)Simplify (frac{44}{12}) to (frac{11}{3}):(frac{(y - 1)^2}{frac{11}{3}} - frac{(x - 2)^2}{44} = 1)So, yes, that seems correct. Therefore, (a^2 = frac{11}{3}) and (b^2 = 44). So, (c^2 = a^2 + b^2 = frac{11}{3} + 44 = frac{143}{3}), so (c = sqrt{frac{143}{3}}), and the distance between the foci is (2c = frac{2sqrt{143}}{sqrt{3}} = frac{2sqrt{429}}{3}).Wait, but I just realized something. When I divided by 44, I should have divided each term by 44, including the constants. Let me check that again.Original equation after moving constants:((x - 2)^2 - 12(y - 1)^2 = -44)Then, to write it in standard form, I need to have 1 on the right side. So, I divide both sides by -44:(frac{(x - 2)^2}{-44} - frac{12(y - 1)^2}{-44} = 1)Which simplifies to:(-frac{(x - 2)^2}{44} + frac{(y - 1)^2}{frac{44}{12}} = 1)Which is the same as:(frac{(y - 1)^2}{frac{11}{3}} - frac{(x - 2)^2}{44} = 1)So, that part is correct. Therefore, my earlier conclusion stands.But wait, another thought: sometimes, when dealing with hyperbolas, it's easy to mix up a and b. In the standard form, the term under y is (a^2) because the hyperbola opens up and down. So, (a^2 = frac{11}{3}) and (b^2 = 44). Therefore, (c^2 = a^2 + b^2 = frac{11}{3} + 44). Let me compute that again:(frac{11}{3} + 44 = frac{11}{3} + frac{132}{3} = frac{143}{3}). So, (c = sqrt{frac{143}{3}}), which simplifies to (frac{sqrt{429}}{3}) because (sqrt{frac{143}{3}} = sqrt{frac{143 times 3}{3 times 3}} = sqrt{frac{429}{9}} = frac{sqrt{429}}{3}).Therefore, the distance between the foci is (2c = 2 times frac{sqrt{429}}{3} = frac{2sqrt{429}}{3}).Wait, but in the initial problem, the equation was (x^2 - 4x - 12y^2 + 24y = -36). When I completed the square, I ended up with a hyperbola that opens up and down. Is that correct?Let me think. The standard form is (frac{(y - k)^2}{a^2} - frac{(x - h)^2}{b^2} = 1), which indeed opens up and down. So, yes, that makes sense because the y term is positive.Alternatively, if the x term were positive, it would open left and right. So, in this case, since the y term is positive after rearrangement, it opens up and down.Therefore, the distance between the foci is (frac{2sqrt{429}}{3}).But wait, another thought: sometimes, when completing the square, signs can be tricky. Let me double-check my steps again.Original equation: (x^2 - 4x - 12y^2 + 24y = -36)Completing the square for x:(x^2 - 4x = (x - 2)^2 - 4). Correct.For y:(-12y^2 + 24y = -12(y^2 - 2y)). Then, (y^2 - 2y = (y - 1)^2 - 1). So, substituting back:(-12[(y - 1)^2 - 1] = -12(y - 1)^2 + 12). Correct.So, substituting back into the equation:((x - 2)^2 - 4 - 12(y - 1)^2 + 12 = -36)Simplify constants: -4 + 12 = 8. So:((x - 2)^2 - 12(y - 1)^2 + 8 = -36)Subtract 8 from both sides:((x - 2)^2 - 12(y - 1)^2 = -44)Multiply both sides by -1:(- (x - 2)^2 + 12(y - 1)^2 = 44)Which is:(12(y - 1)^2 - (x - 2)^2 = 44)Divide by 44:(frac{12(y - 1)^2}{44} - frac{(x - 2)^2}{44} = 1)Simplify:(frac{(y - 1)^2}{frac{11}{3}} - frac{(x - 2)^2}{44} = 1)Yes, that's correct. So, (a^2 = frac{11}{3}), (b^2 = 44), so (c^2 = a^2 + b^2 = frac{11}{3} + 44 = frac{143}{3}), so (c = sqrt{frac{143}{3}} = frac{sqrt{429}}{3}), and the distance between the foci is (2c = frac{2sqrt{429}}{3}).Wait, but I just noticed that in the initial problem, the equation was (x^2 - 4x - 12y^2 + 24y = -36). When I completed the square, I ended up with a hyperbola that opens up and down. Is that correct?Yes, because after rearrangement, the y term is positive, indicating it's a vertical hyperbola. So, the foci are along the vertical axis, which is the y-axis in this case, shifted to the center (2,1). Therefore, the distance between the foci is indeed (2c), which is (frac{2sqrt{429}}{3}).But let me check if I simplified (sqrt{frac{143}{3}}) correctly. (sqrt{frac{143}{3}} = frac{sqrt{143}}{sqrt{3}} = frac{sqrt{143} times sqrt{3}}{3} = frac{sqrt{429}}{3}). Yes, that's correct.Therefore, the distance between the foci is (frac{2sqrt{429}}{3}).Wait, but I just realized that 143 is 11*13, and 429 is 13*33, which is 13*3*11. So, 429 = 3*11*13. There are no square factors, so (sqrt{429}) cannot be simplified further. Therefore, the answer is (frac{2sqrt{429}}{3}).But hold on, in my initial calculation, I had:After completing the square:((x - 2)^2 - 12(y - 1)^2 = -44)Then, I multiplied both sides by -1 to get:(- (x - 2)^2 + 12(y - 1)^2 = 44)Which is the same as:(12(y - 1)^2 - (x - 2)^2 = 44)Then, dividing by 44:(frac{(y - 1)^2}{frac{44}{12}} - frac{(x - 2)^2}{44} = 1)Simplify (frac{44}{12}) to (frac{11}{3}):(frac{(y - 1)^2}{frac{11}{3}} - frac{(x - 2)^2}{44} = 1)So, yes, that's correct. Therefore, (a^2 = frac{11}{3}), (b^2 = 44), so (c^2 = a^2 + b^2 = frac{11}{3} + 44 = frac{143}{3}), so (c = sqrt{frac{143}{3}} = frac{sqrt{429}}{3}), and the distance between the foci is (2c = frac{2sqrt{429}}{3}).I think that's correct. Let me just recap:1. Completed the square for x and y.2. Rewrote the equation in standard hyperbola form.3. Identified (a^2) and (b^2).4. Calculated (c) using (c^2 = a^2 + b^2).5. Found the distance between the foci as (2c).Yes, that seems thorough. I don't see any mistakes in the steps. So, the distance between the foci is (frac{2sqrt{429}}{3}).**Final Answer**The distance between the foci is boxed{dfrac{2sqrt{429}}{3}}.

💡To determine the larger number, let's denote the larger number as ( L ) and the smaller number as ( S ).1. **Set Up the Equations:** - The difference between the two numbers is 2500: [ L - S = 2500 quad text{(1)} ] - When the larger number is divided by the smaller number, the quotient is 6 and the remainder is 15: [ L = 6S + 15 quad text{(2)} ]2. **Express ( S ) in Terms of ( L ):** From equation (1): [ S = L - 2500 quad text{(3)} ]3. **Substitute ( S ) into Equation (2):** [ L = 6(L - 2500) + 15 ] [ L = 6L - 15000 + 15 ] [ L = 6L - 14985 ]4. **Solve for ( L ):** [ L - 6L = -14985 ] [ -5L = -14985 ] [ L = frac{-14985}{-5} ] [ L = 2997 ]Therefore, the larger number is **2997**.

💡Okay, so I have this problem here about a function f(x) which is given as -2a sin(2x + π/6) + 2a + b. And I'm told that a is greater than 0, and that when x is between 0 and π/2, the function f(x) is between -5 and 1. I need to find the values of a and b. Then, there's a second part where I have to consider another function g(x) which is f(x + π/2), and given that the logarithm of [g(x)] is greater than 0, I need to find the monotonic intervals of g(x). Alright, let's start with the first part. I need to find a and b. The function is f(x) = -2a sin(2x + π/6) + 2a + b. Since a is positive, that might help later. The function is a sine function that's been scaled and shifted. I know that the sine function oscillates between -1 and 1, so maybe I can use that to find the maximum and minimum values of f(x).Given that x is in [0, π/2], let's see what the argument of the sine function, which is 2x + π/6, does. When x is 0, 2x + π/6 is π/6, and when x is π/2, 2x + π/6 is 2*(π/2) + π/6 = π + π/6 = 7π/6. So the argument of the sine function goes from π/6 to 7π/6 as x goes from 0 to π/2.Now, sin(θ) where θ is between π/6 and 7π/6. Let me recall the sine curve. At π/6, sin(π/6) is 1/2. Then it goes up to sin(π/2) = 1 at θ = π/2, then decreases to sin(π) = 0, and then goes negative, reaching sin(3π/2) = -1, but wait, 7π/6 is just π/6 past π, so sin(7π/6) is -1/2. So the sine function in this interval goes from 1/2 up to 1, down to -1/2.So sin(2x + π/6) ranges from -1/2 to 1 in this interval. Therefore, the expression -2a sin(2x + π/6) will have its maximum and minimum values when sin(2x + π/6) is at its minimum and maximum.Let me compute that. If sin(2x + π/6) is 1, then -2a*1 = -2a. If sin(2x + π/6) is -1/2, then -2a*(-1/2) = a. So the expression -2a sin(2x + π/6) ranges from -2a to a.Therefore, f(x) = -2a sin(2x + π/6) + 2a + b. So adding 2a + b to the expression above, the minimum value of f(x) is -2a + 2a + b = b, and the maximum value is a + 2a + b = 3a + b.We are told that f(x) is between -5 and 1. So the minimum value is -5 and the maximum is 1. Therefore, we have two equations:1. b = -52. 3a + b = 1Substituting b from the first equation into the second, we get 3a - 5 = 1. So 3a = 6, which gives a = 2.So, a is 2 and b is -5. That seems straightforward.Now, moving on to the second part. We have g(x) = f(x + π/2). So we need to substitute x + π/2 into f(x). Let's write out f(x):f(x) = -2a sin(2x + π/6) + 2a + bSo, f(x + π/2) = -2a sin(2(x + π/2) + π/6) + 2a + bLet me compute the argument of the sine function:2(x + π/2) + π/6 = 2x + π + π/6 = 2x + 7π/6So, g(x) = -2a sin(2x + 7π/6) + 2a + bBut we already found a and b. a is 2 and b is -5, so substituting those in:g(x) = -4 sin(2x + 7π/6) + 4 - 5 = -4 sin(2x + 7π/6) - 1Hmm, that seems a bit complicated. Maybe I can simplify the sine term. Let's recall that sin(θ + π) = -sinθ. So, sin(2x + 7π/6) = sin(2x + π + π/6) = -sin(2x + π/6). So, substituting that in:g(x) = -4*(-sin(2x + π/6)) - 1 = 4 sin(2x + π/6) - 1So, g(x) simplifies to 4 sin(2x + π/6) - 1.Now, the problem says log([g(x)]) > 0. Wait, [g(x)] probably denotes the floor function, but since it's inside a logarithm, which is only defined for positive arguments, maybe it's just g(x) > 1? Because log(something) > 0 implies that something > 1, since log(1) = 0 and log is increasing. So, if log(g(x)) > 0, then g(x) > 1.But wait, the problem says log([g(x)]) > 0. Maybe [g(x)] is the integer part of g(x). So, [g(x)] is the greatest integer less than or equal to g(x). So, [g(x)] > 0 would mean that g(x) is greater than 1, because [g(x)] is at least 1. So, yeah, g(x) must be greater than 1.So, 4 sin(2x + π/6) - 1 > 1. Let's solve this inequality.4 sin(2x + π/6) - 1 > 1Add 1 to both sides:4 sin(2x + π/6) > 2Divide both sides by 4:sin(2x + π/6) > 1/2So, we need to find all x such that sin(2x + π/6) > 1/2.Let me recall where sine is greater than 1/2. The sine function is greater than 1/2 in the intervals (π/6 + 2πk, 5π/6 + 2πk) for any integer k.So, sin(θ) > 1/2 when θ is in (π/6 + 2πk, 5π/6 + 2πk).Therefore, in our case, θ = 2x + π/6. So,π/6 + 2πk < 2x + π/6 < 5π/6 + 2πkSubtract π/6 from all parts:0 + 2πk < 2x < 4π/6 + 2πkSimplify 4π/6 to 2π/3:2πk < 2x < 2π/3 + 2πkDivide all parts by 2:πk < x < π/3 + πkSo, x is in (πk, π/3 + πk) for any integer k.But we need to consider the domain of x. Since the original function f(x) was defined for x in [0, π/2], but g(x) is f(x + π/2), so x + π/2 must be in [0, π/2], which would mean x is in [-π/2, 0]. But that seems contradictory because in the second part, we're probably considering x in some interval, but the problem doesn't specify. Wait, maybe I'm overcomplicating.Wait, actually, g(x) is defined as f(x + π/2), so x can be any real number such that x + π/2 is in the domain of f, which is [0, π/2]. So, x + π/2 ∈ [0, π/2] implies x ∈ [-π/2, 0]. So, x is in [-π/2, 0]. But in the inequality, we have x in (πk, π/3 + πk). So, we need to find k such that (πk, π/3 + πk) overlaps with [-π/2, 0].Let me consider k = 0: (0, π/3). But our x is in [-π/2, 0], so the overlap is empty.k = -1: (-π, -π + π/3) = (-π, -2π/3). The overlap with [-π/2, 0] is (-π/2, -2π/3). Wait, -2π/3 is approximately -2.094, and -π/2 is approximately -1.5708. So, the interval (-π, -2π/3) overlaps with [-π/2, 0] only at (-π/2, -2π/3). But -2π/3 is less than -π/2, so actually, the overlap is (-π/2, -2π/3), but since -2π/3 is less than -π/2, this interval is empty.Wait, maybe I made a mistake. Let me think again. For k = -1, the interval is (-π, -2π/3). Our x is in [-π/2, 0]. So, the overlap is between -π/2 and -2π/3. But since -π/2 is approximately -1.5708 and -2π/3 is approximately -2.094, the overlap is actually (-π/2, -2π/3), but since -2π/3 is less than -π/2, this interval is empty. So, no overlap.k = 1: (π, 4π/3). Doesn't overlap with [-π/2, 0].Wait, maybe I need to consider that x can be any real number, not necessarily restricted to [-π/2, 0]. Because g(x) is defined as f(x + π/2), but f is defined for all x, right? Wait, no, f was given with the condition that x ∈ [0, π/2]. So, f(x) is only defined for x in [0, π/2], so g(x) = f(x + π/2) is only defined when x + π/2 is in [0, π/2], which implies x ∈ [-π/2, 0]. So, x must be in [-π/2, 0].But in our inequality, we have x ∈ (πk, π/3 + πk). So, for x ∈ [-π/2, 0], let's find k such that (πk, π/3 + πk) overlaps with [-π/2, 0].Let me try k = -1: (-π, -2π/3). Overlap with [-π/2, 0] is (-π/2, -2π/3). But as before, since -2π/3 < -π/2, the overlap is empty.k = 0: (0, π/3). Overlap with [-π/2, 0] is empty.k = -2: (-2π, -5π/3). Overlap with [-π/2, 0] is (-π/2, -5π/3). But -5π/3 is approximately -5.23, which is less than -π/2, so overlap is empty.Wait, maybe I'm approaching this wrong. Perhaps the function g(x) is periodic, so even though it's defined for x ∈ [-π/2, 0], the inequality sin(2x + π/6) > 1/2 can be solved for all x, but we need to consider the domain where g(x) is defined, which is x ∈ [-π/2, 0].So, let's solve sin(2x + π/6) > 1/2 for x ∈ [-π/2, 0].Let me set θ = 2x + π/6. Then, θ = 2x + π/6. When x = -π/2, θ = 2*(-π/2) + π/6 = -π + π/6 = -5π/6. When x = 0, θ = 0 + π/6 = π/6.So, θ ranges from -5π/6 to π/6 as x goes from -π/2 to 0.We need to find θ in [-5π/6, π/6] such that sinθ > 1/2.Let me recall where sinθ > 1/2. In the interval [0, 2π], sinθ > 1/2 for θ ∈ (π/6, 5π/6). But since θ is in [-5π/6, π/6], which is a negative angle, we need to consider the sine function in negative angles.We know that sin(-θ) = -sinθ. So, sinθ > 1/2 in [-5π/6, π/6] would correspond to θ where sinθ is positive and greater than 1/2.But in [-5π/6, π/6], the sine function is positive only in (0, π/6). Because from -5π/6 to 0, sine is negative or zero.Wait, let me think again. Let's plot θ from -5π/6 to π/6.At θ = -5π/6, sinθ = -1/2.At θ = -π/2, sinθ = -1.At θ = -π/6, sinθ = -1/2.At θ = 0, sinθ = 0.At θ = π/6, sinθ = 1/2.So, in the interval [-5π/6, π/6], sinθ is greater than 1/2 only at θ = π/6, where it's exactly 1/2. But we need sinθ > 1/2, so actually, there's no θ in [-5π/6, π/6] where sinθ > 1/2. Because the maximum sinθ reaches in this interval is 1/2 at θ = π/6, and it's less than that elsewhere.Wait, that can't be right. Because when θ approaches π/6 from below, sinθ approaches 1/2 from below. So, in the interval [-5π/6, π/6], sinθ never exceeds 1/2. Therefore, the inequality sinθ > 1/2 has no solution in this interval.But that contradicts the problem statement, which says log([g(x)]) > 0, implying that g(x) > 1, which requires sinθ > 1/2. So, perhaps there's a mistake in my reasoning.Wait, let's double-check. We have g(x) = 4 sin(2x + π/6) - 1. We need g(x) > 1, so 4 sin(2x + π/6) - 1 > 1, which simplifies to sin(2x + π/6) > 1/2.But if x is in [-π/2, 0], then 2x + π/6 is in [-5π/6, π/6], as I calculated earlier. In this interval, sinθ is never greater than 1/2. So, does that mean there's no solution? But the problem says log([g(x)]) > 0, so there must be some x where this is true.Wait, maybe I made a mistake in defining the domain of g(x). Because f(x) is defined for x in [0, π/2], so g(x) = f(x + π/2) is defined for x such that x + π/2 ∈ [0, π/2], which implies x ∈ [-π/2, 0]. So, x is in [-π/2, 0].But in this interval, as we saw, sin(2x + π/6) never exceeds 1/2, so g(x) never exceeds 1. Therefore, g(x) > 1 is impossible. But the problem says log([g(x)]) > 0, which implies g(x) > 1. So, this seems contradictory.Wait, maybe I made a mistake in simplifying g(x). Let me go back.We had f(x) = -2a sin(2x + π/6) + 2a + b. With a = 2 and b = -5, f(x) becomes -4 sin(2x + π/6) + 4 - 5 = -4 sin(2x + π/6) - 1.Then, g(x) = f(x + π/2) = -4 sin(2(x + π/2) + π/6) - 1 = -4 sin(2x + π + π/6) - 1.Now, sin(2x + π + π/6) = sin(2x + 7π/6). But sin(θ + π) = -sinθ, so sin(2x + 7π/6) = sin(2x + π + π/6) = -sin(2x + π/6). Therefore, g(x) = -4*(-sin(2x + π/6)) - 1 = 4 sin(2x + π/6) - 1.Wait, that seems correct. So, g(x) = 4 sin(2x + π/6) - 1.But when x is in [-π/2, 0], 2x + π/6 is in [-5π/6, π/6], as before. So, sin(2x + π/6) in this interval is between -1/2 and 1/2. Therefore, 4 sin(2x + π/6) is between -2 and 2, so g(x) is between -3 and 1.Therefore, g(x) never exceeds 1, so g(x) > 1 is impossible. Therefore, the inequality log([g(x)]) > 0 has no solution. But the problem says "and log([g(x)]) > 0", so perhaps I'm misunderstanding the notation.Wait, maybe [g(x)] is not the floor function but the absolute value? Or maybe it's a typo and should be log(g(x)) > 0, which would imply g(x) > 1. But as we saw, g(x) can't be greater than 1 in the domain x ∈ [-π/2, 0]. So, perhaps the problem is intended to have g(x) defined for all x, not just x ∈ [-π/2, 0]. Maybe f(x) is defined for all x, but the condition -5 ≤ f(x) ≤ 1 is only given for x ∈ [0, π/2]. So, perhaps g(x) is defined for all x, and we need to consider the intervals where g(x) > 1.Wait, that makes more sense. Because if f(x) is defined for all x, then g(x) = f(x + π/2) is also defined for all x. So, the condition -5 ≤ f(x) ≤ 1 is only for x ∈ [0, π/2], but f(x) can take other values outside that interval. So, perhaps we need to consider g(x) for all x, not just x ∈ [-π/2, 0].So, let's proceed under that assumption. So, g(x) = 4 sin(2x + π/6) - 1, and we need to find where g(x) > 1, which is 4 sin(2x + π/6) - 1 > 1, so sin(2x + π/6) > 1/2.As before, sinθ > 1/2 when θ ∈ (π/6 + 2πk, 5π/6 + 2πk) for any integer k.So, 2x + π/6 ∈ (π/6 + 2πk, 5π/6 + 2πk)Subtract π/6:2x ∈ (0 + 2πk, 4π/6 + 2πk) = (2πk, 2π/3 + 2πk)Divide by 2:x ∈ (πk, π/3 + πk)So, x is in the intervals (πk, π/3 + πk) for any integer k.Now, we need to find the monotonic intervals of g(x) where g(x) > 1, which is x ∈ (πk, π/3 + πk).But g(x) is a sine function, so it's periodic with period π, since the argument is 2x + π/6, so period is π.To find the monotonic intervals, we need to look at the derivative of g(x).g(x) = 4 sin(2x + π/6) - 1g'(x) = 4 * 2 cos(2x + π/6) = 8 cos(2x + π/6)So, the derivative is 8 cos(2x + π/6). The function g(x) is increasing when g'(x) > 0, i.e., when cos(2x + π/6) > 0, and decreasing when cos(2x + π/6) < 0.So, let's find where cos(2x + π/6) > 0.cosθ > 0 when θ ∈ (-π/2 + 2πk, π/2 + 2πk) for any integer k.So, 2x + π/6 ∈ (-π/2 + 2πk, π/2 + 2πk)Subtract π/6:2x ∈ (-2π/3 + 2πk, π/3 + 2πk)Divide by 2:x ∈ (-π/3 + πk, π/6 + πk)So, g(x) is increasing on intervals (-π/3 + πk, π/6 + πk) and decreasing on intervals (π/6 + πk, 5π/6 + πk), since cosθ < 0 in (π/2 + 2πk, 3π/2 + 2πk), which translates to x ∈ (π/6 + πk, 5π/6 + πk).But we are only interested in the intervals where g(x) > 1, which are x ∈ (πk, π/3 + πk).So, within each interval (πk, π/3 + πk), we need to see where g(x) is increasing or decreasing.Let me consider k = 0: x ∈ (0, π/3)Within this interval, we need to see where g(x) is increasing or decreasing.From the derivative, g(x) is increasing when x ∈ (-π/3 + πk, π/6 + πk). For k = 0, that's (-π/3, π/6). But our interval is (0, π/3). So, the overlap is (0, π/6). Therefore, in (0, π/6), g(x) is increasing.Similarly, g(x) is decreasing when x ∈ (π/6 + πk, 5π/6 + πk). For k = 0, that's (π/6, 5π/6). Overlap with (0, π/3) is (π/6, π/3). So, in (π/6, π/3), g(x) is decreasing.Similarly, for k = 1: x ∈ (π, 4π/3). But since we're considering all k, we can generalize.So, in general, for each k, in the interval (πk, π/3 + πk), g(x) is increasing on (πk, π/6 + πk) and decreasing on (π/6 + πk, π/3 + πk).Therefore, the monotonic intervals of g(x) where g(x) > 1 are:- Increasing on (πk, π/6 + πk)- Decreasing on (π/6 + πk, π/3 + πk)for any integer k.So, putting it all together, the function g(x) is increasing on intervals (πk, π/6 + πk) and decreasing on intervals (π/6 + πk, π/3 + πk) for all integers k, where g(x) > 1.I think that's the solution.

💡Okay, so I have this problem here with two equations: 3a + 7b = 12 and 9a + 2b = 23. I need to find the value of a + b. Hmm, let's see. I remember from school that when you have two equations with two variables, you can solve them using either substitution or elimination. I think elimination might be easier here because the coefficients look like they can be manipulated to eliminate one of the variables.First, let me write down the equations again to make sure I have them right:1. 3a + 7b = 122. 9a + 2b = 23Alright, so if I want to eliminate one variable, I need to make sure that the coefficients of either a or b are the same in both equations. Looking at the coefficients of a, I see 3 and 9. If I multiply the first equation by 3, the coefficient of a will become 9, which matches the second equation. That seems like a good plan.So, multiplying the first equation by 3:3*(3a + 7b) = 3*12Which simplifies to:9a + 21b = 36Now, I have the modified first equation:9a + 21b = 36And the second equation remains:9a + 2b = 23Now, if I subtract the second equation from the modified first equation, the 9a terms will cancel out, and I can solve for b.So, subtracting:(9a + 21b) - (9a + 2b) = 36 - 23Let's do the subtraction step by step:9a - 9a = 021b - 2b = 19b36 - 23 = 13So, after subtracting, I get:19b = 13Now, to find the value of b, I'll divide both sides by 19:b = 13/19Okay, so b is 13 over 19. Now, I need to find a. I can substitute this value of b back into one of the original equations to solve for a. Let's use the first equation because the numbers seem smaller, which might make the calculations easier.So, plugging b = 13/19 into the first equation:3a + 7*(13/19) = 12Let me compute 7*(13/19) first:7*13 = 91So, 91/19Now, the equation becomes:3a + 91/19 = 12To isolate 3a, I'll subtract 91/19 from both sides:3a = 12 - 91/19Hmm, 12 is a whole number, and 91/19 is a fraction. To subtract them, I should express 12 as a fraction with denominator 19. 12 can be written as 12/1, and to get denominator 19, I multiply numerator and denominator by 19:12 = (12*19)/19 = 228/19So, now the equation is:3a = 228/19 - 91/19Subtracting the numerators:228 - 91 = 137So, 3a = 137/19To solve for a, divide both sides by 3:a = (137/19) / 3Dividing by 3 is the same as multiplying by 1/3:a = 137/(19*3) = 137/57Alright, so a is 137 over 57. Now, I need to find a + b. I have a = 137/57 and b = 13/19. To add them together, they need a common denominator. The denominators are 57 and 19. Since 57 is a multiple of 19 (because 19*3=57), I can convert 13/19 to have a denominator of 57.So, 13/19 = (13*3)/(19*3) = 39/57Now, a + b is:137/57 + 39/57Adding the numerators:137 + 39 = 176So, a + b = 176/57Hmm, let me check if this fraction can be simplified. 176 and 57. Let's see, 57 factors into 3*19. Does 176 divide by 3? 1+7+6=14, which is not divisible by 3. Does 176 divide by 19? Let's see, 19*9=171, 19*10=190, so no. So, 176/57 is already in its simplest form.Therefore, the value of a + b is 176/57.Wait, just to make sure I didn't make any mistakes in my calculations. Let me go through the steps again.Starting with the two equations:3a + 7b = 129a + 2b = 23I multiplied the first equation by 3 to get:9a + 21b = 36Subtracting the second equation:(9a + 21b) - (9a + 2b) = 36 - 23Which simplifies to:19b = 13So, b = 13/19Plugging back into the first equation:3a + 7*(13/19) = 123a + 91/19 = 123a = 12 - 91/19Convert 12 to 228/193a = 228/19 - 91/19 = 137/19a = 137/(19*3) = 137/57Then, a + b = 137/57 + 13/19 = 137/57 + 39/57 = 176/57Yes, that seems correct. So, I think I did everything right. The value of a + b is 176/57.