💡To determine the smallest quotient obtained by dividing the numbers on two cards drawn from the set ({-5, -4, 0, +4, +6}), we need to consider all possible pairs and their resulting quotients.First, we note that division by zero is undefined, so we exclude any pair involving the number (0). This leaves us with the numbers (-5), (-4), (+4), and (+6).Next, we calculate the quotients for all possible pairs:1. (-5 div -4 = frac{5}{4} = 1.25)2. (-5 div +4 = -frac{5}{4} = -1.25)3. (-5 div +6 = -frac{5}{6} approx -0.833)4. (-4 div -5 = frac{4}{5} = 0.8)5. (-4 div +4 = -1)6. (-4 div +6 = -frac{2}{3} approx -0.666)7. (+4 div -5 = -frac{4}{5} = -0.8)8. (+4 div -4 = -1)9. (+4 div +6 = frac{2}{3} approx 0.666)10. (+6 div -5 = -frac{6}{5} = -1.2)11. (+6 div -4 = -frac{3}{2} = -1.5)12. (+6 div +4 = frac{3}{2} = 1.5)Among these quotients, the smallest value is (-1.5), which corresponds to (+6 div -4).Therefore, the smallest quotient obtained by dividing the numbers on these two cards is (boxed{-dfrac{3}{2}}).

💡Okay, so I have this problem here about translating a sine function and finding one of its axes of symmetry. Let me try to figure it out step by step. First, the original function is f(x) = sin(2x - π/6). I know that the general form of a sine function is sin(Bx + C), and it can be rewritten as sin(B(x + C/B)). This helps in understanding the phase shift. In this case, B is 2 and C is -π/6, so the phase shift is -C/B, which would be (π/6)/2 = π/12. That means the graph is shifted π/12 units to the right compared to the basic sin(2x) function.Now, the problem says that this graph is translated to the right by an additional π/12 units. So, I need to figure out the new function after this translation. Translating a function to the right by 'a' units involves replacing x with (x - a) in the function. So, applying this to f(x), the new function g(x) should be sin(2(x - π/12) - π/6). Let me expand that:g(x) = sin(2x - 2*(π/12) - π/6) Simplify the terms inside the sine: 2*(π/12) is π/6, so it becomes sin(2x - π/6 - π/6) Combine the constants: -π/6 - π/6 = -π/3 So, g(x) = sin(2x - π/3)Alright, so now the function is sin(2x - π/3). I need to find one of the axes of symmetry for this graph. I remember that for sine functions, the axes of symmetry are the vertical lines passing through the maximum and minimum points. These occur where the function reaches its peak or trough.The general form for the maximum and minimum points of sin(Bx + C) is when Bx + C = π/2 + kπ, where k is any integer. So, setting 2x - π/3 equal to π/2 + kπ:2x - π/3 = π/2 + kπ Let me solve for x:2x = π/2 + kπ + π/3 First, combine the constants π/2 and π/3. To add them, I need a common denominator, which is 6:π/2 = 3π/6 π/3 = 2π/6 So, 3π/6 + 2π/6 = 5π/6 Therefore, 2x = 5π/6 + kπ Divide both sides by 2:x = (5π/6 + kπ)/2 Simplify:x = 5π/12 + kπ/2So, the axes of symmetry are at x = 5π/12 + kπ/2 for any integer k. Let me plug in some values for k to see which one matches the options given.If k = 0: x = 5π/12 If k = 1: x = 5π/12 + π/2 = 5π/12 + 6π/12 = 11π/12 If k = -1: x = 5π/12 - π/2 = 5π/12 - 6π/12 = -π/12Looking at the options provided: A is π/4, B is 3π/8, C is 5π/12, D is 7π/24. So, when k=0, x=5π/12 is one of the axes of symmetry, which is option C.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the original function, translating it right by π/12, so replacing x with x - π/12. Then, expanding that correctly, I got sin(2x - π/3). Then, finding where the function reaches maximum or minimum by setting 2x - π/3 = π/2 + kπ. Solving for x, I got x = 5π/12 + kπ/2. Plugging in k=0 gives 5π/12, which is option C. That seems right.I think I might have confused myself initially with the phase shifts, but breaking it down step by step helped clarify. So, I'm pretty confident that the correct answer is C: x = 5π/12.**Final Answer**boxed{C}

💡Okay, so I have this triangle ABC, and I know that the sides opposite angles A, B, and C are a, b, and c respectively. They've given me that c is 2 and angle C is 60 degrees. There are two parts to this problem.Starting with part (1): I need to find the value of (a + b)/(sin A + sin B). Hmm, okay. I remember something about the Law of Sines, which relates the sides of a triangle to the sines of their opposite angles. Let me recall: the Law of Sines states that a/sin A = b/sin B = c/sin C. So, all these ratios are equal.Given that c is 2 and angle C is 60 degrees, I can write that a/sin A = b/sin B = 2/sin 60°. I know that sin 60° is √3/2, so 2 divided by √3/2 is equal to 4/√3. Wait, but usually, we rationalize the denominator, so that would be (4√3)/3. So, a/sin A and b/sin B both equal (4√3)/3.That means I can express a and b in terms of sin A and sin B. So, a = (4√3)/3 * sin A and b = (4√3)/3 * sin B. Therefore, if I add a and b together, I get a + b = (4√3)/3 * (sin A + sin B). So, when I take (a + b)/(sin A + sin B), it becomes [(4√3)/3 * (sin A + sin B)] / (sin A + sin B). The (sin A + sin B) terms cancel out, leaving me with (4√3)/3. So, that's the value for part (1). That seems straightforward.Moving on to part (2): If a + b = ab, find the area of triangle ABC. Hmm, okay. So, we have a condition that a + b equals ab. I need to find the area, which I know can be calculated using the formula (1/2)*ab*sin C, since we have two sides and the included angle.But first, I need to find the values of a and b or at least find ab because that's what's needed for the area. Let me see. I have a + b = ab. From part (1), I also know that a + b is equal to (4√3)/3 * (sin A + sin B). But maybe it's better to use the Law of Cosines here.Law of Cosines states that c² = a² + b² - 2ab cos C. We know c is 2, so c² is 4. Angle C is 60 degrees, so cos 60° is 0.5. Plugging in, we get 4 = a² + b² - 2ab*(0.5). Simplifying that, 2ab*(0.5) is just ab, so 4 = a² + b² - ab.But from the given condition, a + b = ab. Let me denote ab as some variable to make it easier. Let's say ab = k. Then, a + b = k. Now, I can express a² + b² in terms of k as well. I remember that (a + b)² = a² + 2ab + b², so a² + b² = (a + b)² - 2ab. Substituting, that's k² - 2k.So, going back to the equation from the Law of Cosines: 4 = (k² - 2k) - k. Simplifying that, 4 = k² - 3k. So, bringing everything to one side, we get k² - 3k - 4 = 0. That's a quadratic equation in terms of k.Solving for k, we can use the quadratic formula: k = [3 ± sqrt(9 + 16)] / 2 = [3 ± sqrt(25)] / 2 = [3 ± 5]/2. So, the solutions are (3 + 5)/2 = 4 and (3 - 5)/2 = -1. Since ab can't be negative in the context of triangle sides, we discard -1. So, ab = 4.Now, going back to the area formula: (1/2)*ab*sin C. We have ab = 4, sin 60° is √3/2. So, the area is (1/2)*4*(√3/2) = (2)*(√3/2) = √3. So, the area is √3.Wait, let me double-check my steps to make sure I didn't make a mistake. For part (2), I used the Law of Cosines correctly, substituted a + b = ab, expressed a² + b² in terms of k, solved the quadratic, and found ab = 4. Then, using the area formula, I got √3. That seems right.I think both parts are correct. Part (1) was straightforward with the Law of Sines, and part (2) required using the Law of Cosines and solving a quadratic, which gave me ab = 4, leading to the area √3.**Final Answer**(1) boxed{dfrac{4 sqrt{3}}{3}} (2) boxed{sqrt{3}}

💡First, identify the number of socks of each color: 5 white, 6 brown, 3 blue, and 2 red.To find the number of ways to choose a pair of socks of the same color, calculate the combinations for each color:1. **White Socks:** The number of ways to choose 2 white socks from 5 is calculated using the combination formula ( binom{5}{2} ).2. **Brown Socks:** The number of ways to choose 2 brown socks from 6 is ( binom{6}{2} ).3. **Blue Socks:** The number of ways to choose 2 blue socks from 3 is ( binom{3}{2} ).4. **Red Socks:** The number of ways to choose 2 red socks from 2 is ( binom{2}{2} ).Add all these combinations together to get the total number of ways to choose a pair of socks of the same color.

💡Okay, so I have this geometry problem here about a cyclic quadrilateral ABCD. A circle is circumscribed around it, which means all four vertices lie on the circumference of the circle. The problem gives me two angles: angle ACB is 50 degrees, and angle CAD is 20 degrees. I need to find the sum of angles CAB and ACD. Hmm, let me try to visualize this.First, let me sketch the quadrilateral ABCD inscribed in a circle. Points A, B, C, D are on the circumference. The circle is the circumcircle, so all the vertices lie on it. Now, angle ACB is 50 degrees. That angle is at point C, between points A, C, and B. So, ACB is an inscribed angle subtended by arc AB. Similarly, angle CAD is 20 degrees, which is at point A, between points C, A, and D. So, CAD is another inscribed angle, subtended by arc CD.I remember that in a circle, the measure of an inscribed angle is half the measure of its subtended arc. So, if angle ACB is 50 degrees, then the arc AB it subtends must be 100 degrees. Similarly, angle CAD is 20 degrees, so the arc CD it subtends must be 40 degrees.Now, since ABCD is a quadrilateral inscribed in a circle, the sum of all arcs subtended by its sides should be 360 degrees. So, if I add up the arcs AB, BC, CD, and DA, they should total 360 degrees. I already know arcs AB and CD are 100 degrees and 40 degrees respectively. So, together, they make up 140 degrees. That leaves the remaining arcs BC and DA to add up to 360 - 140 = 220 degrees.Now, the problem asks for the sum of angles CAB and ACD. Let me see where these angles are located. Angle CAB is at point A, between points C, A, and B. So, it's another inscribed angle subtended by arc CB. Similarly, angle ACD is at point C, between points A, C, and D, so it's an inscribed angle subtended by arc AD.Wait, so angle CAB is subtended by arc CB, and angle ACD is subtended by arc AD. So, together, these two angles are subtended by arcs CB and AD. But earlier, I found that arcs CB and DA together make up 220 degrees. Hmm, but arc AD is the same as arc DA, right? So, arcs CB and AD together are 220 degrees.Since both angles CAB and ACD are inscribed angles subtended by these arcs, each angle is half the measure of its respective arc. Therefore, angle CAB is half of arc CB, and angle ACD is half of arc AD. So, the sum of angles CAB and ACD would be half of (arc CB + arc AD), which is half of 220 degrees.Calculating that, half of 220 is 110 degrees. So, the sum of angles CAB and ACD is 110 degrees.Wait, let me double-check to make sure I didn't make a mistake. I know that in cyclic quadrilaterals, opposite angles sum to 180 degrees. But in this case, we're dealing with angles that aren't necessarily opposite. However, the key here is that the sum of the arcs subtended by these angles is 220 degrees, so their inscribed angles should add up to half of that, which is 110 degrees. That seems consistent.Another way to think about it is to consider the entire circle. The arcs AB and CD take up 100 and 40 degrees, leaving 220 degrees for the remaining arcs. Since angles CAB and ACD are each half the measure of their respective arcs, adding them together gives half of 220, which is 110 degrees. Yeah, that makes sense.I think I'm confident with this answer. It all ties back to the properties of inscribed angles and the fact that the total circumference of the circle is 360 degrees. By figuring out the arcs subtended by the given angles, I could find the measure of the remaining arcs and then determine the sum of the desired angles.**Final Answer**The sum of angles CAB and ACD is boxed{110} degrees.

💡Okay, so I have this problem here: If ( n ) is 1 less than a multiple of 75, then what is the remainder when ( n^2 + 2n + 4 ) is divided by 75? Hmm, let me try to figure this out step by step.First, I need to understand what it means for ( n ) to be 1 less than a multiple of 75. That means ( n ) can be expressed as ( 75k - 1 ) where ( k ) is some integer. So, ( n = 75k - 1 ). Got that down.Now, the problem is asking for the remainder when ( n^2 + 2n + 4 ) is divided by 75. So, essentially, I need to compute ( n^2 + 2n + 4 ) modulo 75. Since ( n ) is expressed in terms of 75, maybe I can substitute ( n = 75k - 1 ) into the expression and simplify it modulo 75.Let me write that out:( n^2 + 2n + 4 = (75k - 1)^2 + 2(75k - 1) + 4 )Okay, let's expand each term step by step.First, expand ( (75k - 1)^2 ):( (75k - 1)^2 = (75k)^2 - 2 times 75k times 1 + 1^2 = 5625k^2 - 150k + 1 )Next, expand ( 2(75k - 1) ):( 2(75k - 1) = 150k - 2 )Now, add all the terms together:( 5625k^2 - 150k + 1 + 150k - 2 + 4 )Let's combine like terms:- The ( 5625k^2 ) term stays as is.- The ( -150k ) and ( +150k ) cancel each other out.- The constants: ( 1 - 2 + 4 = 3 )So, after simplifying, we have:( 5625k^2 + 3 )Now, we need to find the remainder when this expression is divided by 75. Let's consider each term modulo 75.First, ( 5625k^2 ) modulo 75. Let's see, 5625 divided by 75 is 75, because 75 times 75 is 5625. So, ( 5625k^2 ) is a multiple of 75, which means ( 5625k^2 equiv 0 pmod{75} ).Then, the constant term is 3, so ( 5625k^2 + 3 equiv 0 + 3 equiv 3 pmod{75} ).Therefore, the remainder when ( n^2 + 2n + 4 ) is divided by 75 is 3.Wait, let me double-check my steps to make sure I didn't make a mistake. So, starting from ( n = 75k - 1 ), I substituted into the expression, expanded, and simplified. The quadratic term became 5625k², which is indeed 75 squared, so that term is a multiple of 75. The linear terms canceled out, and the constants added up to 3. So yes, modulo 75, the entire expression simplifies to 3. That seems correct.Alternatively, maybe I can approach this using modular arithmetic properties without expanding everything. Since ( n equiv -1 pmod{75} ), because ( n = 75k - 1 ), which is the same as ( n equiv -1 pmod{75} ). Then, ( n^2 + 2n + 4 ) modulo 75 can be computed as:( (-1)^2 + 2(-1) + 4 pmod{75} )Calculating each term:( (-1)^2 = 1 )( 2(-1) = -2 )Adding them up with the constant term:( 1 - 2 + 4 = 3 )So, again, I get 3 as the remainder. That confirms my earlier result.I think that's solid. So, the remainder is 3.**Final Answer**The remainder when ( n^2 + 2n + 4 ) is divided by 75 is boxed{3}.