💡To solve the inequality (2 otimes a leqslant 2) using the defined operation (x otimes y = frac{xy}{3} - 2y), follow these steps:1. **Substitute (x = 2) and (y = a) into the operation:** [ 2 otimes a = frac{2a}{3} - 2a ]2. **Simplify the expression:** [ 2 otimes a = frac{2a}{3} - frac{6a}{3} = -frac{4a}{3} ]3. **Set up the inequality:** [ -frac{4a}{3} leqslant 2 ]4. **Solve for (a):** [ -frac{4a}{3} leqslant 2 implies -4a leqslant 6 implies a geqslant -frac{3}{2} ]5. **Determine the smallest integer (a):** Since (a) must be an integer and (a geqslant -frac{3}{2}), the smallest integer satisfying this condition is (-1).Therefore, the smallest integer solution for the inequality (2 otimes a leqslant 2) with respect to (a) is (-1).

💡Okay, so Lucille has this business, and she needs to divide her revenue between employee salaries and stock purchases. The ratio given is 4:11, which means for every 4 parts that go to salaries, 11 parts go to stock. She has a total revenue of 3000, and I need to find out how much she spends on employee salaries.First, I think I need to understand what the ratio 4:11 means. It means that the total revenue is split into 4 parts for salaries and 11 parts for stock. So, the total number of parts is 4 + 11, which is 15 parts.Now, if the total revenue is 3000, and this is divided into 15 equal parts, each part must be worth 3000 divided by 15. Let me calculate that: 3000 ÷ 15 equals 200. So, each part is 200.Since the salaries are 4 parts, I need to multiply the value of one part by 4 to find out how much is spent on salaries. That would be 200 × 4, which equals 800.Wait, let me double-check that. If each part is 200, then 4 parts would indeed be 800. And for stock purchases, 11 parts would be 200 × 11, which is 2200. Adding 800 and 2200 gives 3000, which matches the total revenue. So, that seems correct.I think I've got it. Lucille spends 800 on employee salaries.

💡Okay, so I have this problem here about an arithmetic sequence. It says that the fourth, fifth, and sixth terms are 8, 12, and 16 respectively. I need to find the sum of the first five terms. Hmm, arithmetic sequences, right? They have a common difference between consecutive terms. Let me think about how to approach this.First, let me recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference to the preceding term. So, if I denote the first term as ( a_1 ) and the common difference as ( d ), then the nth term can be expressed as:[a_n = a_1 + (n - 1)d]Okay, so given that, the fourth term is 8, the fifth term is 12, and the sixth term is 16. Let me write down these expressions using the formula.For the fourth term (( n = 4 )):[a_4 = a_1 + 3d = 8]For the fifth term (( n = 5 )):[a_5 = a_1 + 4d = 12]For the sixth term (( n = 6 )):[a_6 = a_1 + 5d = 16]Alright, so I have three equations here. But I only need to find ( a_1 ) and ( d ), so maybe I can use two of these equations to solve for those variables.Looking at the equations for ( a_4 ) and ( a_5 ), I can set up a system of equations:1. ( a_1 + 3d = 8 )2. ( a_1 + 4d = 12 )If I subtract the first equation from the second, I can eliminate ( a_1 ) and solve for ( d ):[(a_1 + 4d) - (a_1 + 3d) = 12 - 8]Simplifying:[a_1 + 4d - a_1 - 3d = 4]Which simplifies further to:[d = 4]Okay, so the common difference ( d ) is 4. Now, I can plug this back into one of the equations to find ( a_1 ). Let's use the first equation:[a_1 + 3(4) = 8]Simplifying:[a_1 + 12 = 8]Subtracting 12 from both sides:[a_1 = 8 - 12 = -4]So, the first term ( a_1 ) is -4. Now that I have both ( a_1 ) and ( d ), I can write out the first five terms of the sequence.Let's list them out:1. First term (( a_1 )): -42. Second term (( a_2 )): ( a_1 + d = -4 + 4 = 0 )3. Third term (( a_3 )): ( a_2 + d = 0 + 4 = 4 )4. Fourth term (( a_4 )): 8 (given)5. Fifth term (( a_5 )): 12 (given)So, the first five terms are: -4, 0, 4, 8, 12.Now, to find the sum of these first five terms, I can add them up:[-4 + 0 + 4 + 8 + 12]Let's compute this step by step:- Start with -4.- Add 0: -4 + 0 = -4.- Add 4: -4 + 4 = 0.- Add 8: 0 + 8 = 8.- Add 12: 8 + 12 = 20.So, the sum of the first five terms is 20.Alternatively, I remember that the sum of the first ( n ) terms of an arithmetic sequence can be calculated using the formula:[S_n = frac{n}{2} times (2a_1 + (n - 1)d)]Let me try using this formula to verify my result. Here, ( n = 5 ), ( a_1 = -4 ), and ( d = 4 ).Plugging in the values:[S_5 = frac{5}{2} times (2(-4) + (5 - 1) times 4)]Simplify inside the parentheses first:[2(-4) = -8][(5 - 1) times 4 = 4 times 4 = 16]So, inside the parentheses:[-8 + 16 = 8]Now, multiply by ( frac{5}{2} ):[S_5 = frac{5}{2} times 8 = frac{40}{2} = 20]Okay, that confirms my earlier calculation. The sum is indeed 20.Just to make sure I didn't make any mistakes, let me double-check the terms:- ( a_1 = -4 )- ( a_2 = -4 + 4 = 0 )- ( a_3 = 0 + 4 = 4 )- ( a_4 = 4 + 4 = 8 )- ( a_5 = 8 + 4 = 12 )Adding them: -4 + 0 + 4 + 8 + 12 = 20. Yep, that's correct.I think I've got it. The sum of the first five terms is 20.

💡Okay, so I need to compute the combination of 10 choose 3, which is written as (binom{10}{3}). I remember that combinations are used when the order doesn't matter, right? Like, if I'm picking 3 items out of 10, and it doesn't matter which order I pick them in. First, I think the formula for combinations is (binom{n}{k} = frac{n!}{k!(n-k)!}). So, plugging in the numbers, it should be (frac{10!}{3!(10-3)!}). Simplifying the denominator, that becomes (frac{10!}{3!7!}).Hmm, factorials can get really big really quickly. I remember that (10! = 10 times 9 times 8 times 7!). Oh, wait, that's helpful because the denominator also has a (7!). So, maybe I can cancel out the (7!) from both the numerator and the denominator. Let me write that out:[binom{10}{3} = frac{10 times 9 times 8 times 7!}{3! times 7!}]Yes, the (7!) cancels out, so now it's just:[frac{10 times 9 times 8}{3!}]And (3!) is (3 times 2 times 1 = 6). So now the equation is:[frac{10 times 9 times 8}{6}]Let me compute the numerator first: (10 times 9 = 90), and then (90 times 8 = 720). So now it's (frac{720}{6}).Dividing 720 by 6, I get 120. So, (binom{10}{3} = 120). Wait, let me double-check that. Maybe I made a mistake in the multiplication or division. So, 10 times 9 is definitely 90, and 90 times 8 is 720. Divided by 6, that's 120. Yeah, that seems right. Alternatively, I could think of it as simplifying before multiplying everything out. Let's see, 10 divided by 3 is not a whole number, but 10 divided by 2 is 5, and 8 divided by 3 is not a whole number either. Hmm, maybe another way to simplify. Wait, 10 divided by 5 is 2, but that's not helpful here. Maybe I can factor out some numbers. Let's see, 10 is 2 times 5, 9 is 3 times 3, and 8 is 2 times 2 times 2. The denominator is 6, which is 2 times 3. So, let's write it all out:Numerator: (2 times 5 times 3 times 3 times 2 times 2 times 2)Denominator: (2 times 3)So, canceling out one 2 and one 3 from both numerator and denominator, we're left with:Numerator: (5 times 3 times 2 times 2 times 2)Which is (5 times 3 = 15), and (15 times 2 = 30), (30 times 2 = 60), (60 times 2 = 120). Yep, same result. So, that confirms it.I guess another way to think about it is using the multiplicative formula for combinations, which is (binom{n}{k} = frac{n times (n-1) times dots times (n - k + 1)}{k!}). So, for (binom{10}{3}), it's (frac{10 times 9 times 8}{3!}), which is exactly what I did earlier. So, that's consistent.I wonder if there's a combinatorial interpretation that could help me understand why it's 120. Like, if I have 10 books and I want to choose 3 to take on a trip, how many different sets can I make? Each set is unique regardless of the order, so that's why combinations are used instead of permutations.If I think about permutations, which do consider order, the number would be higher. For permutations, it's (P(n, k) = frac{n!}{(n - k)!}), so (P(10, 3) = frac{10!}{7!} = 10 times 9 times 8 = 720). But since combinations don't care about order, we divide by the number of ways the 3 books can be arranged, which is (3! = 6), giving us (720 / 6 = 120). That makes sense.Another way to think about it is recursively. I remember that (binom{n}{k} = binom{n-1}{k} + binom{n-1}{k-1}). So, using Pascal's triangle, each number is the sum of the two numbers directly above it. If I build up Pascal's triangle up to the 10th row, the 3rd entry should be 120. Let me try that.Starting from the top, row 0 is just 1. Row 1 is 1 1. Row 2 is 1 2 1. Row 3 is 1 3 3 1. Row 4 is 1 4 6 4 1. Row 5 is 1 5 10 10 5 1. Row 6 is 1 6 15 20 15 6 1. Row 7 is 1 7 21 35 35 21 7 1. Row 8 is 1 8 28 56 70 56 28 8 1. Row 9 is 1 9 36 84 126 126 84 36 9 1. Row 10 is 1 10 45 120 210 252 210 120 45 10 1.Looking at row 10, the third number is indeed 120. So that's another confirmation.I could also use the symmetry property of combinations, which states that (binom{n}{k} = binom{n}{n - k}). So, (binom{10}{3} = binom{10}{7}). If I compute (binom{10}{7}), it should also be 120. Let me check:[binom{10}{7} = frac{10!}{7!3!} = frac{10 times 9 times 8}{3 times 2 times 1} = 120]Yep, same result. So that's consistent too.I think I've verified it enough ways. The answer is definitely 120.

💡Okay, so I have this equation: ( x^4 + kx^2 + 27 = 0 ). And I'm told that 3 is a solution, which means when I plug in 3 for x, the equation should equal zero. I need to find the value of k.Alright, let's start by substituting x with 3 in the equation. So, replacing x with 3, I get:( 3^4 + k cdot 3^2 + 27 = 0 )Now, let's compute each term step by step. First, ( 3^4 ) is 81 because 3 multiplied by itself four times is 81. Then, ( 3^2 ) is 9, so ( k cdot 9 ) is just 9k. The last term is 27, which stays the same.Putting it all together, the equation becomes:( 81 + 9k + 27 = 0 )Now, let's combine the constant terms. 81 plus 27 is 108, so the equation simplifies to:( 108 + 9k = 0 )To solve for k, I need to isolate it. So, I'll subtract 108 from both sides of the equation:( 9k = -108 )Now, to get k by itself, I'll divide both sides by 9:( k = -108 / 9 )Calculating that, -108 divided by 9 is -12. So, k equals -12.Wait, let me double-check my steps to make sure I didn't make any mistakes. I substituted 3 correctly, calculated the exponents properly, combined like terms, and solved for k. Everything seems to add up. So, I think k is indeed -12.

💡Okay, so I have this problem with two parts, both involving a point P(2, -1). Let me try to figure out each part step by step.Starting with part (1): Line m passes through point P, and the intercepts on the two coordinate axes are equal. I need to find the equation of line m.Hmm, intercepts on the axes being equal. That means the x-intercept and y-intercept are the same. Let me recall that the intercept form of a line is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept. Since they are equal, a = b. So, the equation simplifies to x/a + y/a = 1, which can be written as x + y = a.But wait, the line also passes through point P(2, -1). So, I can substitute x = 2 and y = -1 into the equation to find a.Plugging in: 2 + (-1) = a => 1 = a. So, a = 1. Therefore, the equation of the line is x + y = 1, or rearranged as x + y - 1 = 0.But hold on, what if the intercepts are both zero? That is, the line passes through the origin. If a = 0, then both intercepts are zero. Let me check if the line through P(2, -1) and the origin satisfies this condition.The slope would be (-1 - 0)/(2 - 0) = -1/2. So, the equation is y = (-1/2)x. Does this line have equal intercepts? Well, if x = 0, y = 0, and if y = 0, x = 0. So, both intercepts are zero, which are equal. So, that's another possible equation.Therefore, there are two possible lines: one with intercepts at (1, 0) and (0, 1), and another passing through the origin.Moving on to part (2): Line n passes through point P, and the distance from the origin to this line is 2. I need to find the equation of line n.The distance from a point (x0, y0) to the line Ax + By + C = 0 is |Ax0 + By0 + C| / sqrt(A^2 + B^2). Here, the origin is (0, 0), so the distance is |C| / sqrt(A^2 + B^2) = 2.But the line passes through P(2, -1). So, if I let the equation of the line be y = kx + c, it must satisfy -1 = 2k + c. Alternatively, in standard form, it's kx - y + c = 0.Wait, maybe it's better to write it in the form y + 1 = k(x - 2), which is the point-slope form. Then, rearranged, it's y = kx - 2k - 1, or kx - y - 2k - 1 = 0.Now, the distance from the origin to this line is | -2k - 1 | / sqrt(k^2 + 1) = 2.So, setting up the equation: | -2k - 1 | = 2 * sqrt(k^2 + 1).Squaring both sides to eliminate the absolute value and the square root:( -2k - 1 )^2 = 4(k^2 + 1 )Expanding the left side: 4k^2 + 4k + 1 = 4k^2 + 4.Subtracting 4k^2 from both sides: 4k + 1 = 4.Subtracting 1: 4k = 3 => k = 3/4.So, the slope is 3/4. Plugging back into the equation: y + 1 = (3/4)(x - 2).Multiplying both sides by 4 to eliminate the fraction: 4y + 4 = 3x - 6.Rearranging: 3x - 4y - 10 = 0.But wait, what if the line is vertical? A vertical line has an undefined slope. The equation would be x = 2, since it passes through P(2, -1). The distance from the origin to x = 2 is |2| / sqrt(1) = 2, which satisfies the condition.So, there are two possible lines: one with slope 3/4, and another vertical line x = 2.Let me double-check my work.For part (1), I considered both cases where the intercepts are non-zero and zero, leading to two equations: x + y - 1 = 0 and y = (-1/2)x. That seems correct.For part (2), I found the slope-intercept form and solved for k, getting k = 3/4, leading to 3x - 4y - 10 = 0. Then, I considered the vertical line case, which also satisfies the distance condition. That seems thorough.I think I've covered all possibilities for both parts. I don't see any mistakes in my reasoning.**Final Answer**(1) The equation of line m is boxed{y = -frac{1}{2}x} or boxed{x + y - 1 = 0}.(2) The equation of line n is boxed{x = 2} or boxed{3x - 4y - 10 = 0}.