💡Alright, so I have this problem here with two sets, M and N, and I need to find their intersection. Let me take it step by step.First, let's look at set M. It's defined as all real numbers x such that x squared plus x minus 2 is less than zero. So, M = {x | x² + x - 2 < 0}. Hmm, okay, that's a quadratic inequality. I remember that to solve quadratic inequalities, I need to find the roots of the quadratic equation first and then determine the intervals where the quadratic expression is positive or negative.So, let me solve the equation x² + x - 2 = 0. To find the roots, I can use the quadratic formula: x = [-b ± sqrt(b² - 4ac)] / (2a). Here, a = 1, b = 1, and c = -2. Plugging those values in, I get:x = [-1 ± sqrt(1 + 8)] / 2x = [-1 ± sqrt(9)] / 2x = [-1 ± 3] / 2So, the roots are x = (-1 + 3)/2 = 2/2 = 1, and x = (-1 - 3)/2 = -4/2 = -2. Therefore, the roots are x = 1 and x = -2.Now, since the quadratic opens upwards (because the coefficient of x² is positive), the expression x² + x - 2 will be less than zero between the two roots. So, the solution to the inequality x² + x - 2 < 0 is -2 < x < 1. Therefore, set M is the interval (-2, 1).Alright, that takes care of set M. Now, let's move on to set N. It's defined as N = {x | log_{1/2} x > -1}. Hmm, logarithmic inequalities can sometimes be tricky, especially with bases less than 1 because the inequality sign flips when we exponentiate.First, I recall that log_b a > c is equivalent to a < b^c when 0 < b < 1. Since the base here is 1/2, which is less than 1, the inequality will flip when I rewrite it in exponential form.So, starting with log_{1/2} x > -1, I can rewrite this as x < (1/2)^{-1}. Let me compute (1/2)^{-1}. That's the same as 2^1, which is 2. So, x < 2.But wait, I also need to remember the domain of the logarithm function. The logarithm is only defined for positive real numbers, so x must be greater than 0. Therefore, combining these two conditions, we have 0 < x < 2. So, set N is the interval (0, 2).Now that I have both sets M and N, I need to find their intersection, M ∩ N. The intersection of two sets consists of all elements that are common to both sets.Set M is (-2, 1) and set N is (0, 2). So, let me visualize these intervals on the number line.- Set M spans from -2 to 1.- Set N spans from 0 to 2.The overlapping part of these two intervals is from 0 to 1 because that's where both intervals coincide. So, M ∩ N is the interval (0, 1).Looking back at the options provided:A: {x | -2 < x < 1}B: {x | 0 < x < 1}C: {x | x > 2}D: ∅The correct answer should be option B: {x | 0 < x < 1}.Wait, just to make sure I didn't make any mistakes. Let me double-check my steps.For set M:- Quadratic equation x² + x - 2 = 0, roots at x = -2 and x = 1.- Since the quadratic opens upwards, the inequality x² + x - 2 < 0 is satisfied between the roots: -2 < x < 1. That seems correct.For set N:- log_{1/2} x > -1.- Since the base is less than 1, the inequality flips when exponentiating: x < (1/2)^{-1} = 2.- Also, x must be positive because log is only defined for positive numbers. So, 0 < x < 2. That also seems correct.Intersection of (-2, 1) and (0, 2) is indeed (0, 1). So, yes, option B is correct.I think I've covered all the steps thoroughly and checked my work. I don't see any mistakes in my reasoning.

💡Okay, so I have this problem about a right triangle PQR where angle Q is 90 degrees. The equation given is 3 sin P equals 4 cos P, and I need to find sin P. Hmm, right triangles and trigonometric functions. Let me try to remember how these things work.First, in a right triangle, the sine of an angle is the ratio of the length of the opposite side to the hypotenuse, and the cosine is the ratio of the adjacent side to the hypotenuse. So, if I consider angle P, then sin P would be the length of the side opposite to P divided by the hypotenuse, and cos P would be the length of the side adjacent to P divided by the hypotenuse.Let me denote the sides. Since angle Q is 90 degrees, the triangle has sides PQ, QR, and PR, where PR is the hypotenuse. So, for angle P, the opposite side would be QR, and the adjacent side would be PQ. Therefore, sin P is QR/PR, and cos P is PQ/PR.The equation given is 3 sin P = 4 cos P. Substituting the expressions I have for sin P and cos P, this becomes 3*(QR/PR) = 4*(PQ/PR). Hmm, okay, so 3*(QR/PR) equals 4*(PQ/PR). Since both sides have PR in the denominator, I can multiply both sides by PR to eliminate the denominators. That would give me 3 QR = 4 PQ.So, 3 times QR equals 4 times PQ. That means PQ is (3/4) QR. So, PQ is three-fourths of QR. Interesting. So, the sides PQ and QR are related by this ratio.Now, since it's a right triangle, I can apply the Pythagorean theorem, which says that the sum of the squares of the legs equals the square of the hypotenuse. So, PQ squared plus QR squared equals PR squared.But I know that PQ is (3/4) QR, so I can substitute that into the equation. Let me write that out: (3/4 QR)^2 + QR^2 = PR^2.Calculating (3/4 QR)^2, that would be (9/16) QR squared. So, adding that to QR squared, which is the same as (16/16) QR squared, gives me (9/16 + 16/16) QR squared, which is (25/16) QR squared.So, (25/16) QR squared equals PR squared. Therefore, PR squared is (25/16) QR squared. Taking the square root of both sides, PR is (5/4) QR.So, PR is five-fourths of QR. Now, since sin P is QR divided by PR, substituting the value of PR, we get sin P equals QR divided by (5/4 QR). The QR in the numerator and denominator cancels out, leaving 1 divided by (5/4), which is 4/5.So, sin P is 4/5. Let me just double-check that. If sin P is 4/5, then cos P would be 3/5 because in a right triangle, sin squared plus cos squared equals 1. So, (4/5)^2 + (3/5)^2 is 16/25 + 9/25, which is 25/25, or 1. That checks out.Therefore, the value of sin P is 4/5.**Final Answer**The value of sin P is boxed{dfrac{4}{5}}.

💡First, I need to determine the cost of each of the first three pieces of art. Since John paid the same price for each and the total was 45,000 for three pieces, I divide 45,000 by 3.45,000 ÷ 3 = 15,000So, each of the first three pieces of art cost 15,000.Next, I need to find out the cost of the fourth piece of art. The total cost of all four pieces is 67,500. I subtract the total cost of the first three pieces from this amount to find the cost of the fourth piece.67,500 - 45,000 = 22,500The fourth piece of art cost 22,500.Now, I will calculate the percentage increase in price for the fourth piece compared to one of the first three pieces.The increase in price is 22,500 - 15,000 = 7,500.To find the percentage increase, I divide the increase by the original price and multiply by 100.(7,500 ÷ 15,000) × 100 = 0.5 × 100 = 50%The percentage increase in price for the fourth piece of art compared to each of the first three pieces is 50%.

💡First, I need to determine how many times the digits 9 and 8 appear in each digit place (units, tens, and hundreds) from 1 to 700.For the units place:- The digit 9 appears once every 10 numbers (e.g., 9, 19, 29, ..., 699). This happens 70 times (7 hundreds × 10 numbers).- Similarly, the digit 8 appears once every 10 numbers (e.g., 8, 18, 28, ..., 698). This also happens 70 times.For the tens place:- The digit 9 appears in a sequence of 10 numbers (e.g., 90-99, 190-199, ..., 690-699). This happens 70 times.- The digit 8 appears in a sequence of 10 numbers (e.g., 80-89, 180-189, ..., 680-689). This also happens 70 times.For the hundreds place:- The digit 9 does not appear in the hundreds place since we are only going up to 700.- The digit 8 appears in the hundreds place for numbers 800-899, but since we are only going up to 700, it does not appear in the hundreds place either.Adding up all the occurrences:- Digit 9: 70 (units) + 70 (tens) = 140- Digit 8: 70 (units) + 70 (tens) = 140Total occurrences of digits 9 and 8: 140 + 140 = 280

💡Okay, so I have this problem here where I'm given a sequence: (10^{frac{1}{11}}, 10^{frac{2}{11}}, 10^{frac{3}{11}}, ldots, 10^{11}). I need to find the smallest value of (n) such that the product of the first (n) terms of this sequence exceeds 100,000. The options are (A) 7, (B) 8, (C) 9, (D) 10, (E) 11.Alright, let me try to break this down. First, I know that when you multiply terms with exponents that have the same base, you can add the exponents. So, if I have (10^a times 10^b = 10^{a+b}). That seems useful here because all the terms in the sequence are powers of 10.So, the product of the first (n) terms would be (10^{frac{1}{11}} times 10^{frac{2}{11}} times 10^{frac{3}{11}} times ldots times 10^{frac{n}{11}}). Using the property of exponents, this should be equal to (10^{left(frac{1}{11} + frac{2}{11} + frac{3}{11} + ldots + frac{n}{11}right)}).Let me write that out more clearly:[text{Product} = 10^{left(frac{1 + 2 + 3 + ldots + n}{11}right)}]Okay, so the exponent is the sum of the first (n) natural numbers divided by 11. I remember that the sum of the first (n) natural numbers is given by the formula:[sum_{k=1}^{n} k = frac{n(n + 1)}{2}]So, substituting that into the exponent, we have:[text{Product} = 10^{left(frac{frac{n(n + 1)}{2}}{11}right)} = 10^{left(frac{n(n + 1)}{22}right)}]Alright, so the product simplifies to (10^{frac{n(n + 1)}{22}}). Now, we need this product to exceed 100,000. Since 100,000 is (10^5), we can set up the inequality:[10^{frac{n(n + 1)}{22}} > 10^5]Since the base is the same (10), we can compare the exponents directly:[frac{n(n + 1)}{22} > 5]Now, let's solve this inequality for (n). First, multiply both sides by 22 to eliminate the denominator:[n(n + 1) > 110]So, we have:[n^2 + n - 110 > 0]This is a quadratic inequality. To solve it, let's first find the roots of the quadratic equation (n^2 + n - 110 = 0). I can use the quadratic formula:[n = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Here, (a = 1), (b = 1), and (c = -110). Plugging these into the formula:[n = frac{-1 pm sqrt{1^2 - 4(1)(-110)}}{2(1)} = frac{-1 pm sqrt{1 + 440}}{2} = frac{-1 pm sqrt{441}}{2}](sqrt{441}) is 21, so:[n = frac{-1 pm 21}{2}]This gives two solutions:1. (n = frac{-1 + 21}{2} = frac{20}{2} = 10)2. (n = frac{-1 - 21}{2} = frac{-22}{2} = -11)Since (n) represents the number of terms, it can't be negative. So, we discard (n = -11). The critical point is at (n = 10).Now, the quadratic (n^2 + n - 110) is a parabola opening upwards (since the coefficient of (n^2) is positive). Therefore, the inequality (n^2 + n - 110 > 0) holds true for (n < -11) or (n > 10). Again, since (n) must be positive, we consider (n > 10).So, the smallest integer (n) that satisfies this inequality is 11. Therefore, the answer should be (E) 11.Wait, but let me double-check. Maybe I made a mistake somewhere. Let me compute the product for (n = 10) and (n = 11) to see if it actually exceeds 100,000.For (n = 10):The exponent is (frac{10 times 11}{22} = frac{110}{22} = 5). So, the product is (10^5 = 100,000). But the problem says "exceeds" 100,000, so (n = 10) gives exactly 100,000, which doesn't exceed it.For (n = 11):The exponent is (frac{11 times 12}{22} = frac{132}{22} = 6). So, the product is (10^6 = 1,000,000), which is way more than 100,000.Hmm, so actually, (n = 11) is the smallest value where the product exceeds 100,000. But wait, 10^6 is 1,000,000, which is much larger than 100,000. Maybe there's a smaller (n) where the product is just over 100,000?Wait, let me think again. The exponent is (frac{n(n + 1)}{22}). So, for the product to exceed (10^5), we need:[frac{n(n + 1)}{22} > 5]Which simplifies to:[n(n + 1) > 110]So, we need (n(n + 1)) to be greater than 110. Let's compute (n(n + 1)) for (n = 10): 10*11 = 110, which is equal to 110, so the product is exactly 10^5. Since we need it to exceed, we need the next integer, which is 11.But wait, is there a value between 10 and 11 where (n(n + 1)) is just over 110? But (n) has to be an integer because we're talking about the number of terms. So, 10 is the last integer where it's exactly 110, and 11 is the first integer where it's 132, which is way over. So, 11 is indeed the smallest integer where the product exceeds 100,000.But hold on, let me check (n = 10) and (n = 11) again. For (n = 10), the product is exactly 100,000, which doesn't exceed it. For (n = 11), it's 1,000,000, which is way over. So, 11 is the answer.But wait, the options are (A) 7, (B) 8, (C) 9, (D) 10, (E) 11. So, 11 is an option, so that's the answer.But just to make sure, let me compute the product for (n = 10) and (n = 11) step by step.For (n = 10):The exponent is (frac{10 times 11}{22} = 5), so the product is (10^5 = 100,000). Exactly 100,000.For (n = 11):The exponent is (frac{11 times 12}{22} = 6), so the product is (10^6 = 1,000,000). That's way more than 100,000.So, yes, 11 is the smallest (n) where the product exceeds 100,000.Wait, but let me think again. Maybe I can find a smaller (n) where the exponent is just over 5. Let's see.We have:[frac{n(n + 1)}{22} > 5]So,[n(n + 1) > 110]Let me compute (n(n + 1)) for (n = 10): 10*11 = 110.For (n = 9): 9*10 = 90, which is less than 110.For (n = 10): 110, which is equal.For (n = 11): 132, which is greater.So, indeed, 11 is the smallest integer where (n(n + 1)) exceeds 110, making the exponent exceed 5, and thus the product exceed 100,000.Therefore, the answer is (E) 11.

💡First, calculate (4!):[4! = 4 times 3 times 2 times 1 = 24]Next, calculate ((4!)!):[(4!)! = 24!]Now, simplify the expression (frac{(4!)!}{4!}):[frac{(4!)!}{4!} = frac{24!}{24}]This simplifies to:[23 times 22 times 21 times ldots times 2 times 1 = 23!]Therefore, the final answer is:[boxed{A}]