💡First, I will examine the given equation to identify its form. The equation is ((x-3)^2 + 9(y+2)^2 = 144). I notice that both ((x-3)^2) and ((y+2)^2) are present, which suggests that this could be a circle or an ellipse.To determine the type of conic section, I will rewrite the equation in a standard form by dividing both sides by 144:[frac{(x-3)^2}{144} + frac{(y+2)^2}{16} = 1]This matches the standard form of an ellipse, which is:[frac{(x-h)^2}{a^2} + frac{(y-k)^2}{b^2} = 1]Here, (h = 3), (k = -2), (a^2 = 144), and (b^2 = 16). Since both terms are positive and the equation equals 1, it confirms that the conic section is an ellipse.

💡Okay, so I have this problem here: I need to express the product of two repeating decimals, specifically (0.overline{03}) and (0.overline{81}), as a common fraction. Hmm, repeating decimals can sometimes be tricky, but I remember there's a method to convert them into fractions. Let me try to recall how that works.First, I think I should handle each repeating decimal separately and convert them into fractions before multiplying them together. That seems like a logical approach because multiplying decimals directly might be complicated, especially with repeating parts. So, let's start with the first one: 0.overline{03}.Alright, 0.overline{03} means that the digits 03 repeat indefinitely, right? So it's 0.03030303... and so on. To convert this into a fraction, I remember that there's a standard algebraic method where you set the decimal equal to a variable, multiply both sides by a power of 10 to shift the decimal point, and then subtract the original equation to eliminate the repeating part.Let me try that. Let's let x equal 0.overline{03}. So,x = 0.03030303...Now, since the repeating part is two digits long, I should multiply both sides by 100 to move the decimal point two places to the right. That gives me:100x = 3.03030303...Now, if I subtract the original equation (x = 0.03030303...) from this new equation, the repeating decimals should cancel out:100x - x = 3.03030303... - 0.03030303...That simplifies to:99x = 3So, solving for x, I divide both sides by 99:x = 3/99Hmm, 3/99 can be simplified, right? Both numerator and denominator are divisible by 3. Let me divide numerator and denominator by 3:3 ÷ 3 = 199 ÷ 3 = 33So, x = 1/33. Okay, that seems right. So, 0.overline{03} is equal to 1/33 as a fraction.Great, now let's move on to the second repeating decimal: 0.overline{81}. This is similar to the first one, but the repeating part is 81. So, it's 0.81818181... and continues indefinitely.Again, I'll use the same method. Let me set y equal to 0.overline{81}:y = 0.81818181...Since the repeating part is two digits, I'll multiply both sides by 100:100y = 81.81818181...Now, subtract the original equation (y = 0.81818181...) from this:100y - y = 81.81818181... - 0.81818181...Which simplifies to:99y = 81Solving for y, I divide both sides by 99:y = 81/99Hmm, this fraction can also be simplified. Both 81 and 99 are divisible by 9. Let me divide numerator and denominator by 9:81 ÷ 9 = 999 ÷ 9 = 11So, y = 9/11. Okay, so 0.overline{81} is equal to 9/11 as a fraction.Now that I have both decimals converted into fractions, I can multiply them together. So, I need to compute:(1/33) * (9/11)Multiplying fractions is straightforward: multiply the numerators together and the denominators together. So,Numerator: 1 * 9 = 9Denominator: 33 * 11 = 363So, the product is 9/363.Wait a second, can this fraction be simplified further? Let me check if 9 and 363 have any common factors besides 1.First, let's factor 9: 9 = 3 * 3Now, let's factor 363. Hmm, 363 divided by 3 is 121, because 3 * 120 = 360, and 3 * 1 = 3, so 360 + 3 = 363. So, 363 = 3 * 121.Now, 121 is 11 * 11, so 363 = 3 * 11 * 11.So, the prime factors of 9 are 3 and 3, and the prime factors of 363 are 3, 11, and 11.So, the greatest common divisor (GCD) of 9 and 363 is 3.Therefore, I can divide both numerator and denominator by 3 to simplify the fraction:9 ÷ 3 = 3363 ÷ 3 = 121So, the simplified fraction is 3/121.Wait, but 3 and 121 don't have any common factors besides 1, since 121 is 11 squared, and 3 is a prime number that doesn't divide into 11. So, 3/121 is the fraction in its simplest form.Therefore, the product of 0.overline{03} and 0.overline{81} is 3/121.Let me just double-check my work to make sure I didn't make any mistakes. Starting with converting 0.overline{03} to a fraction:x = 0.030303...100x = 3.030303...Subtracting: 99x = 3 => x = 3/99 = 1/33. That seems correct.Then, converting 0.overline{81}:y = 0.818181...100y = 81.818181...Subtracting: 99y = 81 => y = 81/99 = 9/11. That also looks correct.Multiplying 1/33 * 9/11:(1*9)/(33*11) = 9/363. Simplifying by dividing numerator and denominator by 3 gives 3/121. Yep, that seems right.I think I did everything correctly. So, the final answer should be 3/121.

💡Okay, so I have this problem about transforming a sine function and figuring out the range of a parameter ω such that the transformed function has exactly 2 zeros in a specific interval. Let me try to break it down step by step.First, the original function is f(x) = sin(2x + π/6). The problem says that this graph is shifted to the right by π/6 units. I remember that shifting a function to the right by 'a' units means replacing x with (x - a) in the function. So, shifting f(x) to the right by π/6 should give me a new function, let's call it f1(x), which is sin(2(x - π/6) + π/6). Let me simplify that.Expanding the argument inside the sine function: 2(x - π/6) + π/6 = 2x - π/3 + π/6 = 2x - π/6. So, f1(x) = sin(2x - π/6). Got that.Next, the problem says that all the x-coordinates of this resulting graph are changed to 1/ω times their original values, while the y-coordinates remain unchanged. I think this is a horizontal scaling transformation. If you scale the x-coordinates by 1/ω, it's equivalent to replacing x with ωx in the function. So, applying this transformation to f1(x), we get the function g(x) = sin(2(ωx) - π/6) = sin(2ωx - π/6). That makes sense.Now, we need to find the range of ω such that g(x) has exactly 2 zeros in the interval [0, π/4]. A zero of the function is where g(x) = 0, which happens when sin(2ωx - π/6) = 0. The sine function is zero at integer multiples of π, so 2ωx - π/6 = kπ, where k is an integer.Let me solve for x: 2ωx = kπ + π/6 ⇒ x = (kπ + π/6)/(2ω). We need x to be in [0, π/4]. So, (kπ + π/6)/(2ω) should be between 0 and π/4.Let me write that as inequalities: 0 ≤ (kπ + π/6)/(2ω) ≤ π/4.Multiplying all parts by 2ω (since ω > 0, the inequality signs don't change): 0 ≤ kπ + π/6 ≤ (2ω)(π/4) = ωπ/2.So, 0 ≤ kπ + π/6 ≤ ωπ/2.But k is an integer, so let's find the possible values of k such that x is in [0, π/4].First, let's consider the lower bound: kπ + π/6 ≥ 0. Since π/6 is positive, and k is an integer, the smallest k can be is k = 0. If k = 0, then x = (0 + π/6)/(2ω) = π/(12ω). For x to be ≥ 0, this is always true since ω > 0.Now, the upper bound: kπ + π/6 ≤ ωπ/2. Let's solve for k:kπ ≤ ωπ/2 - π/6 ⇒ k ≤ (ωπ/2 - π/6)/π = ω/2 - 1/6.Since k must be an integer, the maximum value of k is the floor of (ω/2 - 1/6). But we need to find the number of zeros, so let's think about how many integer k satisfy 0 ≤ k ≤ (ω/2 - 1/6).Wait, actually, for each k, we get a corresponding x. So, the number of zeros in [0, π/4] corresponds to the number of integers k such that x = (kπ + π/6)/(2ω) is in [0, π/4].So, we need to find the number of integers k for which 0 ≤ (kπ + π/6)/(2ω) ≤ π/4.Let me rearrange the inequality:0 ≤ kπ + π/6 ≤ (2ω)(π/4) = ωπ/2.So, 0 ≤ kπ + π/6 ≤ ωπ/2.Divide all parts by π:0 ≤ k + 1/6 ≤ ω/2.So, 0 ≤ k + 1/6 ≤ ω/2.Subtract 1/6:-1/6 ≤ k ≤ ω/2 - 1/6.But k is an integer, so the number of integers k satisfying this inequality is the number of zeros.We need exactly 2 zeros, so there should be exactly 2 integers k satisfying -1/6 ≤ k ≤ ω/2 - 1/6.But since k is an integer, and -1/6 is approximately -0.1667, the smallest integer k can be is k = 0. So, the possible k values are 0 and 1, because we need exactly 2 zeros.So, k can be 0 and 1. Let's check what this implies.For k = 0: x = (0 + π/6)/(2ω) = π/(12ω). This must be ≤ π/4.So, π/(12ω) ≤ π/4 ⇒ 1/(12ω) ≤ 1/4 ⇒ 1/ω ≤ 3 ⇒ ω ≥ 1/3. But since ω > 0, this is fine.For k = 1: x = (π + π/6)/(2ω) = (7π/6)/(2ω) = 7π/(12ω). This must be ≤ π/4.So, 7π/(12ω) ≤ π/4 ⇒ 7/(12ω) ≤ 1/4 ⇒ 7/12 ≤ ω/4 ⇒ ω ≥ 7/3.Wait, let me check that step again.From 7π/(12ω) ≤ π/4, we can divide both sides by π: 7/(12ω) ≤ 1/4.Then, cross-multiplying: 7*4 ≤ 12ω*1 ⇒ 28 ≤ 12ω ⇒ ω ≥ 28/12 = 7/3 ≈ 2.333.So, ω must be ≥ 7/3 for k=1 to be within the interval.But we also need that k=2 does not yield an x ≤ π/4. Because if k=2 also satisfies x ≤ π/4, then we would have 3 zeros, which is more than we want.So, let's check for k=2:x = (2π + π/6)/(2ω) = (13π/6)/(2ω) = 13π/(12ω). We need this to be > π/4.So, 13π/(12ω) > π/4 ⇒ 13/(12ω) > 1/4 ⇒ 13*4 > 12ω ⇒ 52 > 12ω ⇒ ω < 52/12 = 13/3 ≈ 4.333.Therefore, to have exactly 2 zeros, ω must satisfy 7/3 ≤ ω < 13/3.Wait, let me verify this.If ω = 7/3, then for k=1: x = 7π/(12*(7/3)) = 7π/(28) = π/4. So, x=π/4 is the upper limit. So, at ω=7/3, the zero at k=1 is exactly at π/4. So, does that count as being in the interval [0, π/4]? Yes, because it's inclusive. So, at ω=7/3, we have zeros at k=0 and k=1, both within [0, π/4].If ω increases beyond 7/3, the zero at k=1 moves to the left of π/4, and we still have k=0 and k=1 as zeros.Now, when ω approaches 13/3 from below, the zero at k=2 approaches π/4 from the right. So, for ω < 13/3, the zero at k=2 is outside [0, π/4], meaning we only have k=0 and k=1 as zeros.But at ω=13/3, the zero at k=2 would be exactly at π/4. So, does that mean we have 3 zeros? Wait, let's check.At ω=13/3, x for k=2 is 13π/(12*(13/3)) = 13π/(52) = π/4. So, x=π/4 is included. So, at ω=13/3, we have zeros at k=0, k=1, and k=2, all within [0, π/4]. That's 3 zeros, which is more than we want.Therefore, to have exactly 2 zeros, ω must be less than 13/3. So, the range is 7/3 ≤ ω < 13/3.Looking at the options, option B is [7/3, 13/3), which matches our conclusion.Wait, but let me double-check if there's any other k that could contribute. For example, could k=-1 give a zero within [0, π/4]? Let's see.For k=-1: x = (-π + π/6)/(2ω) = (-5π/6)/(2ω) = -5π/(12ω). Since ω > 0, this x is negative, which is outside our interval [0, π/4]. So, k=-1 doesn't contribute.Similarly, k=2 gives x=13π/(12ω), which as we saw, is greater than π/4 when ω <13/3, so it doesn't contribute. Only k=0 and k=1 are within the interval when 7/3 ≤ ω <13/3.Therefore, the correct range for ω is [7/3, 13/3), which is option B.

💡First, I recognize that the problem involves calculating the rate percent for simple interest. The formula for simple interest is:[text{Simple Interest (SI)} = frac{text{Principal (P)} times text{Rate (R)} times text{Time (T)}}{100}]I am given:- Simple Interest (SI) = ₹400- Principal (P) = ₹1000- Time (T) = 4 yearsMy goal is to find the Rate (R). I will rearrange the formula to solve for R:[R = frac{SI times 100}{P times T}]Plugging in the given values:[R = frac{400 times 100}{1000 times 4}]Simplifying the equation:[R = frac{40000}{4000} = 10]Therefore, the rate percent is 10%.

💡First, I need to determine the rate per kilogram of grapes that Andrew purchased. Andrew bought 14 kg of grapes at an unknown rate per kg, which I'll denote as ( G ). He also bought 10 kg of mangoes at a rate of 62 per kg. The total amount he paid was 1376.To find the rate per kg of grapes, I'll set up an equation based on the total cost:[14G + 10 times 62 = 1376]Next, I'll calculate the cost of the mangoes:[10 times 62 = 620]Substituting this back into the equation:[14G + 620 = 1376]Now, I'll isolate ( G ) by subtracting 620 from both sides:[14G = 1376 - 620][14G = 756]Finally, I'll solve for ( G ) by dividing both sides by 14:[G = frac{756}{14} = 54]Therefore, the rate per kg of grapes is 54.

💡First, calculate the total points obtained over the five tests. We know the mean score is 82, so the sum for all five tests is:[ 5 times 82 = 410 ]The sum of the first three test scores is:[ 92 + 75 + 68 = 235 ]Therefore, the sum needed from the last two tests is:[ 410 - 235 = 175 ]The score limits are below 95, all scores must be different, and at least one score should be in the 60s. Given that we already have a score in the 60s (68), let's consider higher scores for the remaining two, ensuring all scores are unique and less than 95:- Possible combinations for satisfying 175 with unique scores under 95 are: 94 and 81, 93 and 82, 92 and 83, 91 and 84, 90 and 85, 89 and 86, 88 and 87. Not all options are possible since the scores 92 is taken:- The combination 94 and 81 fits the criteria, holding all scores unique and one score in the 60s.Thus, Trisha's five test scores in descending order are:[ boxed{94, 92, 81, 75, 68} ]