💡First, I need to determine the perimeter of the rectangular park. The man cycles at a speed of 6 km/hr and completes one round in 6 minutes. Converting the speed to meters per minute:6 km/hr = 6 * 1000 m / 60 min = 100 m/minThe distance covered in 6 minutes is:Distance = Speed * Time = 100 m/min * 6 min = 600 mThis distance is the perimeter of the park. Let the length be L and the breadth be B. Given the ratio L:B = 1:2, so L = B/2.The perimeter of a rectangle is given by:Perimeter = 2(L + B) = 2(B/2 + B) = 3BSetting the perimeter equal to 600 m:3B = 600 mSolving for B:B = 600 m / 3 = 200 mThen, the length L is:L = B/2 = 200 m / 2 = 100 mFinally, the area of the park is:Area = L * B = 100 m * 200 m = 20,000 m²

💡Alright, so I've got this problem here about a quadratic function and its intersection points with the x-axis. Let me try to figure this out step by step.First, the problem says that a, b, and c form a geometric sequence. Hmm, okay. I remember that in a geometric sequence, each term is multiplied by a common ratio to get the next term. So, if a, b, c are in a geometric sequence, then b is a times some ratio r, and c is b times r, which would make c equal to a times r squared. So, I can write that as:b = a * rc = b * r = a * r^2Got that down. So, now, the quadratic function is given by f(x) = ax^2 + bx + c. I need to find how many times this graph intersects the x-axis. I remember that the number of intersection points depends on the discriminant of the quadratic equation. The discriminant formula is b^2 - 4ac. If the discriminant is positive, there are two real roots; if it's zero, there's exactly one real root; and if it's negative, there are no real roots, meaning the graph doesn't intersect the x-axis.Alright, so let's compute the discriminant for this quadratic function. But since a, b, and c are in a geometric sequence, I can substitute b and c in terms of a and r. So, let's do that.First, b = a * r, and c = a * r^2. So, plugging these into the discriminant:Discriminant, D = b^2 - 4ac = (a * r)^2 - 4 * a * (a * r^2)Let me simplify that step by step.First, (a * r)^2 is a^2 * r^2.Then, 4ac is 4 * a * (a * r^2) = 4 * a^2 * r^2.So, putting it all together:D = a^2 * r^2 - 4 * a^2 * r^2Hmm, that simplifies to:D = a^2 * r^2 - 4a^2 * r^2 = (1 - 4)a^2 * r^2 = -3a^2 * r^2Okay, so the discriminant is -3a^2 * r^2.Now, I need to analyze this discriminant. The discriminant tells us about the nature of the roots. If D > 0, two real roots; D = 0, one real root; D < 0, no real roots.Looking at D = -3a^2 * r^2, let's see what this equals. Well, a^2 is always non-negative because any real number squared is non-negative. Similarly, r^2 is also non-negative for the same reason. So, a^2 * r^2 is non-negative. Then, multiplying by -3, which is negative, the entire discriminant D is non-positive. So, D ≤ 0.But wait, can D be zero? Let's check when D = 0. That would require -3a^2 * r^2 = 0. Since -3 is not zero, we have a^2 * r^2 = 0. But a^2 is zero only if a = 0, and r^2 is zero only if r = 0. However, in a geometric sequence, if a = 0, then all terms would be zero, which is a trivial case. Similarly, if r = 0, then b = 0 and c = 0, again leading to all terms being zero. So, unless a and r are both zero, D is negative.But in a quadratic function f(x) = ax^2 + bx + c, a cannot be zero because then it wouldn't be a quadratic function anymore—it would be linear. So, a ≠ 0. Therefore, a^2 is positive, and r^2 is non-negative. So, D = -3a^2 * r^2 is always negative unless r = 0, which would make D = 0. But if r = 0, then b = 0 and c = 0, making the quadratic function f(x) = ax^2, which is a parabola opening upwards or downwards depending on the sign of a, and it only touches the x-axis at x = 0 if c = 0. Wait, but in this case, c = 0 as well because c = a * r^2, and r = 0, so c = 0.So, if r = 0, then f(x) = ax^2 + 0x + 0 = ax^2. This is a parabola that touches the x-axis at the origin (0,0). So, in this case, the discriminant is zero, and there's exactly one real root at x = 0.But the problem says that a, b, and c form a geometric sequence. If r = 0, then b = 0 and c = 0, which is a valid geometric sequence with common ratio 0. However, in this case, the quadratic function reduces to f(x) = ax^2, which has one real root at x = 0.But wait, the problem doesn't specify whether the geometric sequence is non-trivial or not. It just says that a, b, and c form a geometric sequence. So, technically, r could be zero, leading to one real root, or r could be non-zero, leading to D < 0 and no real roots.But let's think about this. If r = 0, then b = 0 and c = 0, so the quadratic function becomes f(x) = ax^2. This is a parabola that touches the x-axis at the origin. So, it has exactly one intersection point with the x-axis.On the other hand, if r ≠ 0, then D = -3a^2 * r^2 < 0, which means there are no real roots, so the graph doesn't intersect the x-axis at all.Therefore, depending on the value of r, the number of intersection points can be either 0 or 1.But wait, the problem asks for the number of intersection points. It doesn't specify whether it's always the same or can vary. So, in some cases, it's 0, and in other cases, it's 1. Therefore, the number of intersection points is either 0 or 1.But let me double-check. If r = 0, then f(x) = ax^2, which has one real root at x = 0. If r ≠ 0, then D < 0, so no real roots. So, the number of intersection points is either 0 or 1.But wait, the options given are A: 0, B: 1, C: 2, D: 0 or 1.So, D is an option, which says 0 or 1. So, the answer should be D.But hold on, in the case where r = 0, the quadratic function becomes f(x) = ax^2, which is a valid quadratic function as long as a ≠ 0. So, in that case, it has one real root. If r ≠ 0, then D < 0, so no real roots.Therefore, depending on the common ratio r, the number of intersection points can be either 0 or 1. So, the answer is D: 0 or 1.But wait, is r = 0 allowed? Because in a geometric sequence, the common ratio can be zero, but then all subsequent terms are zero. So, a, 0, 0 is a geometric sequence with common ratio 0. So, yes, that's valid.Therefore, the number of intersection points is either 0 or 1, depending on whether the common ratio is zero or not.So, the answer is D: 0 or 1.

💡First, I need to determine which fraction is the largest among the given options: (frac{3}{5}), (frac{2}{3}), (frac{4}{9}), (frac{5}{15}), and (frac{8}{45}).To compare these fractions effectively, I'll find a common denominator. The denominators are 5, 3, 9, 15, and 45. The least common multiple (LCM) of these numbers is 45, which will be the common denominator.Next, I'll convert each fraction to have a denominator of 45:- (frac{3}{5} = frac{3 times 9}{5 times 9} = frac{27}{45})- (frac{2}{3} = frac{2 times 15}{3 times 15} = frac{30}{45})- (frac{4}{9} = frac{4 times 5}{9 times 5} = frac{20}{45})- (frac{5}{15} = frac{5 times 3}{15 times 3} = frac{15}{45})- (frac{8}{45}) remains the same as it already has a denominator of 45.Now, I'll compare the numerators of these equivalent fractions:- (frac{27}{45})- (frac{30}{45})- (frac{20}{45})- (frac{15}{45})- (frac{8}{45})The largest numerator is 30, which corresponds to the fraction (frac{30}{45}). This fraction was originally (frac{2}{3}).Therefore, the largest fraction is (frac{2}{3}).

💡To determine how many rods Joy can choose as the fourth rod to form a quadrilateral with positive area, we need to ensure that the sum of any three sides is greater than the fourth side. The given rods are 4 cm, 9 cm, and 18 cm, and we need to find the possible lengths for the fourth rod, denoted as ( d ).The inequalities we need to check are:1. (4 + 9 + 18 > d)2. (4 + 9 + d > 18)3. (4 + 18 + d > 9)4. (9 + 18 + d > 4)Calculating these:1. (31 > d) (d must be less than 31)2. (13 + d > 18 implies d > 5)3. (22 + d > 9 implies d > -13) (always true for positive ( d ))4. (27 + d > 4 implies d > -23) (always true for positive ( d ))Thus, ( d ) must be between 6 and 30 inclusive. We need to count these integers, which are from 6 to 30. The count is (30 - 6 + 1 = 25). However, we must exclude the lengths already used (4, 9, 18). Therefore, the count is (25 - 3 = 22).Joy can choose the fourth rod from 22 different rods to form a quadrilateral with positive area.

💡First, I need to understand the operation (otimes) defined as (otimes(a, b, c) = frac{a}{b - c}) for any three real numbers (a), (b), and (c) where (b neq c). The problem requires me to compute (otimes(otimes(1, 3, 4), otimes(2, 4, 3), otimes(4, 3, 2))).I'll start by calculating each of the inner (otimes) operations:1. Compute (otimes(1, 3, 4)): [ otimes(1, 3, 4) = frac{1}{3 - 4} = frac{1}{-1} = -1 ]2. Compute (otimes(2, 4, 3)): [ otimes(2, 4, 3) = frac{2}{4 - 3} = frac{2}{1} = 2 ]3. Compute (otimes(4, 3, 2)): [ otimes(4, 3, 2) = frac{4}{3 - 2} = frac{4}{1} = 4 ]Now, I substitute these results into the main (otimes) operation:[otimes(-1, 2, 4) = frac{-1}{2 - 4} = frac{-1}{-2} = frac{1}{2}]So, the value of (otimes(otimes(1, 3, 4), otimes(2, 4, 3), otimes(4, 3, 2))) is (frac{1}{2}).

💡First, I need to determine the donation amounts for each park based on the given ratio and the specific information provided.1. **Identify the Ratio Parts:** - Treetown National Park: 5 parts - The Forest Reserve: 4 parts - Animal Preservation Park: 2 parts - Birds Sanctuary: 3 parts2. **Calculate the Value of One Part:** - Donation to Treetown National Park is 570.00, which corresponds to 5 parts. - One part = 570.00 / 5 = 114.003. **Determine Donations for Each Park:** - Treetown National Park: 5 parts × 114.00 = 570.00 - The Forest Reserve: 4 parts × 114.00 = 456.00 - Animal Preservation Park: 2 parts × 114.00 = 228.00 - Birds Sanctuary: 3 parts × 114.00 = 342.004. **Verify the Additional Information:** - The donation to The Forest Reserve should be 140.00 more than the donation to Animal Preservation Park. - 456.00 (Forest Reserve) - 228.00 (Animal Preservation Park) = 228.00 - This does not match the given 140.00 difference, indicating an inconsistency.5. **Adjust the Calculations:** - Let the donation to Animal Preservation Park be ( x ). - Then, the donation to The Forest Reserve is ( x + 140 ). - According to the ratio, ( x + 140 = 4 times 114 = 456 ). - Solving for ( x ): ( x = 456 - 140 = 316 ).6. **Recalculate Donations with Adjusted Values:** - Animal Preservation Park: 316.00 - The Forest Reserve: 316.00 + 140.00 = 456.007. **Calculate the Total Donation:** - Treetown National Park: 570.00 - The Forest Reserve: 456.00 - Animal Preservation Park: 316.00 - Birds Sanctuary: 342.00 - Total Donation = 570.00 + 456.00 + 316.00 + 342.00 = 1684.00