answer:The circumference (C) of a circle (or wheel in this case) is given by the formula: C = π * d where d is the diameter of the circle and π (pi) is a mathematical constant approximately equal to 3.14159. Given that the diameter (d) of the toy car wheel is 30 cm, we can calculate the theoretical circumference as: C = π * d C = π * 30 cm Using the value of π as 3.14159, we get: C = 3.14159 * 30 cm C = 94.2477 cm (approximately) However, the problem states that the actual measured circumference is 94.2 cm. To find out how many times the circumference is greater than the diameter, we divide the circumference by the diameter: Number of times = Circumference / Diameter Number of times = 94.2 cm / 30 cm Number of times = 3.14 (approximately) So, the circumference of the toy car wheel is approximately boxed{3.14} times greater than its diameter, which is consistent with the value of π.

answer:Let the roots of the equation 3x^2 + 7x + k = 0 be a and b. According to the problem, we require [a^2 + b^2 = 3ab.] Using Vieta's formulas, a + b = -frac{7}{3} and ab = frac{k}{3}. Squaring the sum of the roots, [(a + b)^2 = left(-frac{7}{3}right)^2 = frac{49}{9}.] Therefore, the sum of the squares of the roots can be expressed as [a^2 + b^2 = (a+b)^2 - 2ab = frac{49}{9} - 2 cdot frac{k}{3} = frac{49 - 6k}{9}.] Setting a^2 + b^2 equal to 3ab, we have: [frac{49 - 6k}{9} = 3 cdot frac{k}{3} = k.] Cross-multiplying to clear the fraction, [49 - 6k = 9k,] [49 = 15k,] [k = frac{49}{15}.] Therefore, the value of k is boxed{frac{49}{15}}.

answer:1. Calculate the probability that it does not rain on each day: - Friday: 1 - 0.30 = 0.70 - Saturday: 1 - 0.45 = 0.55 - Sunday: 1 - 0.55 = 0.45 2. Multiply these probabilities to find the probability it does not rain at all: [ P(text{no rain all days}) = 0.70 times 0.55 times 0.45 = 0.17325 ] 3. Subtract this from 1 to find the probability that it rains on at least one day: [ P(text{rains at least one day}) = 1 - 0.17325 = 0.82675 ] 4. Convert to percentage: [ boxed{82.675%} ] Conclusion: Thus, there is an approximately 82.675% chance that it will rain on at least one day during the weekend.

answer:When x < 0, we have 2xf(x) + x^2f'(x) < 0 by multiplying both sides of 2f(x) + xf''(x) > 0 by x. Let g(x) = x^2f(x). Then g'(x) = 2xf(x) + x^2f'(x) < 0, which holds true always. Thus, g(x) is monotonically decreasing on (-infty, 0). From the inequality (x+2018)^2f(x+2018) > 4f(-2), it follows that g(x+2018) > g(-2). This implies x + 2018 < -2, and solving for x we obtain x < -2020. Therefore, the solution set is boxed{(-infty, -2020)}. To solve this problem, we construct an appropriate function based on given information, differentiate it, and utilize the function's monotonicity to find the range of valid x values. This problem assesses the understanding of the relationship between a function's derivative and its monotonicity, with the key to the solution being the construction of a suitable function and the use of its derivative to determine monotonicity.