answer:To find the speed of the train, we can use the formula: Speed = Distance / Time The distance the train travels to cross the electric pole is equal to the length of the train, which is 441 meters. The time taken to cross the pole is given as 21 seconds. Plugging in the values, we get: Speed = 441 meters / 21 seconds Speed = 21 meters/second Therefore, the speed of the train is boxed{21} meters per second.

answer:First, expand the left side of the equation: [ (x+2)(x-3) = x^2 - 3x + 2x - 6 = x^2 - x - 6 ] Set the equation to equal -10: [ x^2 - x - 6 = -10 ] Move -10 to the left side to set the equation to zero: [ x^2 - x - 6 + 10 = 0 Rightarrow x^2 - x + 4 = 0 ] For a quadratic equation of the form ax^2 + bx + c = 0, the product of the roots is frac{c}{a}. For the equation x^2 - x + 4 = 0, a = 1, b = -1, and c = 4. Therefore, the product of the roots is: [ frac{c}{a} = frac{4}{1} = 4 ] Thus, the product of the possible values of x is boxed{4}.

answer:1. **Define the rates and breaks:** Let Rudolph's biking rate be r miles per minute, so Jennifer's rate is frac{3}{4}r miles per minute because she bikes at three-quarters the rate of Rudolph. 2. **Calculate the number of breaks:** Rudolph takes a break at the end of every mile, so for 50 miles, he takes breaks 49 times (no break after the 50th mile). Each break is 5 minutes, so he loses 49 times 5 = 245 minutes in breaks. Jennifer takes a break at the end of every two miles, so for 50 miles, she takes breaks 24 times (no break after the 50th mile). Each break is also 5 minutes, so she loses 24 times 5 = 120 minutes in breaks. 3. **Set up the equations for travel time excluding breaks:** Let t be the total time taken by both to reach the 50-mile mark including breaks. The actual biking time for Rudolph is t - 245 minutes and for Jennifer is t - 120 minutes. 4. **Write the equations for distance traveled:** Since distance = rate × time, we have: [ r = frac{50}{t-245} quad text{(Equation 1 for Rudolph)} ] [ frac{3}{4}r = frac{50}{t-120} quad text{(Equation 2 for Jennifer)} ] 5. **Substitute and solve the equations:** Substitute Equation 1 into Equation 2: [ frac{3}{4} left(frac{50}{t-245}right) = frac{50}{t-120} ] Simplify and solve for t: [ frac{3}{4} cdot frac{50}{t-245} = frac{50}{t-120} ] [ frac{3}{4(t-245)} = frac{1}{t-120} ] Cross-multiply and solve: [ 3(t-120) = 4(t-245) ] [ 3t - 360 = 4t - 980 ] [ t = 620 ] 6. **Conclusion:** It has taken both Rudolph and Jennifer 620 minutes to reach the 50-mile mark at the same time. [boxed{620 text{ minutes}}]

answer:The given gcd information implies: - 40 divides p, both 40 and 45 divide q, both 45 and 60 divide r, and 60 divides s. We can decompose these gcds into their prime factors: begin{align*} 40 &= 2^3 cdot 5, 45 &= 3^2 cdot 5, 60 &= 2^2 cdot 3 cdot 5. end{align*} Hence, we can express p, q, r, s as follows: begin{align*} p &= 2^3 cdot 5 cdot a, q &= 2^3 cdot 3^2 cdot 5 cdot b, r &= 2^2 cdot 3^2 cdot 5 cdot c, s &= 2^2 cdot 3 cdot 5 cdot d, end{align*} where a, b, c, d are positive integers with no common factors with the gcds of the others. As gcd(s, p) = 2^2 cdot 5 cdot gcd(d, a), and 110 < gcd(s, p) < 150, we know gcd(d, a) must contain factors using primes other than 2 and 5, and also a multiplication of the remaining primes leading to a product between 110/20 = 5.5 and 150/20 = 7.5. The most plausible factorization that fits the constraints and uses minimal additional primes for gcd(s, p) would be incorporating 3 and 11 (as smallest primes not involved in the restriction of d and a), giving us 108 = 2^2 cdot 3^3 cdot 5 which satisfies 110 < gcd(s, p) < 150. Hence, 11 must be a divisor of gcd(d, a), and subsequently, a divisor of p. Therefore, the divisor of p from the options provided that must be included is boxed{11}.