answer:Let's denote the original quantity of water in the mixture as ( W ) liters. According to the given information, the ratio of milk to water in the original mixture is 6 : 3, which can be simplified to 2 : 1 (by dividing both parts of the ratio by 3). This means that for every 2 liters of milk, there is 1 liter of water. Since the quantity of milk in the original mixture is 45 liters, we can use the simplified ratio to find the original quantity of water: [ frac{Milk}{Water} = frac{2}{1} ] [ frac{45}{W} = frac{2}{1} ] [ W = frac{45}{2} ] [ W = 22.5 text{ liters} ] So, the original mixture contains 22.5 liters of water. Now, let's denote the added quantity of water as ( X ) liters. After adding ( X ) liters of water, the new ratio of milk to water becomes 6 : 5. The new quantity of water will be ( 22.5 + X ) liters, and the quantity of milk remains the same (45 liters). Using the new ratio, we can write: [ frac{Milk}{Water} = frac{6}{5} ] [ frac{45}{22.5 + X} = frac{6}{5} ] Now, we can solve for ( X ): [ 5 times 45 = 6 times (22.5 + X) ] [ 225 = 135 + 6X ] [ 225 - 135 = 6X ] [ 90 = 6X ] [ X = frac{90}{6} ] [ X = 15 ] Therefore, boxed{15} liters of water were added to the mixture.

answer:First, we need to convert the speed of the train from km/hr to m/s to match the units of the train's length, which is given in meters. Speed in m/s = (Speed in km/hr) * (1000 m / 1 km) * (1 hr / 3600 s) For a speed of 72 km/hr: Speed in m/s = 72 * (1000 / 1) * (1 / 3600) Speed in m/s = 72 * (1000 / 3600) Speed in m/s = 72 * (5 / 18) Speed in m/s = 20 m/s Now, we know the train takes 1.5 minutes to pass through the tunnel. We need to convert this time into seconds: Time in seconds = Time in minutes * (60 seconds / 1 minute) Time in seconds = 1.5 * 60 Time in seconds = 90 seconds The train travels the length of itself plus the length of the tunnel in 90 seconds. We can calculate the total distance traveled by the train during this time: Total distance = Speed * Time Total distance = 20 m/s * 90 s Total distance = 1800 meters The total distance is the sum of the length of the train and the length of the tunnel. We know the length of the train is 100 meters, so we can find the length of the tunnel by subtracting the length of the train from the total distance: Length of the tunnel = Total distance - Length of the train Length of the tunnel = 1800 meters - 100 meters Length of the tunnel = 1700 meters Finally, we need to convert the length of the tunnel from meters to kilometers: Length of the tunnel in km = Length of the tunnel in meters / 1000 Length of the tunnel in km = 1700 / 1000 Length of the tunnel in km = 1.7 km The length of the tunnel is boxed{1.7} kilometers.

answer:There must exist a polynomial Q(x) such that [P(x) - a = (x-1)(x-2)(x-3)(x-4)Q(x).] Then, plugging in values of -1, -2, -3, and -4, we get: [P(-1) - a = (-2)(-3)(-4)(-5)Q(-1) = 120Q(-1) = -2a,] [P(-2) - a = (-3)(-4)(-5)(-6)Q(-2) = 360Q(-2) = -2a,] [P(-3) - a = (-4)(-5)(-6)(-7)Q(-3) = -840Q(-3) = -2a,] [P(-4) - a = (-5)(-6)(-7)(-8)Q(-4) = 1680Q(-4) = -2a.] That is, [-2a = 120Q(-1) = 360Q(-2) = -840Q(-3) = 1680Q(-4).] Thus, a must be a multiple of text{lcm}(120, 360, 840, 1680) = 1680. Now we show there exists a Q(x) such that a = 1680. Setting this value for a would give: [Q(-1) = -28, quad Q(-2) = -10, quad Q(-3) = 4, quad Q(-4) = -2.] Given Q(-2) = -10 and Q(-3) = 4, Q(x) must be a polynomial such as R(x)(x+2)(x+3) - 10 for some R(x), satisfying the other conditions. You can take R(x) = 16-x to meet values at -4 and -1. Therefore, the smallest possible value of a with such conditions is boxed{1680}.

answer:First, verify if the given values form a Pythagorean triple. Given the hypotenuse (c = 13 meters) and one leg (a = 5 meters), we can find the other leg (b) using the equation: [ c^2 = a^2 + b^2 ] [ 13^2 = 5^2 + b^2 ] [ 169 = 25 + b^2 ] [ b^2 = 144 ] [ b = 12 text{ meters} ] The other leg b is 12 meters. **Area Calculation:** [ text{Area} = frac{1}{2} times a times b = frac{1}{2} times 5 times 12 = 30 text{ square meters} ] [ boxed{30 text{ square meters}} ] **Perimeter Calculation:** [ text{Perimeter} = a + b + c = 5 + 12 + 13 = 30 text{ meters} ] [ boxed{30 text{ meters}} ]