answer:Since the central angle of the sector is frac{pi}{3}, and the chord length opposite to it is 3 cm, it follows that the radius r=3. Therefore, the area of this sector is =frac{1}{2} times 3 times frac{pi}{3} times 3 = frac{3pi}{2} cm². Hence, the answer is: boxed{frac{3pi}{2}}. This can be obtained by using the formula for the area of a sector. This question tests the formula for the area of a sector, the formula for arc length, and examines reasoning and computational skills, making it a basic question.

answer:Let's first calculate the initial amount of water in the 70-liter mixture. Since water makes up 10% of the mixture, we have: Initial water = 10% of 70 liters Initial water = 0.10 * 70 Initial water = 7 liters Now, we add 14 liters of water to the mixture, so the total amount of water becomes: Total water after adding = Initial water + Added water Total water after adding = 7 liters + 14 liters Total water after adding = 21 liters The total volume of the new mixture is the initial volume plus the added water: Total volume of new mixture = Initial volume + Added water Total volume of new mixture = 70 liters + 14 liters Total volume of new mixture = 84 liters Now, we can calculate the new percentage of water in the mixture: New percentage of water = (Total water after adding / Total volume of new mixture) * 100 New percentage of water = (21 / 84) * 100 New percentage of water = 0.25 * 100 New percentage of water = 25% So, the new percentage of water in the mixture is boxed{25%} .

answer:Let's denote the rate at which the inlet pipe fills the tank as ( R ) litres per hour and the rate at which the tank leaks as ( L ) litres per hour. Given that the leak can empty the full tank in 9 hours, we can write: [ L = frac{Volume , of , the , tank}{Time , to , empty} = frac{12960}{9} , litres/hour ] Now, when the inlet is opened, the tank is empty in 12 hours. This means that the combined effect of the inlet filling and the leak emptying results in the tank being emptied in 12 hours. We can write this as: [ R - L = frac{Volume , of , the , tank}{Time , to , empty , with , inlet , open} = frac{12960}{12} , litres/hour ] First, let's calculate ( L ): [ L = frac{12960}{9} = 1440 , litres/hour ] Now, let's calculate the rate at which the tank is emptied with the inlet open: [ R - L = frac{12960}{12} = 1080 , litres/hour ] Now we can solve for ( R ): [ R = L + 1080 ] [ R = 1440 + 1080 ] [ R = 2520 , litres/hour ] So, the rate at which the inlet pipe fills water is boxed{2520} litres per hour.

answer:To find all integer solutions ( (x, y) ) satisfying the equation [ x^2 + x = y^4 + y^3 + y^2 + y, ] we will analyze and simplify the given polynomial equation and explore specific ranges for ( y ) to determine integer solutions. 1. **Factorize the Left and Right Side of the Equation**: The left-hand side of the equation ( x^2 + x ) can be rewritten as: [ x^2 + x = x(x + 1). ] The right-hand side is already expressed as: [ y^4 + y^3 + y^2 + y. ] 2. **Simplify and analyze ranges for ( y )**: To understand the behavior of ( y^4 + y^3 + y^2 + y ), we start by examining the inequality and identifying suitable ranges for ( y ). 3. **Use Inequality bounds**: Consider the product form of the following inequalities: [ (y^2 + frac{y}{2} - frac{1}{2})(y^2 + frac{y}{2} + frac{1}{2}) < y^4 + y^3 + y^2 + y, ] This holds true except for ( -1 leq y leq -frac{1}{3} ). Similarly, [ (y^2 + frac{y}{2})(y^2 + frac{y}{2} + 1) > y^4 + y^3 + y^2 + y, ] This is valid unless ( 0 leq y leq 2 ). 4. **Check Integer Values of ( y ) in the range ( -1 leq y leq 2 )**: Since non-integer values do not satisfy our condition for integer solutions, we check integer values for ( y ) in the range from ( -1 ) to ( 2 ): - For ( y = -1 ): [ y^4 + y^3 + y^2 + y = (-1)^4 + (-1)^3 + (-1)^2 + (-1) = 1 - 1 + 1 - 1 = 0. ] This gives: [ x^2 + x = 0 implies x(x + 1) = 0 implies x = 0 text{ or } x = -1. ] Thus, for ( y = -1 ), the pairs ( (x, y) ) are ( (0, -1) ) and ( (-1, -1) ). - For ( y = 0 ): [ y^4 + y^3 + y^2 + y = 0^4 + 0^3 + 0^2 + 0 = 0. ] This gives: [ x^2 + x = 0 implies x(x + 1) = 0 implies x = 0 text{ or } x = -1. ] Thus, for ( y = 0 ), the pairs ( (0, 0) ) and ( (-1, 0) ). - For ( y = 1 ): [ y^4 + y^3 + y^2 + y = 1^4 + 1^3 + 1^2 + 1 = 1 + 1 + 1 + 1 = 4. ] This gives: [ x^2 + x = 4 implies x^2 + x - 4 = 0 implies x = frac{-1 pm sqrt{1 + 16}}{2} = frac{-1 pm sqrt{17}}{2}. ] Since (sqrt{17}) is not an integer, there are no integer ( x ) solutions for ( y = 1 ). - For ( y = 2 ): [ y^4 + y^3 + y^2 + y = 2^4 + 2^3 + 2^2 + 2 = 16 + 8 + 4 + 2 = 30. ] This gives: [ x^2 + x = 30 implies x^2 + x - 30 = 0 implies x = frac{-1 pm sqrt{1 + 120}}{2} = frac{-1 pm sqrt{121}}{2} implies x = frac{-1 pm 11}{2}. ] So, ( x = 5 ) or ( x = -6 ). Thus, for ( y = 2 ), the pairs ( (x, y) ) are ( (5, 2) ) and ( (-6, 2) ). # Conclusion: The integer solutions satisfying ( x^2 + x = y^4 + y^3 + y^2 + y ) are: [ boxed{(0, -1), (-1, -1), (0, 0), (-1, 0), (5, 2), (-6, 2)} ]