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question:Our club has 20 members, X boys and 10 girls. In how many ways can we choose a president and a vice-president if they must be of the same gender? Assume no one can hold both offices. If we know the answer to the above question is 180, what is the value of unknown variable X?
answer:We want to find the value of X in the given situation. To choose a president and a vice-president from the club members, we need to consider two cases: choosing from the group of boys and choosing from the group of girls. In the case of choosing from the group of boys, we have X options for the president and X-1 options for the vice-president (since the same person cannot hold both offices). In the case of choosing from the group of girls, we have 10 options for the president and 9 options for the vice-president. To find the total number of ways to choose a president and a vice-president, we can add the number of ways from each case: Total number of ways = (Number of ways for boys) + (Number of ways for girls) Total number of ways = X cdot (X-1) + 10 cdot 9 We are given that the total number of ways is 180, so we can write: 180 = X cdot (X-1) + 10 cdot 9 Simplifying, we have: 180 = X^2 - X + 90 Rearranging the equation, we get: X^2 - X + 90 - 180 = 0 X^2 - X - 90 = 0 Now, we can solve this quadratic equation to find the value of X. Factoring the equation, we have: (X-10)(X+9) = 0 Setting each factor equal to zero and solving for X, we get two possible values: X = 10 and X = -9. However, since we are looking for the number of boys in the club, X must be a positive value. Therefore, the value of X is 10. The answer is: 10
question:f x cdot (x+y) = x^2 + 8, what is the value of xy?
answer:Expanding the left side of the equation, we have x^2 + xy = x^2 + 8. Simplifying, we get xy = 8. Therefore, the value of xy is boxed{8}.The answer is: 8
question:The water pressure of a sink has a steady flow of 2 cups per 10 minutes for the first 30 minutes. It still flows at 2 cups per 10 minutes for the next 30 minutes after. For the next hour, the water pressure maximizes to 4 cups per 10 minutes and stops. Shawn now has to dump half of the water away. How much water is left?
answer:For the first 30 minutes, the water flows at a rate of 2 cups per 10 minutes, so in total it flows for 30/10 = 3 sets of 10 minutes. So, in the first 30 minutes, the sink fills with 2 cups * 3 sets = 6 cups. For the next 30 minutes, the water still flows at a rate of 2 cups per 10 minutes, so again it flows for 30/10 = 3 sets of 10 minutes. So, in the next 30 minutes, the sink fills with 2 cups * 3 sets = 6 cups. For the next hour, the water pressure maximizes to 4 cups per 10 minutes, so it flows for 60/10 = 6 sets of 10 minutes. So, in the next hour, the sink fills with 4 cups * 6 sets = 24 cups. In total, the sink initially fills with 6 cups + 6 cups + 24 cups = 36 cups of water. Shawn then dumps half of the water away, so he dumps 36 cups / 2 = 18 cups of water. So, there are 36 cups - 18 cups = 18 cups of water left. 18 The answer is: 18
question:Let A = (1,1) be a point on the parabola y = x^2. The normal to the parabola at A is drawn, intersecting the parabola again at B. Find B. [asy] unitsize(1 cm); pair A, B; A = (1,1); B = (-3/2,9/4); real parab (real x) { return(x^2); } draw(graph(parab,-2,2)); draw((A + (-1,-2))--(A + (1,2))); draw((A + (1,-0.5))--(A + (-3,1.5))); draw(rightanglemark(A + (-3,1.5), A, A + (1,2), 10)); dot("A", A, S); dot("B", B, SW); [/asy] Note: The normal at a point P on a curve mathcal{C} is the line passing through P that is perpendicular to the tangent to mathcal{C} at P.
answer:oint on the parabola y = x^2. Then the slope of line AB is [frac{x^2 - 1}{x - 1} = frac{(x - 1)(x + 1)}{x - 1} = x + 1.]The slope of the tangent line to the parabola at A is 2, so the slope of the normal to the parabola at A is -frac{1}{2}. Therefore, we must have [x + 1 = -frac{1}{2}.]Solving, we find x = -frac{3}{2}. Hence, B = boxed{left( -frac{3}{2}, frac{9}{4} right)}.The answer is: (-frac{3}{2},frac{9}{4})