answer：First, I need to determine the total number of ways to choose 2 quitters out of the 18 contestants. This can be calculated using the combination formula:[binom{18}{2} = frac{18 times 17}{2} = 153]Next, I'll calculate the number of ways to choose both quitters from the same tribe. Each tribe has 6 members, so the number of ways to choose 2 quitters from one tribe is:[binom{6}{2} = frac{6 times 5}{2} = 15]Since there are 3 tribes, the total number of ways to have both quitters from any one tribe is:[3 times 15 = 45]Finally, the probability that both quitters are from the same tribe is the ratio of the number of favorable outcomes to the total number of possible outcomes:[frac{45}{153} = boxed{frac{15}{51}} = boxed{frac{5}{17}}]

answer：Okay, so I have this problem here: "Suppose that x is inversely proportional to y. If x = 5 when y = -4, find x when y = 10." Hmm, inversely proportional... I remember that means as one goes up, the other goes down, right? But I need to be precise about it.Let me recall what inverse proportionality means. If x is inversely proportional to y, then their product is a constant. So, mathematically, that should be something like x * y = k, where k is a constant. Yeah, that sounds right. So, if I know x and y at one point, I can find k, and then use that to find x when y changes.Alright, so given that x = 5 when y = -4, I can plug those into the equation to find k. Let's do that:x * y = k5 * (-4) = kSo, 5 times -4 is -20. Therefore, k = -20.Got it. So now I know that for this relationship, the constant k is -20. Now, the problem asks me to find x when y = 10. Using the same relationship, x * y = k, I can plug in y = 10 and k = -20 to find the new x.Let me write that out:x * 10 = -20To solve for x, I need to divide both sides by 10:x = -20 / 10x = -2Wait, so x is -2 when y is 10? That seems to make sense because as y increased from -4 to 10, which is a positive increase, x decreased from 5 to -2, which is a negative decrease. So, they're moving in opposite directions, which fits the idea of inverse proportionality.Let me double-check my calculations to make sure I didn't make a mistake. First, when x = 5 and y = -4, their product is indeed 5 * (-4) = -20. Then, when y = 10, x = -20 / 10 = -2. Yep, that looks correct.I think I got it right. So, the answer should be x = -2.

answer：Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. I know that cos B is 4/5 and side b is 2. There are two parts to this problem.Starting with the first part: Find the value of a when angle A is π/6. Hmm, π/6 is 30 degrees, right? So angle A is 30 degrees. I remember that in a triangle, if I know two angles, I can find the third one because the sum of angles in a triangle is 180 degrees or π radians. But here, I only know angle A and have information about angle B through its cosine. Maybe I can use the Law of Sines here.Law of Sines says that a/sin A = b/sin B = c/sin C. I know angle A is π/6, so sin A is sin(π/6) which is 1/2. I also know side b is 2, and I need to find sin B. Since cos B is 4/5, I can find sin B using the Pythagorean identity: sin² B + cos² B = 1. So sin² B = 1 - (16/25) = 9/25, which means sin B is 3/5. Since angle B is between 0 and π, sin B is positive.Now, applying the Law of Sines: a/(1/2) = 2/(3/5). Let me compute that. 2 divided by (3/5) is 2*(5/3) = 10/3. So a/(1/2) = 10/3, which means a = (10/3)*(1/2) = 5/3. So a is 5/3. That seems straightforward.Wait, let me double-check. If angle A is 30 degrees, and angle B has cos 4/5, which is approximately 36.87 degrees, so angle C would be 180 - 30 - 36.87 = 113.13 degrees. Then, using Law of Sines, a/sin(30) = 2/sin(36.87). Sin(36.87) is 3/5, so 2/(3/5) is 10/3, and a is (10/3)*sin(30) which is (10/3)*(1/2) = 5/3. Yep, that's correct.Moving on to the second part: Find the value of a + c when the area of triangle ABC is 3. Okay, the area of a triangle can be given by (1/2)*a*c*sin B. So, plugging in the known values, 3 = (1/2)*a*c*(3/5). Let me write that equation: (1/2)*a*c*(3/5) = 3. Simplifying, (3/10)*a*c = 3. So, multiplying both sides by 10/3, we get a*c = 10. So the product of a and c is 10.Now, I need to find a + c. I remember that in the Law of Cosines, b² = a² + c² - 2ac*cos B. We know b is 2, so 4 = a² + c² - 2ac*(4/5). Let's compute that: 4 = a² + c² - (8/5)ac. We already know that ac = 10, so substituting that in, we get 4 = a² + c² - (8/5)*10. Calculating (8/5)*10 is 16, so 4 = a² + c² - 16. Therefore, a² + c² = 20.Now, I have a² + c² = 20 and ac = 10. I need to find a + c. I recall that (a + c)² = a² + 2ac + c². So, substituting the known values, (a + c)² = 20 + 2*10 = 20 + 20 = 40. Therefore, a + c is the square root of 40, which simplifies to 2*sqrt(10).Wait, let me make sure I didn't make any mistakes. So, area is 3, which gives ac = 10. Then, using Law of Cosines, I found a² + c² = 20. Then, (a + c)² = a² + 2ac + c² = 20 + 20 = 40, so a + c = sqrt(40) = 2*sqrt(10). That seems right.Is there another way to check this? Maybe using the Law of Sines again. If I can find angles A and C, but I don't have enough information. Alternatively, since I know a*c = 10 and a + c = 2*sqrt(10), maybe I can solve for a and c individually. Let me set up the equations:Let a + c = 2*sqrt(10) and a*c = 10. These are the sum and product of a and c, so they are the roots of the quadratic equation x² - (2*sqrt(10))x + 10 = 0. Let me solve this quadratic: x = [2*sqrt(10) ± sqrt((2*sqrt(10))² - 4*1*10)] / 2. Calculating discriminant: (4*10) - 40 = 40 - 40 = 0. So x = [2*sqrt(10)] / 2 = sqrt(10). So both a and c are sqrt(10). Wait, that can't be right because in the first part, a was 5/3, which is approximately 1.666, and sqrt(10) is about 3.162. So that contradicts the first part. Hmm, maybe I made a mistake.Wait, no, in the first part, angle A was given as π/6, but in the second part, the area is given as 3, which might correspond to a different triangle configuration. So in the first part, a was 5/3, but in the second part, a and c are both sqrt(10). So they are different scenarios. So in the second part, a + c is 2*sqrt(10). That makes sense because when the area is given, it's a different triangle, so a and c can be different from the first part.Wait, but in the second part, are we still given that cos B is 4/5 and b is 2? Yes, so the triangle is still defined by those parameters, but with a different angle A or C? Or is it the same triangle? Wait, no, in the first part, angle A was given as π/6, but in the second part, the area is given as 3. So it's a different triangle with the same side b = 2 and cos B = 4/5, but different angles A and C. So in the second part, a and c are both sqrt(10), so a + c is 2*sqrt(10). That seems consistent.Wait, but if a and c are both sqrt(10), then using Law of Sines, a/sin A = c/sin C = b/sin B. So sqrt(10)/sin A = sqrt(10)/sin C = 2/(3/5) = 10/3. So sin A = sin C = sqrt(10)*(3/10) = (3*sqrt(10))/10 ≈ 0.9487. But sin A can't be more than 1, so that's okay because 3*sqrt(10)/10 ≈ 0.9487 < 1. So angles A and C would both be arcsin(3*sqrt(10)/10). Let me compute that: arcsin(0.9487) ≈ 71.57 degrees. So angles A and C are both approximately 71.57 degrees, and angle B is arccos(4/5) ≈ 36.87 degrees. Adding up, 71.57 + 71.57 + 36.87 ≈ 180 degrees. That checks out.So in the second part, a and c are both sqrt(10), so a + c is 2*sqrt(10). That makes sense.I think I've covered all the steps and checked for consistency. So the answers should be 5/3 for the first part and 2*sqrt(10) for the second part.

answer：First, I need to determine how many small bottles of 35 milliliters are required to fill a large bottle of 500 milliliters.I'll start by dividing the capacity of the large bottle by the capacity of one small bottle:[frac{500}{35} approx 14.2857]Since Jasmine cannot purchase a fraction of a bottle, I'll round up to the nearest whole number. This means she needs 15 small bottles to completely fill the large bottle.Therefore, the minimum number of small bottles Jasmine must buy is 15.

answer：Okay, so I have this function f(x) that's defined piecewise. It's 2^{-x} - 1 when x is less than or equal to 0, and it's log base 2 of x when x is greater than 0. I need to find the range of x values where f(x) is less than 1. The options are A: (-1, 2), B: (-1, infinity), C: {x | x > 0 or x < -2}, and D: {x | x > 2 or x < -1}.Alright, let's break this down. Since the function is piecewise, I should handle each piece separately.First, let's consider the case when x is less than or equal to 0. The function here is 2^{-x} - 1. I need to find when this is less than 1. So, set up the inequality:2^{-x} - 1 < 1Let me solve this step by step. Add 1 to both sides:2^{-x} < 2Hmm, 2^{-x} is the same as (1/2)^x. So, (1/2)^x < 2. Since 2 is 2^1, maybe I can write this as:(1/2)^x < 2^1But (1/2)^x is equal to 2^{-x}, so:2^{-x} < 2^1Since the bases are the same and greater than 1, the inequality of the exponents will hold. So, -x < 1. Multiply both sides by -1, remembering to flip the inequality:x > -1But wait, in this case, x is less than or equal to 0. So combining these, x must be greater than -1 and less than or equal to 0. So, the interval here is (-1, 0].Okay, that's the first part. Now, let's move on to the second piece where x is greater than 0. The function here is log base 2 of x. We need to find when this is less than 1.So, set up the inequality:log_2(x) < 1To solve this, I can rewrite it in exponential form. Remember that log_b(a) < c is equivalent to a < b^c when b > 1. Since 2 is greater than 1, this applies here.So, x < 2^1Which simplifies to:x < 2But in this case, x is greater than 0. So combining these, x must be greater than 0 and less than 2. So, the interval here is (0, 2).Now, to find the overall solution, I need to combine the two intervals from both cases. The first case gave me (-1, 0] and the second case gave me (0, 2). So, combining these, the overall interval is (-1, 2).Looking back at the options, option A is (-1, 2), which matches my solution. Let me just double-check to make sure I didn't make any mistakes.For the first case, when x ≤ 0, solving 2^{-x} - 1 < 1 led me to x > -1. That seems right because as x increases from -infinity towards 0, 2^{-x} decreases from infinity to 1, so subtracting 1 would go from just below infinity to 0. So, setting it less than 1 would mean x has to be greater than -1.For the second case, when x > 0, solving log_2(x) < 1 gives x < 2. That makes sense because log base 2 of 2 is 1, so anything less than 2 will give a log less than 1.Combining both intervals, I get (-1, 2), which is option A. So, I think that's the correct answer.

answer：Alright, so I have this polar equation: ( r = 3 tan theta sec theta ). I need to figure out what kind of curve this represents. The options are Line, Circle, Parabola, Ellipse, or Hyperbola. Hmm, okay, let's break this down step by step.First, I remember that in polar coordinates, ( r ) is the distance from the origin, and ( theta ) is the angle made with the positive x-axis. To identify the curve, it might help to convert this polar equation into Cartesian coordinates because I'm more familiar with identifying curves in that system.I recall that ( tan theta = frac{sin theta}{cos theta} ) and ( sec theta = frac{1}{cos theta} ). So, substituting these into the equation, I get:[ r = 3 times frac{sin theta}{cos theta} times frac{1}{cos theta} ]Simplifying that, it becomes:[ r = 3 times frac{sin theta}{cos^2 theta} ]Now, I want to convert this into Cartesian coordinates. I know that in Cartesian coordinates, ( x = r cos theta ) and ( y = r sin theta ). Also, ( r^2 = x^2 + y^2 ). Maybe I can manipulate the equation to express it in terms of ( x ) and ( y ).Starting from:[ r = 3 times frac{sin theta}{cos^2 theta} ]I can multiply both sides by ( cos^2 theta ) to get rid of the denominator:[ r cos^2 theta = 3 sin theta ]Now, let's express ( r cos^2 theta ) in terms of ( x ) and ( y ). Since ( x = r cos theta ), then ( cos theta = frac{x}{r} ). Therefore, ( cos^2 theta = left( frac{x}{r} right)^2 = frac{x^2}{r^2} ).Substituting back into the equation:[ r times frac{x^2}{r^2} = 3 sin theta ]Simplifying the left side:[ frac{x^2}{r} = 3 sin theta ]But ( sin theta = frac{y}{r} ), so substituting that in:[ frac{x^2}{r} = 3 times frac{y}{r} ]Multiplying both sides by ( r ) to eliminate the denominator:[ x^2 = 3y ]Okay, so now I have the equation ( x^2 = 3y ). This looks familiar. In Cartesian coordinates, the general form of a parabola that opens upward is ( x^2 = 4py ), where ( p ) is the focal length. Comparing that to my equation, ( x^2 = 3y ), it seems like ( 4p = 3 ), so ( p = frac{3}{4} ). That means the parabola opens upward with its vertex at the origin.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the original equation:1. Expressed ( tan theta ) and ( sec theta ) in terms of sine and cosine.2. Multiplied both sides by ( cos^2 theta ) to get ( r cos^2 theta = 3 sin theta ).3. Replaced ( cos^2 theta ) with ( frac{x^2}{r^2} ) and ( sin theta ) with ( frac{y}{r} ).4. Simplified to get ( x^2 = 3y ).Everything seems to check out. So, the equation ( x^2 = 3y ) is indeed a parabola. Therefore, the original polar equation represents a parabola.I think I'm confident with this conclusion. It's always good to verify by plotting a few points or considering the general shape, but given the algebraic manipulation and the resulting equation, it's clear that it's a parabola.