If x + 2xy + 3y2 = 1, prove that (x + 3y)3 d2y/dx2 + 2 + 0

Here are the solutions to the questions from your image: Probability Question: There are 11 blue balls and 7 red balls, making a total of 11 + 7 = 18 balls. We are picking two balls without replacement. Step 1: Calculate the probability of picking a blue ball first, then a red ball second. P(1st Blue) = (11)/(18) After picking one blue ball, there are 17 balls left, with 7 red balls remaining. P(2nd Red | 1st Blue) = (7)/(17) P(Blue then Red) = (11)/(18) × (7)/(17) = (77)/(306) Step 2: Calculate the probability of picking a red ball first, then a blue ball second. P(1st Red) = (7)/(18) After picking one red ball, there are 17 balls left, with 11 blue balls remaining. P(2nd Blue | 1st Red) = (11)/(17) P(Red then Blue) = (7)/(18) × (11)/(17) = (77)/(306) Step 3: Add the probabilities of these two mutually exclusive events to find the total probability of picking balls of different colours. P(Different Colours) = P(Blue then Red) + P(Red then Blue) P(Different Colours) = (77)/(306) + (77)/(306) = (154)/(306) Step 4: Simplify the fraction. (154)/(306) = (77)/(153) The probability that both balls are of different colours is (77)/(153). Implicit Differentiation Proof: Given the equation x^2 + 2xy + 3y^2 = 1. We need to prove that (x+3y)^3 (d^2y)/(dx^2) + 2 = 0. Step 1: Differentiate the given equation implicitly with respect to x. (d)/(dx)(x^2) + (d)/(dx)(2xy) + (d)/(dx)(3y^2) = (d)/(dx)(1) Using the product rule for 2xy and the chain rule for 3y^2: 2x + (2 · y + 2x · (dy)/(dx)) + (3 · 2y · (dy)/(dx)) = 0 2x + 2y + 2x (dy)/(dx) + 6y (dy)/(dx) = 0 Step 2: Solve for (dy)/(dx). (2x + 6y) (dy)/(dx) = -2x - 2y (dy)/(dx) = (-2x - 2y)/(2x + 6y) = (-2(x+y))/(2(x+3y)) (dy)/(dx) = -(x+y)/(x+3y) Step 3: Differentiate (dy)/(dx) again with respect to x using the quotient rule. Let u = -(x+y) and v = x+3y. Then (du)/(dx) = -(1 + (dy)/(dx)) and (dv)/(dx) = 1 + 3(dy)/(dx). The quotient rule is (d^2y)/(dx^2) = (v du)/(dx) - u (dv)/(dx)v^2. (d^2y)/(dx^2) = ((x+3y) ( -(1 + dy)/(dx)) ) - (-(x+y)) ( 1 + 3(dy)/(dx) )(x+3y)^2 (d^2y)/(dx^2) = (-(x+3y)(1 + dy)/(dx)) + (x+y)(1 + 3(dy)/(dx))(x+3y)^2 Step 4: Substitute (dy)/(dx) = -(x+y)/(x+3y) into the numerator. Numerator: -(x+3y)(1 - (x+y)/(x+3y)) + (x+y)(1 - 3(x+y)/(x+3y)) = -(x+3y)((x+3y - (x+y))/(x+3y)) + (x+y)((x+3y - 3(x+y))/(x+3y)) = -(x+3y)((2y)/(x+3y)) + (x+y)((x+3y - 3x - 3y)/(x+3y)) = -2y + (x+y)((-2x)/(x+3y)) = (-2y(x+3y) - 2x(x+y))/(x+3y) = (-2xy - 6y^2 - 2x^2 - 2xy)/(x+3y) = (-2x^2 - 4xy - 6y^2)/(x+3y) = (-2(x^2 + 2xy + 3y^2))/(x+3y) From the original equation, we know that x^2 + 2xy + 3y^2 = 1. So, the numerator simplifies to (-2(1))/(x+3y) = (-2)/(x+3y). Step 5: Substitute the simplified numerator back into the expression for (d^2y)/(dx^2) and rearrange. (d^2y)/(dx^2) = (-2)/(x+3y)(x+3y)^2 (d^2y)/(dx^2) = (-2)/((x+3y)^3) Multiply both sides by (x+3y)^3: (x+3y)^3 (d^2y)/(dx^2) = -2 Add 2 to both sides: (x+3y)^3 (d^2y)/(dx^2) + 2 = 0 This proves the statement. Trigonometry Question: The question asks to "find the value of (4 y)/( y - y)" given 0^ y 90^. This question is incomplete as there is no additional information or equation provided to determine a specific value for y or the expression. Without more context, the expression cannot be simplified to a numerical value. Last free one today — make it count tomorrow, or type /upgrade for unlimited.