answer:To solve the system of equations, we can use the method of substitution or elimination. Here, we will use the method of elimination. First, we can eliminate z by adding the second and third equations: (15 x-19 y+11 z)+(-17 y-4 z)=(-17)+20 15 x-36 y+7 z=3 Next, we can eliminate z again by adding the first and third equations: (-21 x-15 y-19 z)+(-17 y-4 z)=18+20 -21 x-32 y-23 z=38 Now, we have two equations with two variables: 15 x-36 y=3 -21 x-32 y=38 We can solve this system of equations using the method of substitution or elimination. Here, we will use the method of substitution. From the first equation, we can express x in terms of y: 15 x=3+36 y x=frac{3+36 y}{15} Substituting this expression for x into the second equation, we get: -21 left(frac{3+36 y}{15}right)-32 y=38 -7+126 y-32 y=38 94 y=45 y=frac{45}{94} Substituting this value of y back into the expression for x, we get: x=frac{3+36 left(frac{45}{94}right)}{15} x=frac{3+180}{15} x=frac{183}{15} Finally, we can substitute the values of x and y into the third equation to solve for z: -17 left(frac{45}{94}right)-4 z=20 -frac{765}{94}-4 z=20 -4 z=20+frac{765}{94} -4 z=frac{1940+765}{94} -4 z=frac{2705}{94} z=-frac{2705}{94 cdot 4} z=-frac{2705}{376} Therefore, the solution to the system of equations is: x=frac{183}{15}, y=frac{45}{94}, z=-frac{2705}{376} The answer is x=frac{13}{2}, y=frac{238}{263}, z=-frac{4653}{526}

answer:Using the Lagrange Multipliers method, we have: begin{align*} nabla f &= lambda nabla g leftlangle {2x - 2,12y} rightrangle &= lambda leftlangle {2left( {x - 1} right),2y} rightrangle end{align*} Solving for x, y, and λ, we get: begin{align*} x &= 1 y &= pm frac{1}{2} lambda &= 6 end{align*} Therefore, the critical points are {eq}(1, frac{1}{2}){/eq} and {eq}(1, -frac{1}{2}).{/eq}

answer:Let there be a box containing n+1 balls. We label the balls with the numbers 1,2,dots,n+1. We take out randomly m+1 balls one by one and denote the drawn numbers by X_{0},X_{1},dots,X_{m}. After drawing the first ball, the balls with a number less than X_{0} are painted white and all other remaining balls are painted black. Then we go on with drawing the other balls. If Y denotes the number of white balls among the balls with labels X_{1},dots,X_{m} then by symmetry it is evident that Y has uniform distribution on left{ 0,1,dots,mright} . Then: Pleft(Y=mright)=frac{1}{m+1} and the event left{ Y=mright} corresponds with the event W_{m} mentioned in the original answer. Therefore, the probability that the next ball drawn is white is frac{m+1}{m+2}.

answer:begin{split} int_{0}^{2 pi} (128 cos t sin t + 64 cos^2 t) , mathrm{d}t &= 128 int_{0}^{2 pi} cos t sin t , mathrm{d}t + 64 int_{0}^{2 pi} cos^2 t , mathrm{d}t &= 64 int_{0}^{2 pi} sin 2t , mathrm{d}t + 64 int_{0}^{2 pi} frac{1 + cos 2t}{2} , mathrm{d}t &= 64 left[ -frac{1}{2} cos 2t right]_{0}^{2 pi} + 32 left[ t + frac{1}{2} sin 2t right]_{0}^{2 pi} &= 64 left( -frac{1}{2} cos 4pi + frac{1}{2} cos 0 right) + 32 left( 2pi + frac{1}{2} sin 4pi - 0 - frac{1}{2} sin 0 right) &= 64 pi end{split} Therefore, int_{0}^{2 pi} (128 cos t sin t + 64 cos^2 t) , mathrm{d}t = 64 pi