Rom

\(x_{n+1} = \dfrac{(x_n)^3-1}{4},~x_1 = -1\\ x_2 = \dfrac{(x_1)^3-1}{4} = \dfrac{(-1)^3-1}{4} = -\dfrac 1 2\\ x_3 = \dfrac{(x_2)^3-1}{4} = \dfrac{\left(-\dfrac 1 2\right)^3 - 1}{4} = -\dfrac{9}{32}\)

\(\text{The solution occurs when the recursion has stabilized on a single value}\\ \text{so just run the iteration until }x_{n+1} = x_n \text{ to 6 decimal places}\\ \text{Use a spreadsheet or something}\\ \text{You should get }x = -0.254101 \)

Now just copy this method for problem 2 which has a slightly different recursion relation.

\(r^2 = (-1 - 2)^2 + (6-3)^2 = 18\\ (x-2)^2 + (y-3)^2 = 18\\ x^2 - 4x + 4 + y^2 - 6y + 9 = 18\\ x^2 + y^2 -4x -6y-5 = 0\\ (-4)(-6)(-5) = -120\)

\(\text{let }s_n = \sum \limits_{k=1}^n~a_k = n(n+1)(n+2)\\ s_{10}=s_9+a_{10}\\ a_{10} = s_{10}-s_9 = (10)(11)(12) - (9)(10)(11) = 330\)

\(a+b+c=5,~b = 5-a-c\\ \dfrac{1}{a+b} + \dfrac{1}{b+c} = \dfrac{1}{5-c}+\dfrac{1}{5-a}\\ \text{If we want the minimum of this expression it seems clearly that }a,~c\\ \text{should be made as small as possible}\\ \text{Further, as the problem is symmetric in }a \text{ and }c\\ \text{we expect these two to be equal}\\ (a,b,c)=\left(\dfrac 3 2,~2,~\dfrac 3 2\right)\\ \dfrac{1}{a+b}+\dfrac{1}{a+c} =\dfrac 4 7\)

\(\text{If you grind through it using }a_1,~a_2 \text{ as the first two sequence elements you get}\\ s =\left\{a_1,a_2,a_1+a_2,a_1+2 a_2,2 a_1+3 a_2,3 a_1+5 a_2,5 a_1+8 a_2,8 a_1+13 a_2,13 a_1+21 a_2,21 a_1+34 a_2\right\}\\ \text{Summing we get }\\ \sum \limits_{n=1}^{10}~s_k = 55 a_1+88 a_2\\ s_7=5 a_1+8 a_2 = 6\\ a_2 = \dfrac{1}{8} \left(6-5 a_1\right)\\ \text{dumping this into the sum we find (like magic!)}\\ \sum \limits_{n=1}^{10}~s_k = 66\)

\(c = \lambda \nu = (0.3m)(100s^{-1}) = 30\dfrac m s\)

\(\text{odds of doing X} = \dfrac{P[\text{doing X}]}{P[\text{Not doing X}]}=\dfrac{P[\text{doing X}]}{1-P[\text{doing X}]}\)

\(\text{odds of pulling prize }=\dfrac{3}{4} \Rightarrow P[\text{pulling prize}] = \dfrac{3}{3+4}= \dfrac 3 7\)

\(P[\text{not pulling prize}]=1-P[\text{pulling prize}] = \\ 1 - \dfrac 3 7 = \dfrac 4 7\)

\(\text{if we've got vertex at }(0,-1) \text{ then}\\ y = a x^2-1\\ \text{we're told that }(3,3) \text{ is on the parabola}\\ 3 = a 3^2 - 1\\ 4 = 9a\\ a = \dfrac{4}{9}\\ y = \dfrac 4 9 x^2 - 1\)

just plug -2 in

if not for measuring the weight how else would i be able to know if it has evaporated?

No idea.

The water is going to be mixed in with everything, and absorbed by the sponge etc.

There's no visual measurement that will let know accurately know the water level.

Is weighing them a problem?