# Tag Info

I'm on a brief break so forgive any mistakes or any lack of utility, but here is an attempt to apply some formalization to the problem. Let $X$ and $Y$ be two sets. In this case let $X = \{ x | x \text{ is a possible month} \}$ and $Y = \{ y|y \text{ is a possible day} \}$. Then the choices Cheryl gives is $Z \subset X \times Y$. Lets also say an answer to ...
For a slightly systematized solution consider the following table which is known to both A and B: $$\matrix{&&{\rm B:}\cr &&14&15&16&17&18&19\cr {\rm A:}&{\rm May}&&\bullet&\bullet&&&\bullet\cr &{\rm Jun}&&&&\bullet&\bullet\cr &{\rm ... 2 As explained in a comment, typing there the line streamplot[{0.01*7*x - 0.0001*4*x*y,-0.01*8*y + 0.0001*5*x*y},{x, -0.1, 600}, {y, -0.1, 600}] produces Lazy people would rather type the line streamplot[{7x - 4xy,-8y + 5xy},{x, 0, 6}, {y, 0, 6}] which, for good reasons, produces the same diagram, only rescaled: 1 No the initial ambient is the outside temperature, you have:$$70=C_2e^{c\times 0}+10=C_2+10$$so C_2=60. Also the way you have set up your equations k<0 since you need dT/dt<0 when T>T_m assuming T_m is the ambient temperature. 1 What percentage of a wall would they paint together in an hour? Each A-class painter would paint \frac{3}{5} of a wall, since he would need \frac{5}{3} hour (or 1h40) for a full wall. Each B-class painter would paint a full wall. Each C-class would paint \frac12 of a wall. So, all the nine painters would paint, in an hour: ... 1 You are confusing two different types of problems. Cheryl's Birthday Problem is not a variation of the Zebra Problem. In the Zebra Problem, you are given just one big list of statements. In the Cheryl's Birthday Problem, you have two independent agents/players (Albert and Bernard), with two separate states of knowledge. The Zebra Problem is a classic ... 1 Differentiating the first equation,$$ R''=-aR'+bJ'. $$Substituting from the second equation and then the first equation,$$ R''+aR'=b(-aJ+bR) = b^2 R -a(bJ)=b^2R - a(R'+aR) $$Hence$$ R''+2aR'+(a^2-b^2)R=0, which has general solution Ae^{-at}\cosh{bt} + Be^{-at} \sinh{bt} Since the equations are linear, you can insert this into the J equation to ... 1 Your ODE is one-dimensional, thus also the phase portrait is one-dimensional. You can of course embed this one axis into the graph of the system function. The essential information is to determine the roots of the system function, those are the stationary points, and the sign of the system function on the intervals generated by the root, those determine the ... 1 Start from begin{align} d(O, P) = \sqrt{\frac{\hat{x_1^2}}{\hat{s_{11}}} + \frac{\hat{x_2^2}}{\hat{s_{22}}}}. \end{align} You want to convert that to begin{align} d(O, P) = \sqrt{a_{11}x_1^2 + 2a_{12}x_1x_2 + a_{22}x_2^2} \end{align} and work out the values of the a_{ij}. To do so, I'll show that the things under the square roots are equal, that is, I'll ... 1 The first one is simple, you just read off the answer from the problem statement. I will leave it to you. The second one is a little bit harder. The tangent line to g(x) at x=x_0 passes through (x_0,g(x_0)) with slope g'(x_0). The normal line passes through (x_0,g(x_0)) and is perpendicular to the tangent line, so it has slope -\frac{1}{g'(x_0)} ... 1 here is another way to look at question (b). look at the tangent at (x,y). since the normal goes through (0,1), it has slope \frac{y-1}x. therefore the tangent at (x, y) has slope -\frac{y-1}x. we can write the differential equations as\frac{dy}{dx} = -\frac{y-1}x \to y-1 = \frac Cx\to y = 1+ \frac Cx. $$1 First, your equation is correct, m\dot v=mg-bv. The solution of which with v(0)=0 is$$ v(t)-\frac{mg}b= e^{-b/m·t}\left(v_0-\frac{mg}b\right)\iff v(t)=\frac{mg}b\left(1-e^{-b/m·t}\right) \approx g·t-\frac{gb}{2m}·t^2,  the latter approximation for small $t$, corresponding in first order to the free fall without friction. In the approximation, the ...