Create your own 2 × 2 payoff matrix for a two-person zero-sum game with v − < v+. Solve your game (find a saddle point and the value of the game).

(Relativity) Draw the world line of a particle moving in the XY plane:

a) Describing a circle with constant speed V

b) Accelerating from the rest until reaching certain speed V

Complete the proof for the claim that any open ball B(x₀,r) in Euclidean space Rⁿ is homeomorphic to Rⁿ.

proof is given below the theorem. Show that suggested map g is in fact homeomorphism.

Theorem: Let X₀, X₁, and X₂ be topological spaces and let f: X₀ -> X₁ and g : X₁ -> X₂ be continuous functions. Then g∘f : X₀ -> X₂ is continuous.

proof : Suppose that V is open in X₂. Since g is continuous, g^-1(V) is open in X₁. Since f is continuous, f^-1(g^-1(V)) = (g∘f)^-1(V) is open in X₀. It follows that g∘f is continuous.

Let V = R4 and let U = hu1, u2i, where u1 =    1 2 0 −3    , u2 =     1 −1 1 0    . 1. Determine dimU and dimV/U. 2. Let v1 =    1 0 0 −3    , v2 =     1 2 0 0    , v3 =     1 3 −1 −6    , v4 =     −2 2 0 9    . For any two of the vectors v1,...,v4, determine whether they are in the same coset of U in V or not. 3. Find a basis of V that contains a basis of U. Hence, determine a basis of V/U. 4. Find two (distinct) elements of the coset e1 + U.

Drag measurements were taken for a 5 cm diameter sphere in water at 20 °C to predict the drag force of a 1 m diameter balloon rising in air with standard temperature and pressure. Given kinematic viscosity of water (v) = 1.0 X 10^-6 m²/s and kinematic viscosity of air (v) = 1.46 X 10^-5 m²/s.

1. Perform the Buckingham Pi theorem to generate a relationship for F_D as a function of the independent variables. Assume the drag F_D is a function of diameter (d), velocity (V), fluid density (ρ) and kinematic viscosity (v)

2.Determine the sphere velocity if the balloon was rising at 3 m/s

For S and S' in standard configuration, the Galilean transformations are:

x' = x - vt, y' = y, z' = z, t' = t

From the Lorentz transformations for v << c:

x' = x - vt, y' = y, z' = z, t' = t - vx/c^2

So it looks as if the Galilean transformations become increasingly accurate for:

vx -> 0, v << c

And exact for v = 0 for all x.

Yet, all text books I've come across state that the Galilean transformatons become more accurate for the condition v << c only.

So what are the conditions under which the Galilean transformations become more accurate and why?

A 3.0 m length of wire is made by welding the end of a 100 cm long silver wire to the end of a 200 cm long copper wire . Each piece of wire is 0.80 mm in diameter. The wire is at room temperature, so the resistivities are as given in the table below. A potential difference of 5.0 V is maintained between the ends of the 3.0 m composite wire.

(a) What is the current in the copper section?. A

(b) What is the current in the silver section? A

(c) What is the magnitude of vector E in the copper? V/m

(d) What is the magnitude of vector E in the silver? V/m

(e) What is the potential difference between the ends of the sliver section of wire? V

The velocity of an enzymatic reaction (expressed in nmoles x liter-1 x min-1 ) for different concentrations of substrate (S, expressed in moles x liter–1 )) are summarized in the following table.

1. The Hanes-Woolf plot represents [S]/v (y axis) versus [S] (x axis). Rearrange the Lineweaver-Burk to determine the linear equation for the Hanes-Woolf plot (y = ax + b, with y= [S]/v and x = [S]). What do the intercepts with the [S]/v and the [S] axis, and the slope of the straight line represent? (For example in the Lineweaver-Burk plot, the intercept of the straight line and the y-axis represent 1/Vmax) (6 pts)

2. The Woolf-Augustinsson-Hofstee plot represents v versus v/[S]. Rearrange the HenriMichaelis-Menten equation to determine the linear equation for the Woolf-AugustinssonHofstee plot. What do the intercepts with the v and the v/[S] axis, and the slope of the straight line represent? (6 pts)

3. Compare the numerical values of Km and Vmax calculated from each plot. (3 pts)

Answer True or False

1. For graph representation, adjacency Matrix is more efficiency than adjacency list in term of searching for edge.

2. Topological sort runs in O(|V| + |E|) where |V| is the number of vertices, and |E| is the number of edges in the input graph.

3. If vertex u can reach vertex v, and vertex v can reach vertex u, then vertices u and v are in the same Strongly-connected component (SCC).

4. The Bellman-Ford algorithm will run forever if the input graph has negative weights on the edges.

5. For a graph with only positive edge weights, Dijkstra's algorithm solves the single-source shortest path (SSSP) problem faster than Bellman-Ford on a graph.

6. Dynamic programming depends on the input problem having an optimal substructure.

7. The longest-common subsequence problem on strings of length n and m can be solved in time O(nm).

8. The adjacency matrix’s space complexity is O(|V|+|E|), for a graph G = .

9. Given any two strings S1 and S2, there is only one longest common subsequence (that is, the LCS is unique).

10. Depth-First Search runs in O(|V| + |E|) where |V| is the number of vertices, and |E| is the number of edges in the input graph.

Breadth-First Search finds the shortest distance --- in terms of the number of hops --- from source vertex to each other reachable vertex in a graph.

Kruskal's algorithm is a greedy algorithm.

For any graph G with positive edge weights, there is only 1 minimum-spanning-tree (MST) for G.

The time complexity of rod-cutting problem is Θ(n2)

2^(n+1)= O(2^n)

What would have to be changed in the code if the while statement were changed to:

while (menu == 5);

Code is as follows

1. #include <stdio.h>

2. void printHelp ()

3. {

4. printf ("\n");

5. printf ("a: a(x) = x*x\n");

6. printf ("b: b(x) = x*x*x\n");

7. printf ("c: c(x) = x^2 + 2*x + 7\n");

8. printf ("d: shrink(x) = x/2\n");

9. printf ("q: quit\n");

10. }

11. void a(float x)

12. {

13. float v = x*x;

14. printf (" a(%.2f) = %.2f^2 = %.2f\n", x, x, v);

15. } // end function a

16. void b(float x)

17. {

18. float v = x*x*x;

19. printf (" b(%.2f) = %.2f^3 = %.2f\n", x, x, v);

20. } // end function b

21. void c(float x)

22. {

23. float v = x*x + 2*x + 7;

24. printf (" c(%.2f) = %.2f^2 + 2*%.2f + 7 = %.2f\n",

25. x, x, x, v);

26. } // end function c

27. void shrink(float x){

28. float v = x/2;

29. printf("shrink(%.2f) = %.2f/2 = %.2f\n", x, x, v);

30. }//end of function shrink

31. int menu ()

32. {

33. char selection;

34. float x;

35. printHelp ();

36. scanf ("%s", &selection);

37. if (selection == 'q')

38. return 1;

39. scanf ("%f", &x);

40. if (selection == 'a')

41. a(x);

42. if (selection == 'b')

43. b(x);

44. if (selection == 'c')

45. c(x);

46. if(selection == 'd')

47. shrink(x);

48. return 0;

49. } // end function menu

50. int main()

51. {

52. while (menu() == 0);

53. printf ("... bye ...\n");

54. return 0;

55. } // end main