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- CashOrNothingOption = function(S, k, Time, r, q, sigma) {d1 = (log(S / k) + (r - q + sigma ^ 2 / 2) * Time) / (sigma * sqrt(Time))d2 = d1 - sigma * sqrt(Time)if(k>S){return (k * exp (-r * Time) * (-d2))}elseprint("zero")}int functionC (int n) { int i, j, sumC = 0; for (i=n; i > 0; i=i-5) for (j=1; j 0) { if (functionC(n) % 2 == 0) { for (i=m; i > 0; i=i/3) sumE++; } else (10) Asymptotic runtime of functionE { for (i=m; i > 0; i=i-3) sumE--; } n--; } return sumE;5. F has one-tó-one correspondence. F {(-5, p), (10, w), (15, n), (20, m), (25, u), (30, b)}. Draw an arrow diagram for F. %3D
- TM M = (Q, E, I, 6, 90, 9a, qr), where Q = {90, 91, 92, 9a, 9r}, Σ = {0, 1}, r = {0, 1, L}, and 8 is: 8(qo, U) = (qr, U, R) 8(90, 0) = 8(go, 1) (91, 0, R) (qo, 1, R) = 8(g1, L) = (ga, U, R) 8(91,0) = (91, 0, R) 8(91, 1) = (92, 1, R) = (92, U, R) 8(92, U) 8(92, 0) = (90, 0, R) 8(92, 1) = (92, 1, R) i. Prove that M is NOT a decider. ii. Mathematically describe the language A that M recognises. Prove that A ≤ L(M). iii. Prove A = L(M). iv. Is A Turing-decidable? [Give clear reasons for your answer. No need for a formal proof.]SOLVE IN PYTHON A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [],{},and ().A matching pair of brackets is not balanced if the set of brackets it encloses are not matched. For example, {[(])} is not balanced because the contents in between { and } are not balanced. Write a program that take string of bracket pairs from user and reply Yes or No after evaluate. Your program should working like This.{[()]} Yes{[(])} No{{[[(())]]}} YesPEARHEAEGANEHNA 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 let x = "me"; let y = 3; let z = myFun (x, y) alert(z); z=plusOne(plusOne (y)); alert(x); function myFun (a,b){ alert (b); moreFun(); return b* a. length; } fucntion moreFun () { } alert(y); function plusOne (c) return c + 1
- Use set builder notation to define the set of points (x-y coordinates/ ordered pairs) on the curve defined by y = x^2. Where x and y are integersGiven A={1,2,3,4,56}, B= {2,4,5,6} and C= (1,6,7} 1. B' U C'=HW2 B-spline equation is : P(u) =PN (u), %3D j=0 If the order is 4, and p0= [1; 2] p1= [-2 ;2] P2=[ 5;-5] P3=[6; -2] P4=[0; 2] Write matlab code to find N1,4 at u=0.5
- Q4/Full the following blanks (1101 1010)BCD(5211) = ( )BCD(3321)=()EX-3=()16 * (1101 1010)BCD(5211) = (1110 1100)BCD(3321)=(1010110)EX-3= (A6)16 (1101 1010)BCD(5211) = (1110 0111)BCD(3321)=(1011001)EX-3= (56)16 (1101 1010)BCD(5211) = (1101 0111)BCD(3321)=(1011001)EX-3= (2B)16 (1101 1010)BCD(5211) = (1011 1110)BCD(3321)=(1000111)EX-3= (44)16 None of them IIQuadratic Root Solver For a general quadratic equation y = ax? + bx + c, the roots can be classified into three categories depending upon the value of the discriminant which is given by b2 - 4ac First, if the discriminant is equal to 0, there is only one real root. Then, if the discriminant is a positive value, there are two roots which are real and unequal. The roots can be computed as follows: -b+ Vb? – 4ac 2a Further, if the discriminant is a negative value, then there are two imaginary roots. In this case, the roots are given by b ь? - 4ас 2a 2a Programming tasks: A text file, coeff.txt has the following information: coeff.txt 3 4 4 4 1 4 Each line represents the values of a, b and c, for a quadratic equation. Write a program that read these coefficient values, calculate the roots of each quadratic equation, and display the results. Your program should perform the following tasks: • Check if the file is successfully opened before reading • Use loop to read the file from main…Need help with changing the following code to meet the request in the image for MATLAB. THank you. syms z numerator = z^2 + 2*z; denominator = z^2 + 0.4*z - 0.12; zeros = solve(numerator == 0, z); poles = solve(denominator == 0, z); figure; hold on; plot(real(zeros), imag(zeros), 'o', 'MarkerSize', 10); plot(real(poles), imag(poles), 'x', 'MarkerSize', 10); grid on; xlabel('Real part'); ylabel('Imaginary part'); title('Zero-Pole Plot'); legend('Zeros', 'Poles'); if all(abs(poles) < 1) disp('The system is stable.'); else disp('The system is unstable.'); end