Recent content by physics=world

1. Simplify the expression: F = xyz' + xy'z' + x'yz + xyz2. Postulates and theoremsThe Attempt at a Solution F = xyz' + xy'z' + x'yz + xyz = x(yz' + y'z') + yz(x' + x) (Distributive) = x(yz' + y'z') + yz.1 (Complement) = x(yz' + y'z') + yz (identity) This is where I need...

1. prove that: X'Y'Z + X'YZ' + XY'Z' + XYZ = (X⊕Y)⊕Z Homework Equations Use postulates and theorems. The Attempt at a Solution X'Y'Z + X'YZ' + XY'Z' + XYZ (original expression) X'Y'Z + X'YZ' + X(Y'Z' + YZ) (distributive) X'Y'Z + X'YZ' + X.1 (complement) X'Y'Z + X'YZ' + X (identity)[/B]...

I see what your talking about. The "t" in both equation is not suppose to be the value taken from the user. Thanks for helping me! :)

Are you saying get rid of my input "t"?

#include <iostream> #include <iomanip> #include <cmath> using namespace std; int main() { cout << "Enter the total time: "; double t; cin >> t; cout << "Enter the step-size: "; double step; cin >> step; double steps = ceil(t/step); const double a = 9.806...

________________a The 6 A is 9 A for the problem I'm solving for. Homework Equations Only using KCL The Attempt at a Solution I tried using KCL @ node a: 9 = io + i0/4 I get io = 7.2 A. The answer in the book is 6A. What am I doing wrong?

denominator: x4 + y2 = r4cos4(theta) + sin2(theta)

Can I choose y = 0 and find the limit. Then, y = x2 and find the limit. Will that show that it approaches different limits and Therefore, it would not exist?

After solving I get cos2(theta) / sin(theta)

Homework Statement Find the limit: the limit of (x2y)/(x4 + y2) as (x,y) approaches (0,0) Homework Equations The Attempt at a Solution I took the limit of the numerator and denominator separately. The numerator equals to 0 as well as the denominator. So, I get the...

sorry. typo. the ln(x) was after integration of (1/x)

1. Find a second solution of the differential eq. by using the formula. xy" + y' = 0 ; y1 = ln(x) 2. y2 = y1(x) ( ∫(e-∫P(x)dx) / (y1(x)2) )dx 3. I found the p(x): p(x) = 1/x and then I plug in everything into the formula: y2 = ln(x) ∫(e-∫((1)/(x))dx) /...

After integrating I get: ce(x) + 2ln(x) Is this the answer? What about the value for c?

1. Use Abel's Formula to find the Wronskian of two solutions of the given differential equation without solving the equation. x2y" - x(x+2)y' + (t + 2)y = 0 2. Abel's Formula W(y1, y2)(x) = ce-∫p(x)dx3. I put it in the form of y" + p(x)y' + q(x)y = 0 to find my p(x) to use for Abel's...

okay now at x = 1 y = a + b + c y' = 5a + +3b + c