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What is the firing order for a 1983 4 cylinder Nissan pickup truck with 8 spark plugs?
CCW rotation 1-3-4-2 / 1 ex, 1 int / 3 int, 3 ex / 4 int, 4 ex / 2 ex, 2 int
How do you do matrix multiplication in c using structure?
You took an example of The Product AB is determined as the dot products of the ith row in A and the jth column in B,placed in ith row and jth column of the resulting m x p matrix C. so: this may help you. #include <stdio.h> #include <stdlib.h> // function prototypes void Matrix_Mult( int a1[][3], int a2[][4], int a3[][4] ); void Matrix_MultAlt( int a1[][3], int a2[][4], int a3[][4] ); int dot3(const int a1[][3], const int a2[][4], int row, int col); void PrnNx4 (int ar[][4], int n); //--------------------------------------------------------------------------------- // Function: main(void) // Description: // demonstration of Matrix Multiplication // // Programmer: Paul Bladek // // Date: 10/31/2001 // // Version: 1.0 // // Environment: Hardware:IBM Pentium 4 // Software: Microsoft XP with .NET framework for execution; // Compiles under Microsoft Visual C++.Net 2005 // // Calls: Matrix_Mult(int a1[][3], int a2[][4], int a3[][4]) // Matrix_MultAlt(int a1[][3], int a2[][4], int a3[][4]) // PrnNx4(int ar[][4] // // // Parameters: int a1[][3] -- left matrix // int a2[][4] -- right matrix // int a3[][4] -- answer matrix // // Returns: EXIT_SUCCESS // ------------------------------------------------------------------------------ int main(void) { int A[2][3] = {{1, 3, 4}, {2, 0, 1}}, B[3][4] = {{1, 2, 3, 1}, {2, 2, 2, 2}, {3, 2, 1, 4}}, C[2][4] = {{0, 0, 0, 0}, {0, 0, 0, 0}}; Matrix_Mult(A, B, C); PrnNx4(C, 2); Matrix_MultAlt(A, B, C); // alternate form that calls dot3 PrnNx4(C, 2); return EXIT_SUCCESS; } //--------------------------------------------------------------------------------- // Function: Matrix_Mult(int a1[][3], int a2[][4], int a3[][4]) // Description: // multiplies a 2X3 matrix by a 3X4 matrix // // Programmer: Paul Bladek // // Date: 10/31/2001 // // Version: 1.0 // // Environment: Hardware:IBM Pentium 4 // Software: Microsoft XP with .NET framework for execution; // Compiles under Microsoft Visual C++.Net 2005 // // Calls: None // // Called By: main() // // Parameters: int a1[][3] -- left matrix // int a2[][3] -- right matrix // int a3[][3] -- answer matrix // ------------------------------------------------------------------------------ void Matrix_Mult(int a1[][3], int a2[][4], int a3[][4]) { int i = 0; int j = 0; int k = 0; for(i = 0; i < 2; i++) for( j = 0; j < 4; j++) for( k = 0; k < 3; k++) a3[i][j] += a1[i][k] * a2[k][j]; } //--------------------------------------------------------------------------------- // Function: Matrix_MultAlt(int a1[][3], int a2[][4], int a3[][4]) // Description: // multiplies a 2X3 matrix by a 3X4 matrix -- Alternate Form // // Programmer: Paul Bladek // // Date: 10/31/2001 // // Version: 1.0 // // Environment: Hardware:IBM Pentium 4 // Software: Microsoft XP with .NET framework for execution; // Compiles under Microsoft Visual C++.Net 2005 // // Calls: dot3(const int a1[][3], const int a2[][4], int row, int col) // // Called By: main() // // Parameters: int a1[][3] -- left matrix // int a2[][3] -- right matrix // int a3[][3] -- answer matrix // ------------------------------------------------------------------------------ void Matrix_MultAlt(int a1[][3], int a2[][4], int a3[][4]) { int i = 0; int j = 0; for( i = 0; i < 2; i++) for( j = 0; j < 4; j++) a3[i][j] = dot3(a1, a2, i, j); } //--------------------------------------------------------------------------------- // Function: dot3(const int a1[][3], const int a2[][4], int row, int col) // Description: // dot product of a1 row and a2 col // // Programmer: Paul Bladek // // Date: 10/31/2001 // // Version: 1.0 // // Environment: Hardware:IBM Pentium 4 // Software: Microsoft XP with .NET framework for execution; // Compiles under Microsoft Visual C++.Net 2005 // // Calls: None // // Called By: Matrix_MultAlt(int a1[][3], int a2[][4], int a3[][4]) // // Parameters: int a1[][3] -- left matrix // int a2[][3] -- right matrix // int row -- the row number // int col -- the column number // // Returns: the dot product // ------------------------------------------------------------------------------ int dot3(const int a1[][3], const int a2[][4], int row, int col) { int k = 0; int sum = 0; for( k = 0; k < 3; k++) sum += a1[row][k] * a2[k][col]; return sum; } //--------------------------------------------------------------------------------- // Function: PrnNx4(int ar[][4], int n) // Description: // prints out an NX4 matrix // // Programmer: Paul Bladek // // Date: 10/31/2001 // // Version: 1.0 // // Environment: Hardware:IBM Pentium 4 // Software: Microsoft XP with .NET framework for execution; // Compiles under Microsoft Visual C++.Net 2005 // // Called By: main() // // Parameters: int ar[][4] -- matrix to print // int n -- number of elements // ------------------------------------------------------------------------------ void PrnNx4 (int ar[][4], int n) { int i = 0; int j = 0; for(i = 0; i < n; i++) { for( j = 0; j < 4; j++) printf("%4d", ar[i][j]); putchar('\n'); } }
What is the integral of sin x cubed?
= cos(x)-(cos3(x))/3 * * * * * Right numbers, wrong sign! Int(sin3x)dx = Int(sin2x*sinx)dx = Int[(1-cos2x)*sinx]dx = Int(sinx)dx + Int[-cos2x*sinx]dx Int(sinx)dx = -cosx . . . . . (I) Int[-cos2x*sinx]dx Let u = cosx, the du = -sinxdx so Int(u2)du = u3/3 = 1/3*cos3x . . . . (II) So Int(sin3x)dx = 1/3*cos3x - cosx + C Alternatively, using the multiple angle identities, you can show that sin3x = 1/4*[3sinx - sin3x] which gives Int(sin3x)dx = 1/4*{1/3*cos(3x) - 3cosx} + C
How many times will 4 go int 6040?
1510 times
How do you initialise arrays?
int a[] = {1, 2, 3, 4, 5};
How yo create a 3D array?
1D:int[] a = new int[5];2Dint[][] a = new int[5][5];3D:int[][][] a = new int[5][5][5];The 'a' is the name of the array, and 5 is the size of the arrayExample:int[][][] a = new int[5][3][7];a[3][1][4] = 18;System.out.println(a[3][1][4]);The number '18' will be printed on screen.
How much can 4 go into 14?
4 times
Slope equals -4 and y-int equals 3?
-7
How much can 4 go into 12?
3 times
How much does 3 go into 4?
3 goes into 4 one time, with 1/3 left over.
Write a java program to show following pattern 5 5 4 5 4 3 5 4 3 2 5 4 3 2 1?
// num = the highest number to start with final int num = 5; // Go through each value from num through 1. for (int n = num; n > 0; --n) { // For each value, print each value from num all the way down to the value. for (int m = num; m >= n; --m) { System.out.print(m); System.out.print(' '); } }
When do you use call by reference?
You cannot do that in C, but you can pass pointers to objects, like this: void swap (int *a, int *b) { int tmp=*a; *a=*b; *b= tmp; } int x=3, y=4; swap (&x, &y);
