All the multiples of 650 - numbers of the form k*650 in which k is any integer.
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#include<stdio.h> #include<stdlib.h> #include<math.h> #include<conio.h> void main(void) { int K, P, C, J; double A[100][101]; int N; int Row[100]; double X[100]; double SUM, M; int T; do { printf("Please enter number of equations [Not more than %d]\n",100); scanf("%d", &N); } while( N > 100); printf("You say there are %d equations.\n", N); printf("From AX = B enter elements of [A,B] row by row:\n"); for (K = 1; K <= N; K++) { for (J = 1; J <= N+1; J++) { printf(" For row %d enter element %d please :\n", K, J); scanf("%lf", &A[K-1][J-1]); } } for (J = 1; J<= N; J++) Row[J-1] = J - 1; for (P = 1; P <= N - 1; P++) { for (K = P + 1; K <= N; K++) { if ( fabs(A[Row[K-1]][P-1]) > fabs(A[Row[P-1]][P-1]) ) { T = Row[P-1]; Row[P-1] = Row[K-1]; Row[K-1] = T; } } if (A[Row[P-1]][P-1] 0) { printf("The matrix is SINGULAR !\n"); printf("Cannot use algorithm --- exit\n"); exit(1); } X[N-1] = A[Row[N-1]][N] / A[Row[N-1]][N-1]; for (K = N - 1; K >= 1; K--) { SUM = 0; for (C = K + 1; C <= N; C++) { SUM += A[Row[K-1]][C-1] * X[C-1]; } X[K-1] = ( A[Row[K-1]][N] - SUM) / A[Row[K-1]][K-1]; } for( K = 1; K <= N; K++) printf("X[%d] = %lf\n", K, X[K-1]); getch(); }
p = r - c r - c = p r - c - r = p - r -(-c) = -(p) c = -p
Isocost Lines Recall that a universally accepted objective of any firm is to maximise profit. If the firm maximises profit, it will necessarily minimise cost for producing a given level of output or maximise output for a given level of cost. Suppose there are 2 inputs: capital (K) and labour (L) that are variable in the relevant time period. What combination of (K,L) should the firm choose in order to maximise output for a given level of cost? If there are 2 inputs, K,L, then given the price of capital (P k ) and the price of labour (P L ), it is possible to determine the alternative combinations of (K,L) that can be purchased for a given level of expenditure. Suppose C is total expenditure, then C= P L * L + P k * K This linear function can be plotted on a graph. ISOCOST K C/P k A B O C/P L L • N • P Figure 7.7: Isocost line If only capital is purchased, then the maximum amount that can be bought is C/P k shown by point A in figure 7.7. If only labour is purchased, then the maximum amount of labour that can be purchased is C/P L shown by point B in the figure. The 2 points A and B can be joined by a straight line. This straight line is called the isocost line or equal cost line. It shows the alternative combinations of (K,L) that can be purchased for the given expenditure level C. Any point to the right and above the isocost is not attainable as it involves a level of expenditure greater than C and any point to the left and below the isocost such as P is attainable, although it implies the firm is spending less than C. You should verify that the slope of the isocost is 1 - k L k L P P P C * P C ∆ L ∆ K EXAMPLE : Consider the following data: P L = 10, P k = 20 Total Expenditure = 200. Let us first plot the various combinations of K and L that are possible. We 1 The nagative sign is due to the fact that the slope of the isocost is negative.15 consider only the case when the firm spends the entire budget of 200. The alternative combinations are shown in the figure (7.8). P r o d u c t i o n F u n c t i o n K A 10 9 8 7 6 5 4 3 2 1 O B L C 2 4 6 8 1 0 1 2 1 4 1 6 1 8 2 0 Figure 7.8: Shifting of Isocost The slope of this isocost is -½. What will happen if labour becomes more expensive say P L increases to 20? Obviously with the same budget the firm can now purchase lesser units of labour. The isocost still meets the Y-axis at point A (because the price of capital is unchanged), but shifts inwards in the direction of the arrow to meet the X-axis at point C. The slope therefore changes to -1. You should work out the effect on the isocost curve on the following: (i) decrease in the price of labour (ii) increase in the price of capital (iii) decrease in the price of capital (iv) increase in the firms budget with no change in the price of labour and capital
K. P. A. C. Lalitha was born on 1947-02-25.
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All the multiples of 650 - numbers of the form k*650 in which k is any integer.
Assume Plaintext = P, Ciphertext = C, and the Key = K. C = P * K Therefore, multiply both sides by the inverse of P and you will get: C * P^(-1) = K Or, (ciphertext) * (inverse plaintext) = key If the size of the key is known as well, then use that same size when creating P and C matrices.
MagnesiumMelting point : 650 °C , 923 K , 1202 °FBoiling point : 1091 °C, 1363 K, 1994 °F
The melting point of magnesium is approximately 650 degrees Celsius (1202 degrees Fahrenheit).
Melting Point: 923 K (650 °C, 1202 °F)
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C-a-m-p r-o-c-k
The letters c e k o p t can be rearranged to make the word pocket.
I'm going to assume you mean combinations - the unique set of these letters in any order with no sequence repeated. With these letters, there are 60 possible combinations. To see the maths behind this, try typing "permutations of {c,c,c,p,p,k}" into wolfram alpha.