Want this question answered?
Be notified when an answer is posted
Chat with our AI personalities
r=0 is the solution...
The concept of a distance, or metric, between the elements of a set is well defined in topology. Let B be any set with x, y, z Є B, and let D be a function from the Cartesian product, B X B, into the set of real numbers, R. D is called a metric on B if the following four statements hold:1) D(x,y) ≥ 0 for all x, y2) D(x,y) = D(y,x) for all x, y3) D(x,y) = 0 if and only if x = yfor all x, y4) D(x,y) + D(y,z) ≥ D(x,z) for all x, y, zIn this case, the set B having metric D is called a metric space, and is often notated as B, D.For more information and related definitions, see the links below.
Divide both sides by 6: 3 - r = 0, so r = 3 satisfies the equation.
R2 - 16 = 0 R2 = 16 Take the square root of each side to get rid of the square on R. R = 4
int matrix[][]; // the matrix to find the max in int max = matrix[0][0]; int r,c; for(r = 0; r < 3; ++r) { for(c = 0; c < 3; ++c) { if(matrix[r][c] > max) { max = matrix[r][c]; } } } // max is now the maximum number in matrix