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r = 5z 15z = 3y (divide by 3 to both sides) 5z = y r = y
Closure: If x and y are any two elements of Rthen x*y is an element of R.Associativity: For and x, y and z in R, x*(y*z) = (x*y)*z and so, without ambiguity, this may be written as x*y*z.Identity element: There exists an element 1, in R, such that for every element x in R, 1*x = x*1 = x.Inverse element: For every x in R, there exists an element y in R such that x*y = y*x = 1. y is called the inverse of x and is denoted by x^-1.The above 4 properties determine a group.
A circle centre (0, 0) and radius r has equation x² + y² = r² The circle x² + y² = 36 has: r² = 36 → radius = 6
There is no such number. For suppose x was a number such that it keft the same remainder r, when divided by 99 and 204. Then x = 99a + r and x = 204b +r for some integers a and b. Consider y = x + 99*204 then y > x Now y = (99a + r) + 99*204 = 99*(a+204) + r so y leaves a remainder of r when divided by 99. Also y = (204b + r) + 99*204 = 204*(b+99) + r so y leaves a remainder of r when divided by 204. This contradicts the proposition that x is the maximum such number.
(x,y) -> (x-3, y+4)