This problem asks us to find 2 numbers, n1 and n2, with the following relations between them: * n2 = 4 n1 * n1 + n2 = 45 Substituting the first equation into the second one gives us: * n1 + 4n1 = 45 which gives us * n1 = 9. We can now use this solution to find n2 with the first equation * n2 = 4 n1 = 36 So the first number is 4, the second number is 36.
If you have a number x, then twice that number is 2*x and so the expression means 42 - 2*x
The equation is: 0.5*(n2-3n) = diagonals whereas n is the number of sides of the polygon
The number 2 means that there are 2 chlorine atoms attached to one magnesium atom in a molecule of magnesium chloride
The formula for finding the number of distinguishable permutations is: N! -------------------- (n1!)(n2!)...(nk!) where N is the amount of objects, k of which are unique.
1/2*(n2-3n) = 665 diagonals where n means number of sides
0 in N2
Each shell has a total of n2 orbitals, where n is the principal quantum number. For N shells the total orbitals is therefore :- N2 + (N-1)2 + (N-2)2 +....+1
1/2*(n2-3n) where n is the number of sides of the polygon.
Let n be the number of sides: 1/2*(n2-3n) = diagonals 1/2*(n2-3n) = 902 Multiply both sides by 2 and form a quadratic equation: n2-3n-1804 = 0 Solving the above by means of the quadratic equation formula gives a positive value for n as 44 Therefore the polygon has 44 sides
The sum of any two successive triangular number is a square number.The nth triangular number is n(n+1)/2 = (n2 +n)/2The next triangular number is (n+1)*(n+2)/2 = (n2 + 3n + 2)/2Their sum is (2n2 + 4n + 2)/2 = n2 + 2n + 1 = (n + 1)2This is easy to visualise. For example T3 + t4 = 42[T3, the 3th triangular number represented by X,t4, the 4th triangular number represented by x]XXX_____xXX_____xxX_____xxx_____xxxxBring them together and you have:XXXxXXxxXxxxxxxxThe sum of any two successive triangular number is a square number.The nth triangular number is n(n+1)/2 = (n2 +n)/2The next triangular number is (n+1)*(n+2)/2 = (n2 + 3n + 2)/2Their sum is (2n2 + 4n + 2)/2 = n2 + 2n + 1 = (n + 1)2This is easy to visualise. For example T3 + t4 = 42[T3, the 3th triangular number represented by X,t4, the 4th triangular number represented by x]XXX_____xXX_____xxX_____xxx_____xxxxBring them together and you have:XXXxXXxxXxxxxxxxThe sum of any two successive triangular number is a square number.The nth triangular number is n(n+1)/2 = (n2 +n)/2The next triangular number is (n+1)*(n+2)/2 = (n2 + 3n + 2)/2Their sum is (2n2 + 4n + 2)/2 = n2 + 2n + 1 = (n + 1)2This is easy to visualise. For example T3 + t4 = 42[T3, the 3th triangular number represented by X,t4, the 4th triangular number represented by x]XXX_____xXX_____xxX_____xxx_____xxxxBring them together and you have:XXXxXXxxXxxxxxxxThe sum of any two successive triangular number is a square number.The nth triangular number is n(n+1)/2 = (n2 +n)/2The next triangular number is (n+1)*(n+2)/2 = (n2 + 3n + 2)/2Their sum is (2n2 + 4n + 2)/2 = n2 + 2n + 1 = (n + 1)2This is easy to visualise. For example T3 + t4 = 42[T3, the 3th triangular number represented by X,t4, the 4th triangular number represented by x]XXX_____xXX_____xxX_____xxx_____xxxxBring them together and you have:XXXxXXxxXxxxxxxx
in a math equation "n" is normally the unidentified number in the equation and 2 is the number squared :)
n2-3n+2
n(n + 2) = n2 + 2n
1/2*(n2-3n) = number of diagonals where n is the number of sides of the polygon.
Number of diagonals = 1/2*(n2-3n) where n = the number of sides of the polygon.
declare n1 number; n2 number; n3 number; begin n1:=3 n2 :=5 n3:= sum(n1,n2); dbms_output.put_line( n3); end