decreased
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.
The product of n multiplied by itself is written as n^2, which is read as "n squared." This is because when you multiply a number by itself, you are essentially squaring that number. So, n times n is equal to n^2.
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
1/2*(n2-3n) = 665 diagonals where n means number of sides
There are 9.06 x 10^22 atoms of nitrogen in 0.755 mol of N2. This is calculated by multiplying Avogadro's number (6.022 x 10^23) by the number of moles of N2 (.755 mol) and the number of nitrogen atoms in one molecule of N2 (2 atoms).
It has two atoms. Hence the two after the N.
0 in N2
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
The oxidation number of N in N2 is 0 since it is in its elemental form. In a diatomic molecule like N2, each nitrogen atom has an oxidation number of 0.
4*n2 = 16 Divide both sides by 4: n2 = 4 Take square roots: n = -2 or n = 2
double the number n2 2n 2xn nx2