(n2 + 2n - 1) (n2 + 2n - 1) = n4 + 2n3 - n2 + 2n3 + 4n2 - 2n - n2 - 2n + 1 = n4 + 4n3 + 2n2 - 4n + 1 try with n = 5: (5 squared + 10 - 1) squared = 34 squared = 1156 with formula (5^4) + (4 *(5^3)) + (2 * (5^2)) - (4 * 5) + 1 = 625 + 500 + 50 - 20 + 1 = 1156
n x n = n2
n2-3n+2
No, N2 + 3H2 -> 2NH is the formula to make Ammonia, a gas that is present in urine.
If you mean: n2+16n+64 then it is (n+8)(n+8) when factored
2NH4NO3->2N2+O2+4H2O
The balance is 2N2 +2O2= 4NO
Yes I can.
To balance the chemical equation N2 + O2 = N2O5, start by counting the atoms of each element on both sides. You would need to put a coefficient of 2 in front of N2 to balance the nitrogen atoms. Then, add a coefficient of 5 in front of O2 to balance the oxygen atoms. The balanced equation is 2N2 + 5O2 = 2N2O5.
The balanced equation for the reaction between nitrogen (N2) and oxygen (O2) to form nitrogen dioxide (NO2) is: 2N2 + 4O2 -> 4NO2
2(n2+3n-54) 2(n+9)(n-6)
The integers are 43 and 44. The answer can be found by representing the two consecutive numbers as "n" and "n+1". n2 + (n+1)2 = 3785 n2 + (n2 + 2n + 1) = 3785 2n2 + 2n + 1 = 3785 2n2 + 2n - 3784 = 0 n2 + n - 1892 = 0 (n + 44) (n-43) = 0 n = -44, +43 The answer is restricted to positive values, so n = 43 and n+1 = 44
The balanced equation for the reaction of nitrogen gas (N2) and oxygen gas (O2) to form nitric oxide (NO) is: 2N2 + O2 → 2NO
The sequence is too short to be certain but it could be tn = (n-1)2 + n2 = 2n2 -2n + 1
2nh4no3 ---> 2n2 + o2 + 4h2o
Even numbers : they can be written 2n 2n1+2n2+....+2nm = 2(n1+n2+....+nm) so it's always an even number 2n1x 2n2x....x2nm = 2(n1+n2+....+nm) which is always an even number Odd numbers: they canbe written 2n+1 (2n1+1) + (2n2+1) + ....+ (2nm +1) = 2(n1+n2+....+nm) + m which is even if m is even and odd if m is odd, so it depend of the number of terms (2n1+1) x (2n2+1) x ....x (2nm +1) is always odd cause the last term of the expansion is always +1 and other terms have at least 2 as factor
An odd number is written 2n+1 with n any number (2n1+1)+(2n2+1)+...+(2n9+1) = 2(n1+n2+...+n9)+9 = 50 is not possible as 2(n1+n2+...+n9) = 41 is not possible with integers