k is less than or equal to 7
2k - 3(h - 5k) = 2k - 3h + 15k = 17k - 3h
Is the following what you are claiming? 2k = 2k+1 -1 20 not equal 20+1 - 1 21 not equal 21+1 -1
2k - 3l
For n = 4, 4! = 4*3*2*1 = 24 and 24 = 2*2*2*2 = 16 So the statement is true for n = 4. Suppose it is true for n = k, that is, k! > 2k Then (k+1)! = (k+1)*k! > (k+1)*2k (since k! > 2k) > 2*2k (since k >= 4 > 2) = 2k+1 So if the statement is true for n = k then it must be true for n = k+1. Therefore, since it is true for n = 4 it must be true for all n > 4.
Consider prime numbers p1, p2 greater than 2. Since p1 and p2 are prime and greater than 2, they are both necessarily odd. Hence, they are of the form: p1 = 2k+1, p2 = 2j+1, where j and k are positive integers. Their sum is then: p1+p2 = (2k+1)+(2j+1) = 2k+2j+2 = 2 (k+j+1). So 2 divides their sum, hence the sum can't be prime
A 2000 metres walk.
9 - 2k - 3 = k Add 2k to both sides: 9 - 3 = k + 2k Combine like terms: 6 = 3k Divide both sides by 3: 2 = k
2k = 5k-30 Subtract 5k from both sides: -3k = -30 Divide both sides by -3 to find the value of k remembering that a minus divided into a minus becomes a plus: k = 10
Let even be of the form 2k and odd be of the form 2k+1. Then odd * even becomes 2k*2k+1, or 4k^2 +2k. This can be written as 2(k^2 + k), which is of the form 2k. Therefore, odd X even equals even.
20
The equation 2K + Br2 -> 2KBr is balanced as there are equal numbers of atoms of each element on both sides of the reaction.
If you mean: 10-2k = 4 then k = 3