They are -2, 2, 6, 10 and 14.
The expression "n plus 3" can be represented as ( n + 3 ). To find the first five terms, we can substitute the values ( n = 1, 2, 3, 4, ) and ( 5 ) into the expression. The first five terms are: ( 1 + 3 = 4 ) ( 2 + 3 = 5 ) ( 3 + 3 = 6 ) ( 4 + 3 = 7 ) ( 5 + 3 = 8 ) Thus, the first five terms are 4, 5, 6, 7, and 8.
To find the first three terms of an arithmetic sequence with a common difference of -5, we first need the last term. If we denote the last term as ( L ), the terms can be expressed as ( L + 10 ), ( L + 5 ), and ( L ) for the first three terms, since each term is derived by adding the common difference (-5) to the previous term. Thus, the first three terms would be ( L + 10 ), ( L + 5 ), and ( L ).
To simplify the expression (12r + 5 + 3r - 5), combine like terms. First, combine the (r) terms: (12r + 3r = 15r). Then, combine the constant terms: (5 - 5 = 0). Thus, the simplified expression is (15r).
5
8
"4n6" is a common abbreviation for "forensics," typically used in the context of digital forensics or forensic science. It represents the combination of the four letters in "forensics" (4) and the two letters in "n" and "6".
+4n6!n4 +0 ..
5
it is 8.
To find the first three terms of an arithmetic sequence with a common difference of -5, we first need the last term. If we denote the last term as ( L ), the terms can be expressed as ( L + 10 ), ( L + 5 ), and ( L ) for the first three terms, since each term is derived by adding the common difference (-5) to the previous term. Thus, the first three terms would be ( L + 10 ), ( L + 5 ), and ( L ).
To simplify the expression (12r + 5 + 3r - 5), combine like terms. First, combine the (r) terms: (12r + 3r = 15r). Then, combine the constant terms: (5 - 5 = 0). Thus, the simplified expression is (15r).
5 first terms in n²+3
2
20, 15, 10, 5, 0, -5, -10, -15, -20 and so on.
5
To find the first 5 terms, plug 1, 2, 3, 4 and 5 in for n:3*1-3 = 03*2-3 = 33*3-3 = 63*4-3 = 93*5-3 = 12The first five terms are 0, 3, 6, 9 and 12.
5, 11, 17, 23, 29