17+4 = 21 the answer is 21
5 + 6 hahahahaha hahhahhahaha
Answer:
(a)18
(b)1089
(c)Sunday
Step-by-step explanation:
The problem presented is an arithmetic sequence where:
- First Sunday, a=1
- Common Difference (Every subsequent Sunday), d=7
We want to determine the number of Sundays in the 120 days before Christmas.
(a)In an arithmetic sequence:

Since the result is a whole number, there are 18 Sundays in which Aldsworth advertises.
Therefore, Aldsworth advertised 18 times.
(b)Next, we want to determine the sum of the first 18 terms of the sequence
1,8,15,...

The sum of the numbers of days published in all the advertisements is 1089.
(c)SInce the 120th day is the 18th Sunday, Christmas is on Sunday.
Answer:
Plus 2 (+2)
Step-by-step explanation:
The rate of change is always a constant change from the given first term to the next one. In this case, you are adding 2 each time:
3 + 2 = 5
5 + 2 = 7
7 + 2 = 9
etc.
~
Answer:
n = 5
Step-by-step explanation:
100,000 = 10^n
Take the log of both sides
log 100,000 = log 10^n
log 100,000 = n * log 10
Dividing both sides by log 10
log 100,000 / log 10 = n * log 10 / log 10
n = log 100,000 / log 10
= 5 / 1
n = 5