The first term is 138
The difference is 55
The iterative rule for the amount of money Mr Speas has after n weeks is
55/2 n² + 221/2 n
During the first week she has $138 in his bank account. At the end of each week she deposited $55 into her bank account.
The first term will be 138 .
The common difference is 55 because her bank always increase by $55 dollars every week. The sequence will be 138, 193, 248, 303, 358...…
The difference = 193 - 138 = 55.
The iterative rule for the amount of money Mr Speas has after n weeks can be represented below
n = number of weeks
a = first term = 138
d = common difference = 55
Using AP formula,
sₙ = n/2(2a + (n - 1)d)
sₙ = n/2 (2(138)+ (n - 1)55)
sₙ = n /2(276 + 55n -55)
sₙ = n /2(221 + 55n)
sₙ = 55/2 n² + 221/2 n
read more: brainly.com/question/20373665?referrer=searchResults
Try solving it on your own
you must factor the integers
0.00017 is the answer though
Answer:
- <u><em>The solution to f(x) = s(x) is x = 2012. </em></u>
Explanation:
<u>Rewrite the table and the choices for better understanding:</u>
<em>Enrollment at a Technical School </em>
Year (x) First Year f(x) Second Year s(x)
2009 785 756
2010 740 785
2011 690 710
2012 732 732
2013 781 755
Which of the following statements is true based on the data in the table?
- The solution to f(x) = s(x) is x = 2012.
- The solution to f(x) = s(x) is x = 732.
- The solution to f(x) = s(x) is x = 2011.
- The solution to f(x) = s(x) is x = 710.
<h2>Solution</h2>
The question requires to find which of the options represents the solution to f(x) = s(x).
That means that you must find the year (value of x) for which the two functions, the enrollment the first year, f(x), and the enrollment the second year s(x), are equal.
The table shows that the values of f(x) and s(x) are equal to 732 (students enrolled) in the year 2012,<em> x = 2012. </em>
Thus, the correct choice is the third one:
- The solution to f(x) = s(x) is x = 2012.
Lolz lolz is lolz lolz 782
