Answer: the tuition in 2020 is $502300
Step-by-step explanation:
The annual tuition at a specific college was $20,500 in 2000, and $45,4120 in 2018. Let us assume that the rate of increase is linear. Therefore, the fees in increasing in an arithmetic progression.
The formula for determining the nth term of an arithmetic sequence is expressed as
Tn = a + (n - 1)d
Where
a represents the first term of the sequence.
d represents the common difference.
n represents the number of terms in the sequence.
From the information given,
a = $20500
The fee in 2018 is the 19th term of the sequence. Therefore,
T19 = $45,4120
n = 19
Therefore,
454120 = 20500 + (19 - 1) d
454120 - 20500 = 19d
18d = 433620
d = 24090
Therefore, an
equation that can be used to find the tuition y for x years after 2000 is
y = 20500 + 24090(x - 1)
Therefore, at 2020,
n = 21
y = 20500 + 24090(21 - 1)
y = 20500 + 481800
y = $502300
Answer:
- 1 < x < 4
Step-by-step explanation:
Answer:
7/6
Step-by-step explanation:
The width to length ratio is ...
... width / length = (4 2/3 yards)/(4 yards) = (14/3)/4 = 14/12
... width/length = 7/6
I believe it's increasing on all fronts, because if you start from the right, you see that the y values always increase, hence they are increasing. They do it for when x<0 and when x>0. So, it should be increasing on all real numbers
Answer:
Follows are the solution to this question:
Step-by-step explanation:
Following are the differential equation:
In equation:
