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:
x= 2 and y = -4
Step-by-step explanation:
8x + 3y = 4 ---------------------------------(1)
-7x + 5y = -34 -----------------------------(2)
Multiply through equation (1) by 5 and multiply through equation(2) by 3
40x + 15y = 20 ----------------------------(3)
-21x + 15y =-102----------------------------(4)
Subtract equation (4) from equation (3)
61x = 122
Divide both-side of the equation by 61
61x/61 = 122/61
(At the left-hand side of the equation 61 will cancel-out 61 leaving us with just x, while at the left-hand side of the equation 122 will be divided by 61)
x = 122/61
x=2
Substitute x= 2 into equation (1)
8x + 3y = 4
8(2) + 3y = 4
16 + 3y = 4
Subtract 16 from both-side of the equation
16-16 + 3y = 4-16
3y = -12
Divide both-side of the equation by 3
3y/3 = -12/3
y = -4
x= 2 and y = -4
LET X= ADDITIONAL AMOUNT
<u>EQAUTION=</u> 5+2x=17
-5 -5
__________
2x=12
_______
2 2
X=6 IS THE ANSWER. HOPE THIS HELPS!!!!!
Answer:
24
Step-by-step explanation:
-27 - x = -51
+27 +27
-x = -24
x = 24
