We know that
In any quadrilateral where the vertices are on the circumference of a circle, the opposite angles add up to
degrees.
So,
in this problem
therefore
the answer is
Answer:
Subtract 3 and one-fifth from both sides of the equation
Step-by-step explanation:
Well to find n you gotta separate it.
3 1/5 + n = 9
-3 1/5
n = 5.8
<em>Thus,</em>
<em>to seperate it you subtract 3 1/5 from both sides.</em>
Answer:
FV= $3,725.07
Step-by-step explanation:
Giving the following information:
Initial investment (PV)= $3,294
Number of periods (n)= 5 years
Interest rate (i)= 2.6% = 0.026
<u>To calculate the future value (FV), we need to use the following formula:</u>
FV= PV*(1+i)^n
FV= 3,294*(1.026^5)
FV= $3,725.07
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
So, you can do this one of two ways.
The first is knowing what the numbers in your equation represent:
For example: y=mx+b in this case "b" is 9. b is the y-value when x=0. So the first point on our graph is (0,9). Next we have to pick an xvalue to solve for another y-coordinate.
I chose x= -3. Plug x into the equation to get y.
y=3x+9
y=3(-3)+9
y=(-9)+9
y=0
So, our second point is (-3,0).
connect the points with a ruler to graph the line.
