We are given with the equationL
V(t) = 320 e^(-3.1t)
At V = 200
200 = 320 e^(-3.1t)
t = 0.15 s
100 = 320 e^(-3.1t)
t = 0.36 s
It takes 0.36 - 0.15 = 0.21 s for the voltage to drop from 200 to 100 volts.
Answer:
<x = 20°
Step-by-step explanation:
<x = 1/2(arc UT - arc SV)
<x = 1/2(65° - 25°)
<x = 1/2 (40°)
<x = 20°
That would be 423 total. :)
The cost of parking is an initial cost plus an hourly cost.
The first hour costs $7.
You need a function for the cost of more than 1 hour,
meaning 2, 3, 4, etc. hours.
Each hour after the first hour costs $5.
1 hour: $7
2 hours: $7 + $5 = 7 + 5 * 1 = 12
3 hours: $7 + $5 + $5 = 7 + 5 * 2 = 17
4 hours: $7 + $5 + $5 + $5 = 7 + 5 * 3 = 22
Notice the pattern above in the middle column.
The number of $5 charges you add is one less than the number of hours.
For 2 hours, you only add one $5 charge.
For 3 hours, you add two $5 charges.
Since the number of hours is x, according to the problem, 1 hour less than the number of hours is x - 1.
The fixed charge is the $7 for the first hour.
Each additional hour is $5, so you multiply 1 less than the number of hours,
x - 1, by 5 and add to 7.
C(x) = 7 + 5(x - 1)
This can be left as it is, or it can be simplified as
C(x) = 7 + 5x - 5
C(x) = 5x + 2
Answer: C(x) = 5x + 2
Check:
For 2 hours: C(2) = 5(2) + 2 = 10 + 2 = 12
For 3 hours: C(3) = 5(3) + 2 = 15 + 2 = 17
For 4 hours: C(3) = 5(4) + 2 = 20 + 2 = 22
Notice that the totals for 2, 3, 4 hours here
are the same as the right column in the table above.
