#1.) The first row of an amphitheater contains 50 seats, and each row there after contains an additional 8 seats. If the theater seats 938 people, find the number of rows in the amphitheater.
#2.) At oceanside deck the first high tide today occurs at 2:00 am with depth 6 meters and the first low tide occur4s at 8:30 am with depth 2 meters.
---> Find the equation of the graph showing the depth of water d at the dock as a function of time t.
---> Suppose a tanker that requires at least 4 meters of water depth is planning to dock after 9:00 AM. Using your graph in part "a," determine the earliest possible time that the tanker can dock and the longest period of time that the tanker can be docked before the water level becomes too low.
total seats= 50+(n-1)8 for n=1,2,3...
solve for n.
The depth of water will be a cosine function.
I think I would graph it.
Answers:
#1.) n = 10.6 rows
#2.) f(t) = 2cos(2pi/13)(t-2)+4
#1.) To find the number of rows in the amphitheater, we can set up an equation using the information given. Let's denote the number of rows as 'n'.
The first row contains 50 seats, and each subsequent row contains an additional 8 seats. So, the total number of seats in the amphitheater can be expressed as:
Total seats = 50 + (n-1)*8
Given that the theater seats 938 people, we can equate the total seats to 938 and solve for 'n':
938 = 50 + (n-1)*8
Subtracting 50 from both sides, we have:
938 - 50 = (n-1)*8
888 = (n-1)*8
To solve for 'n', we divide both sides by 8:
888/8 = n-1
111 = n-1
Adding 1 to both sides, we find that:
n = 111 + 1 = 112
Therefore, there are 112 rows in the amphitheater.
#2.) To find the equation of the graph showing the depth of water 'd' at the dock as a function of time 't', we can use a cosine function.
In the given information, the first high tide occurs at 2:00 am with a depth of 6 meters. Let's note this as (2, 6) on the graph.
The first low tide occurs at 8:30 am with a depth of 2 meters. Let's note this as (8.5, 2) on the graph.
The cosine function in its general form is:
f(t) = A*cos(B(t-C)) + D
In this case, since the depth at high tide is 6 meters and the depth at low tide is 2 meters, we have:
A = (6-2) = 4
The period of the tide is 12 hours (from high tide to high tide), so we can set B = 2pi/12 = pi/6.
To find C, we can use the time of the first high tide at 2:00 am:
2 = C
Finally, D represents the average depth, which is halfway between the high and low tide depths:
D = (6+2)/2 = 4
Putting all these values into the general equation, we have:
f(t) = 4*cos((pi/6)*(t-2)) + 4
Thus, the equation of the graph showing the depth of water 'd' at the dock as a function of time 't' is:
f(t) = 4*cos((pi/6)*(t-2)) + 4
To determine the earliest possible time that the tanker can dock after 9:00 am, we need to find the first value of 't' where the depth is at least 4 meters.
The depth reaches 4 meters at low tide, which happens at 8:30 am. After that, the depth starts increasing and reaches 4 meters again at high tide, which occurs 12 hours later at 8:30 pm.
So, the earliest possible time that the tanker can dock after 9:00 am is at the next high tide, which is at 8:30 pm.
To determine the longest period of time that the tanker can be docked before the water level becomes too low, we need to find the next low tide after the earliest possible docking time.
The next low tide occurs 6 hours after the high tide, which is at 2:30 am.
Therefore, the longest period of time that the tanker can be docked before the water level becomes too low is from 8:30 pm to 2:30 am the next day, which is 6 hours.
To solve the first question:
The number of seats in each row can be represented by a pattern. The first row has 50 seats, and each subsequent row has 8 more seats than the previous row. So, we can create an equation to find the number of seats in each row, given the row number.
Let's assume that the number of rows is represented by 'n'. The equation to find the number of seats in the amphitheater can be written as:
Total number of seats = 50 + (n - 1) * 8
We know that the total number of seats in the amphitheater is 938. So, we can substitute this value into the equation and solve for 'n'.
938 = 50 + (n - 1) * 8
Subtract 50 from both sides:
938 - 50 = (n - 1) * 8
888 = (n - 1) * 8
Divide both sides by 8:
111 = n - 1
Add 1 to both sides:
112 = n
So, there are 112 rows in the amphitheater.
For the second question:
To find the equation of the graph showing the depth of water 'd' at the dock as a function of time 't', we need to know the relationship between the depth and time. In this case, the depth of water follows a cosine function.
We are given two points on the graph: (2:00 am, 6 meters) and (8:30 am, 2 meters). These two points can be used to determine the equation of a cosine function.
A cosine function can be written as:
f(t) = A * cos(B * (t - C)) + D
Where A is the amplitude, B affects the period, C shifts the graph horizontally, and D shifts the graph vertically.
From the given information, we can identify the following values:
Amplitude (A) = (6 - 2) / 2 = 2
Frequency (B) = 2π / (12h + 30min) = 2π / 13
Phase shift (C) = 2:00 am = 2
Vertical shift (D) = 4 meters (minimum depth requirement)
Therefore, the equation of the graph showing the depth of water 'd' at the dock as a function of time 't' is:
f(t) = 2 * cos((2π/13) * (t - 2)) + 4
To determine the earliest possible time that the tanker can dock, we need to find the first time after 9:00 AM when the depth of water is at least 4 meters. We can solve this by substituting 't' with values after 9:00 AM in the equation and finding the first value that satisfies the condition.
To find the longest period of time that the tanker can be docked before the water level becomes too low, we need to determine the interval during which the depth of water remains at least 4 meters. This can be done by graphing the equation and identifying the time intervals where the graph is above or equal to 4.
Note: The accuracy of these values may depend on the time resolution used and any additional information not mentioned in the question.