A rectangular playing field will be divided by fences as shown here. A total of 800 metres of fence is
available. Since we don’t know the width, we can call it x. (6 marks)
a. How many “x” lengths of fencing will be used? How
much of the 800 metres of fencing will that leave for
the two long sides? How much fence will that leave for
each one of the long sides?
x
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b. Using x as your width and your last answer in a) as your length, write an equation that
represents the area of this field. Remember that area of a rectangle = length x width.
c. This equation for the area should be a quadratic function. Find the maximum area.
d. What are the dimensions—length and width—for each third of this playing field?
a. The number of "x" lengths of fencing that will be used is 4. This is because there are 4 sides, and each side has a width of x.
So the total length of the sides with width x is 4x.
This leaves 800 - 4x meters of fencing for the two long sides.
Each long side will have half of the remaining fence length, so each one will have (800 - 4x)/2 = 400 - 2x meters of fence.
b. The equation that represents the area of the field is:
Area = length x width
Substituting x for the width and the answer from part a) for the length, the equation becomes:
Area = (400 - 2x) * x
c. To find the maximum area, we need to find the vertex of the quadratic function. The vertex can be found using the formula:
x = -b / 2a
In this case, a = -2 and b = 400.
x = -400 / (2 * -2) = -400 / -4 = 100
Substituting x = 100 back into the equation for the area, we get:
Area = (400 - 2*100) * 100
= 200 * 100
= 20000
So the maximum area of the field is 20000 square meters.
d. Each third of the playing field will have the same width, which is x.
The length of each third can be found by dividing the total length (400 - 2x) by 3. So each third will have a length of (400 - 2x)/3.
n 1 hour, Nate paddles his canoe 6 km upstream and then returns back downstream to the dock. The
current in the river is 3 km/h.
a) Create a simplified quadratic equation to model the situation. (3 marks)
b) What is Nate’s speed in still water?
a) To create a simplified quadratic equation to model the situation, we can use the formula:
Time = Distance / Speed
The time taken to paddle upstream is given as 1 hour, so we can set up the equation:
1 = 6 / (Nate's speed - 3)
The time taken to paddle downstream is also given as 1 hour, so we can set up the equation:
1 = 6 / (Nate's speed + 3)
To simplify this, we can cross-multiply:
Nate's speed - 3 = 6 / 1
Nate's speed + 3 = 6 / 1
Simplifying further, we get:
Nate's speed - 3 = 6
Nate's speed + 3 = 6
b) To find Nate's speed in still water, we can solve the two equations simultaneously.
Solving the first equation:
Nate's speed - 3 = 6
Nate's speed = 6 + 3
Nate's speed = 9 km/h
Solving the second equation:
Nate's speed + 3 = 6
Nate's speed = 6 - 3
Nate's speed = 3 km/h
Since Nate's speed cannot be both 9 km/h and 3 km/h, there seems to be an error in the problem setup or data provided. Please double-check the information given.
What is the initial height of the rocket? (1 mark)
b) What is the maximum height of the rocket? (1 mark)
c) How long after launch does the rocket reach the maximum height? (1 mark)
d) How long does it take for the rocket to reach the ground? (3 marks)
e) State the domain and range for this situation.
Unfortunately, the initial height of the rocket, as well as the equation or information needed to calculate it, is not mentioned in the given question. Therefore, it is not possible to determine the initial height of the rocket.
Similarly, without the equation or information regarding the rocket's flight path, it is not possible to determine the maximum height of the rocket, the time it takes to reach the maximum height, or the time it takes to reach the ground. These values would depend on the specific data provided in the question.
The domain and range for this situation would also depend on the specific data and context given in the question. Without that information, it is not possible to accurately state the domain (possible inputs) and range (possible outputs) for the situation.
a. To find out how many "x" lengths of fencing will be used, we need to consider that the rectangular field consists of two "x" lengths for the short sides and two long sides. This gives us a total of 4x.
To find out how much fencing will be left for the two long sides, we subtract the total length of the short sides from the total available fencing:
800m - 4x = Remaining fencing for the two long sides
To find out how much fence will be left for each one of the long sides, we divide the remaining fencing by 2:
(800m - 4x) / 2 = Remaining fencing for each long side
b. The equation representing the area of this field can be written as:
Area = length x width
Area = x(800m - 4x)
c. To find the maximum area, we need to determine the vertex of the quadratic equation. In this case, the equation is in the form of y = ax^2 + bx + c, where "a" is -4, "b" is 800, and "c" is 0.
The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
x = -800 / (2*(-4))
x = -800 / (-8)
x = 100
To find the maximum area, we substitute the x-coordinate of the vertex back into the equation:
Area = 100(800 - 4*100)
Area = 100(800 - 400)
Area = 100(400)
Area = 40,000 square metres
d. To find the dimensions for each third of the playing field, we can divide the length and width by 3:
Length of each third = x / 3
Width of each third = (800 - 4x) / 3
Therefore, the dimensions for each third of the playing field are (x/3) and [(800 - 4x)/3].
a. To determine how many "x" lengths of fencing will be used, we need to calculate the perimeter of the rectangular playing field. The perimeter of a rectangle is given by the formula: P = 2L + 2W, where L is the length and W is the width.
In this case, since we don't know the width, we can call it x. So the perimeter becomes: P = 2L + 2x.
Given that the total available fence is 800 meters, we can set up the equation: 800 = 2L + 2x.
To find the amount of fence remaining for the two long sides, we subtract the fence used for the "x" lengths: 800 - 2x.
Finally, to find the amount of fence remaining for each long side, we divide the above result by 2: (800 - 2x) / 2.
b. To write an equation that represents the area of the field, we use the formula for the area of a rectangle: A = L * W.
Since the width is given as x and the length is the value we found in part a), our equation becomes: A = L * x.
c. To find the maximum area, we need to maximize the quadratic function A = L * x. Since the length of the field was determined in part a) as (800 - 2x) / 2, we substitute this value into the equation for A: A = [(800 - 2x) / 2] * x.
We can expand and simplify the equation: A = 400x - x^2.
To find the maximum area, we can take the derivative of the equation with respect to x, set it equal to zero, and solve for x. However, since this is a quadratic function, we can also recognize that the maximum occurs at the vertex of the parabola. The x-coordinate of the vertex can be found using the formula: x = -b / (2a), where a is the coefficient of x^2 and b is the coefficient of x in the quadratic equation.
In this case, a = -1 and b = 400. So the x-coordinate of the vertex is x = -400 / (2*(-1)) = 200.
Substituting this value back into the equation, we find the maximum area: A = 400 * 200 - 200^2 = 40,000 square units.
d. To find the dimensions (length and width) for each third of the playing field, we can divide the length found in part a) by 3.
So the length of each third is (800 - 2x) / 6. The width remains the same, which is x.
Therefore, the dimensions for each third of the playing field are: length = (800 - 2x) / 6 and width = x.