Question

PLEASE HELP ME I NEED THIS DONE AS SOON AS POSIBLE

a. Liam wants to know how the function would need to change to reflect the amount of money the business will take in. Use this function to model a function for gross sales. (Gross sales are the total amount of money the flowers sell for.)

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b. Liam knows that his cost (c) in terms of the units of mums sold (x) can be modeled by the function Replace the variable x in this function with the function that gives the number of units sold (x) in terms of price p. What does this turn the function c into?

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c. Liam wants to calculate the profit his shop would earn from the sale of mums. He knows that it can be done by deducting the costs from the money that the store makes selling mums. Now that he has the functions for the cost and gross sales, can he subtract the function for cost from the function representing the gross sales to arrive at a function for profit? Why?

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d. Find the store’s profit function for selling mums by subtracting the function for cost from the function for gross sales. Consider pr(p) as the profit function.

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e. Is the profit function linear or quadratic?

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f. Liam wants to use the profit function to determine the range of prices at which he can sell mums and make a profit. How can he determine the price range?

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g. Use the quadratic formula to find the factors of the profit function you wrote.

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h. Using the factors you found in part g, find the zeros of the quadratic profit function.

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i. Liam needs to find the range of prices at which he can sell mums and make a profit. Based on the equation, what are the points between which he will earn a profit on his mums?

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j. Liam wants to determine the most profitable price point for the mums. Use the completing the square technique to find this price.

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k. Liam is interested in determining the maximum profit possible for the mums. Find the maximum profit corresponding to the most profitable price point.

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Task 2: A Boat Ride
Jake went on a riverboat tour. After returning from the tour, he was curious to know the speed at which the boat was going. During the tour, he was told that he would travel 10 miles upstream and then 10 miles downstream. He also knows that the river was flowing at a rate of 3 miles/hour and that the tour took about three hours.

The boat moved at a constant speed for the whole journey, although its actual speed varied with the current.

a. Jake wants to know the boat’s actual speed going upstream and actual speed coming downstream. If the boat was moving at a constant speed of x miles per hour, determine the boat’s relative speed in each direction.

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b. Was the time it took the boat to go upstream the same as the time it took to come downstream?

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c. Jake first reasoned that because the boat ride took 3 hours, it must have taken 1.5 hours to go upstream and 1.5 hours to go downstream. Thinking about it further, however, he realized that the ride upstream had to take longer than the trip downstream. What is the reason it took longer to travel one direction than the other?

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d. The time it took the boat to go upstream is U, and the time it took to come downstream is D. What is U + D?

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e. Jake knows that the relationship between distance (d), velocity (v), and time taken (t) is given by d = vt. Rewrite this equation for t.

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f. Using the equation you rewrote in part e, write an equation for the time taken for the upstream trip.

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g. Now construct an equation for the time taken for the downstream trip.

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h. Jake already has the equation U + D = 3, which represents the total time taken for the whole journey. Now that he has individual equations for U and D, he needs to combine these three equations to get a single equation. Construct a combined equation for U + D in terms of x, the constant speed at which the boat was going.

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i. Jake now has an equation with x as the only unknown variable. He needs to solve for x. Transform the equation you found in part h into a quadratic equation, and find the value of x, the constant speed at which the boat traveled.

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Answer 1

Answer::b.

Step-by-step explanation:

Answer 2

a. The boat's actual speed going upstream is
`\(x - 3\)` miles per hour, and its actual speed coming downstream is
`\(x + 3\)` miles per hour.

b. No, the time it took the boat to go upstream was not the same as the time it took to come downstream.

c. The reason it took longer to travel upstream than downstream is the opposition of the river current, which adds to the effective speed when going downstream but subtracts from the effective speed when going upstream.

d.
`\(U + D = 3\)` hours.

e.
`\(t = (d)/(v)\).`

f.
`\(U = (10)/(x - 3)\).`

g.
`\(D = (10)/(x + 3)\).`

h. Combining the equations,
`\(U + D = (10)/(x - 3) + (10)/(x + 3) = 3\)`.

i. Simplifying and rearranging the equation in part h, we get the quadratic equation
`\((x - 5)(x + 1) = 0\),` and solving it yields
`\(x = 5\) or \(x = -1\).`

Step-by-step explanation:

a. The boat's actual speed going upstream is
`\(x - 3\)` miles per hour, and its actual speed coming downstream is
`\(x + 3\)` miles per hour. This is determined by considering the effect of the river current, which either opposes or aids the boat's speed.

b. The time it took the boat to go upstream was not the same as the time it took to come downstream. The opposition or assistance of the river current creates different effective speeds, leading to unequal travel times.

c. The longer time taken to travel upstream is a result of the opposition of the river current. When going against the current, the boat faces a stronger effective current speed, causing the upstream journey to take more time than the downstream journey.

d. The total time equation is
`\(U + D = (10)/(x - 3) + (10)/(x + 3) = 3\),` representing the sum of the time taken to go upstream
`(\(U\))` and downstream
`(\(D\)).`

e. The equation
`\(t = (d)/(v)\)` relates distance, velocity, and time. Rearranging it,
`\(U = (10)/(x - 3)\) and \(D = (10)/(x + 3)\)` represent the times taken for upstream and downstream trips.

f. Combining the equations for
`\(U\) and \(D\)` gives the equation
`\(U + D = (10)/(x - 3) + (10)/(x + 3) = 3\).` Simplifying and rearranging, we get the quadratic equation
`\((x - 5)(x + 1) = 0\).`

g. Solving the quadratic equation
`\((x - 5)(x + 1) = 0\)` yields two solutions:
`\(x = 5\) or \(x = -1\).` However, since speed cannot be negative,
`\(x = 5\)` is the valid solution.