You may have to convert from feet to miles several times in this assignment. You can use 1 mile = 5,280 feet for your conversions.

1.Many people know that the weight of an object varies on different planets, but did you know that the weight of an object on earth also varies according to the elevation of the object? In particular, the weight of an object follows this equation: , where C is a constant, and r is the distance that the object is from the center of the earth.
a. Solve w=Cr^-2the equation for r.
My Answer: r = sqrt(C/w)
b. Suppose that an object is 100 pounds when it is at sea level. Find the value of C that makes the equation true. (Sea level is 3,963 miles from the center of the earth.)
My answer: C = 1,570,536,900
c. Use the value of C you found in the previous question to determine how much the object would weigh in
i. Death Valley (282 feet below sea level)
My answer: w = 100.15 pounds
ii. The top of Mt McKinley (20,430 feet above sea level)
My Answer: w = 99.96 pounds
2. The equation D=1.2(sqrt)h which should read D equals 1.2 square root of h gives the distance, D, in miles that a person can see to the horizon from a height, h, in feet.
a. Solve this equation for h.
My Answer: (D / 1.2)^2 = h
b. Long’s Peak in the Rocky Mountain National Park, is 14,255 feet in elevation. How far can you see to the horizon from the top of Long’s Peak? Can you see Cheyenne, Wyoming (about 89 miles away)? Explain your answer.
My Answer: A person can see 143.27 miles from the top and yes you can see Cheyenne because 89 is less than 143.

ii. The top of Mt McKinley (20,430 feet above sea level)

My Answer: w = 99.96 pounds
2. The equation D=1.2(sqrt)h which should read D equals 1.2 square root of h gives the distance, D, in miles that a person can see to the horizon from a height, h, in feet.
a. Solve this equation for h.
My Answer: (D / 1.2)^2 = h
b. Long’s Peak in the Rocky Mountain National Park, is 14,255 feet in elevation. How far can you see to the horizon from the top of Long’s Peak? Can you see Cheyenne, Wyoming (about 89 miles away)? Explain your answer.
My Answer: A person can see 143.27 miles from the top and yes you can see Cheyenne because 89 is less than 143.

ii.The top of Mt McKinley (20,430 feet above sea level)

My Answer: w = 99.96 pounds
2. The equation D=1.2(sqrt)h which should read D equals 1.2 square root of h gives the distance, D, in miles that a person can see to the horizon from a height, h, in feet.
a. Solve this equation for h.
My Answer: (D / 1.2)^2 = h
b. Long’s Peak in the Rocky Mountain National Park, is 14,255 feet in elevation. How far can you see to the horizon from the top of Long’s Peak? Can you see Cheyenne, Wyoming (about 89 miles away)? Explain your answer.
My Answer: A person can see 143.27 miles from the top and yes you can see Cheyenne because 89 is less than 143.

I don't know how you got the answers to the Death Valley and Mt. McKinley questions

I would assume that in the original formula the radius r is measured in miles, so you would have to convert 282 feet below seaslevel to -282/5280 miles
and the radius would be 3963-282/5280

then w = C/r^2 gave me an answer of 100.0027

for the McKinley mountain

r = 3963 + 20430/5280
for a weight of 99.81

i have a problem with your basic solutions:

if w=cr²

then w/c=cr²/c

then w/c=r²

if 100=c*3963²

then 100/3963²=(c*3963²)/3963²

then 100/3963²=c
this result is a small number divided by a very large number: 1,570,536,900 can't be correct

Ticket Sales

Living in or near a metropolitan area has some advantages. Entertainment opportunities are almost endless in a major city. Events occur almost every night, from sporting events to the symphony. Tickets to these events are not available long and can often be modeled by quadratic equations.

Answer the following questions..

1. Suppose you are an event coordinator for a large performance theater. One of the hottest new Broadway musicals has started to tour and your city is the first stop on the tour. You need to supply information about projected ticket sales to the box office manager. The box office manager uses this information to anticipate staffing needs until the tickets sell out. You provide the manager with a quadratic equation that models the expected number of ticket sales for each day x. ( is the day tickets go on sale).


a. Does the graph of this equation open up or down? How did you determine this?

b. Describe what happens to the tickets sales as time passes.

c. Use the quadratic equation to determine the last day that tickets will be sold.

Note. Write your answer in terms of the number of days after ticket sales begin.

d. Will tickets peak or be at a low during the middle of the sale? How do you know?

e. After how many days will the peak or low occur?

f. How many tickets will be sold on the day when the peak or low occurs?

g. What is the point of the vertex? How does this number relate to your answers in parts e. and f?

h. How many solutions are there to the equation ? How do you know?

i. What do the solutions represent? Is there a solution that does not make sense? If so, in what ways does the solution not make sense?

To solve part a of question 1, you need to solve the equation w = Cr^(-2) for r.

Here's how to do it:
1. Start by isolating r on one side of the equation. Divide both sides of the equation by w to get 1 = C/w * r^(-2).
2. Take the reciprocal of both sides to get r^2 = w/C.
3. Take the square root of both sides to get r = sqrt(w/C).

So, the solution to part a is r = sqrt(w/C).

In part b, the problem states that the object weighs 100 pounds at sea level, which is 3,963 miles from the center of the earth. We need to find the value of C that makes the equation true.

To do this, we substitute the given values into the equation:
100 = C/(3,963)^2.

Now we can solve for C:
C = 100 * (3,963)^2.

Calculating this, we find that C = 1,570,536,900.

So the value of C that makes the equation true is C = 1,570,536,900.

Moving on to part c, we use the value of C from part b to determine how much the object would weigh at Death Valley (282 feet below sea level) and the top of Mt McKinley (20,430 feet above sea level).

To calculate the weight at Death Valley, substitute the values into the equation:
w = (1,570,536,900) / (282)^2.

Calculating this, we find that w is approximately 100.15 pounds. So in Death Valley, the object would weigh about 100.15 pounds.

To calculate the weight at the top of Mt McKinley, substitute the values into the equation:
w = (1,570,536,900) / (20,430)^2.

Calculating this, we find that w is approximately 99.96 pounds. So at the top of Mt McKinley, the object would weigh about 99.96 pounds.

Moving on to question 2, let's solve part a, where we need to solve the equation D = 1.2(sqrt)h for h.

Here's how to do it:
1. Isolate h on one side of the equation. Square both sides of the equation to get D^2 = 1.2^2 * h.
2. Divide both sides of the equation by 1.44 to get D^2 / 1.44 = h.

So, the solution to part a is h = D^2 / 1.44.

In part b, we are given the elevation of Long's Peak as 14,255 feet. We need to find how far you can see to the horizon from the top of Long's Peak and determine if you can see Cheyenne, Wyoming (about 89 miles away).

Using the formula from part a, we plug in the given elevation into the equation:
D = 1.2 * sqrt(14,255).

Calculating this, we find that D is approximately 143.27 miles. This means that you can see approximately 143.27 miles to the horizon from the top of Long's Peak.

Since Cheyenne, Wyoming is about 89 miles away and 89 miles is less than 143.27 miles, you would be able to see Cheyenne from the top of Long's Peak.

Therefore, the answer is that you can see Cheyenne, Wyoming from the top of Long's Peak.