Sunday

December 21, 2014

December 21, 2014

Posted by **frank** on Saturday, March 1, 2008 at 10:34am.

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.

- algebra -
**frank**, Saturday, March 1, 2008 at 10:34amii. 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.

- algebra -
**frank**, Saturday, March 1, 2008 at 10:35amii.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.

- algebra -
**Reiny**, Saturday, March 1, 2008 at 11:13amI 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

- qcbd qglez -
**qcbd qglez**, Sunday, February 1, 2009 at 11:40amvumzlej iaclwgeu wumcfdhr hjxwkv brvfz mxtjlz unvjcb

- algebra -
**bill**, Thursday, April 23, 2009 at 4:11pmi 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

- algebra -
**lynn**, Thursday, June 24, 2010 at 10:08amTicket 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?

**Answer this Question**

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