# Algebra 2

A parent leaves his house during rush-hour each weekday to get her husband from work and pick up her child from school. Her house, husband work and school are equal distance of 5 mile from each other. (On an imaginary triangle.) she usually makes the whole trip in 20 minutes. Because of heavy traffic at she averages 10 miles/hour slower between her husband work and the school. How fast is her speed between the house and the and her husband work?.

I am suppose to us rational equation to solve this problem I have no Idea where to start can some one please explain this to me.

are they any where I can go with work problem for rational equation that I can get more help and PRACTICE.

A parent leaves his house during rush-hour each weekday to get her husband from work and pick up her child from school. Her house, husband work and school are equal distance of 5 mile from each other. (On an imaginary triangle.) she usually makes the whole trip in 20 minutes. Because of heavy traffic at she averages 10 miles/hour slower between her husband work and the school. How fast is her speed between the house and the and her husband work?.

I am suppose to us rational equation to solve this problem I have no Idea where to start can some one please explain this to me.

are they any where I can go with work problem for rational equation that I can get more help and PRACTICE.

You say because of heavy traffic at (at what) she averages 100 mph slower between work and school.

How fast is her speed from the house too work?

Something is missing in your problem description. The only thing I can conclude as a logical question is what would her speed have to be between home and work and between the school and home for her to make the total trip in the same 20 minutes.

If true, from the given information, her average speed under normal circumstances is V = 15 miles/(1/3)hour = 45 mph.

If she is slowed down to 35 mph between work and school, the time for that leg becomes Tws = 5/35 = .1428h = 8.571 minutes.

To complete the trip in the same 20 minutes, she would have to travel the other 10 miles in 20 - 8.971 = 11.429 min = .190047=8 hr.

Therefore, she would have to average a speed of V = 10/.19048h = 52.5 mph from home to work and school too home.

If this is not the correct interpretation of your question, please clarify your problem statement.

I left my name off of the earlier response. Sorry about that.

A parent leaves his house during rush-hour each weekday to get her husband from work and pick up her child from school. Her house, husband work and school are equal distance of 5 mile from each other. (On an imaginary triangle.) she usually makes the whole trip in 20 minutes. Because of heavy traffic at she averages 10 miles/hour slower between her husband work and the school. How fast is her speed between the house and the and her husband work?.

You say because of heavy traffic at (at what) she averages 10 mph slower between work and school.

How fast is her speed from the house to work?

Something is missing in your problem description. The only thing I can conclude as a logical question is what would her speed have to be between home and work and between the school and home for her to make the total trip in the same 20 minutes.

If true, from the given information, her average speed under normal circumstances is V = 15 miles/(1/3)hour = 45 mph.

If she is slowed down to 35 mph between work and school, the time for that leg becomes Tws = 5/35 = .1428h = 8.571 minutes.

To complete the trip in the same 20 minutes, she would have to travel the other 10 miles in 20 - 8.971 = 11.429 min = .190047=8 hr.

Therefore, she would have to average a speed of V = 10/.19048h = 52.5 mph from home to work and school too home.

Another interpretation is if she travels from home to work at her normal 45mph, work to school at 35mph, how fast must she travel to arrive home in the same 20 minute time period?

Now, Thw = 5(60)/45 = 6.666 minutes
Tws = 5/35 = 5(60)/35 = 8.571 minutes
Tsh must equal 20 - 6.666 - 8.571 = 4.763 minutes making the travel speed from school back home equal to Tsh = 5/(4.763/60) = 62.98 mph.

If neither of these is not the correct interpretation of your question, please clarify your problem statement.

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