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.
To solve this problem, we need to analyze the given information and use rational equations to find the answer.
Let's assume the parent's speed from the house to her husband's work is "x" miles per hour.
We know that the distance between the house and the husband's work is 5 miles.
Using the formula Speed = Distance / Time, we can calculate the time it takes for the parent to travel from the house to the husband's work.
Time = Distance / Speed
Time = 5 / x hours
Now, let's consider the speed at which the parent travels between the husband's work and the school. We are told that due to heavy traffic, she averages 10 miles per hour slower on this leg of the journey.
Therefore, her speed from the husband's work to the school is (x - 10) miles per hour.
The distance between the husband's work and the school is also 5 miles.
Using the same formula, the time it takes for the parent to travel from the husband's work to the school is:
Time = Distance / Speed
Time = 5 / (x - 10) hours
The total time for the entire trip is given as 20 minutes, which is equal to 1/3 hour.
Therefore, we can set up the equation:
Time from house to work + Time from work to school + Time from school to house = 1/3
5 / x + 5 / (x - 10) + 5 / (x - 10) = 1/3
Now, we can solve this rational equation for the value of "x".
To simplify the equation, we can multiply through by 3(x)(x - 10) to eliminate the denominators:
(3(x)(x-10)) * (5 / x) + (3(x)(x-10)) * (5 / (x - 10)) + (3(x)(x-10)) * (5 / (x - 10)) = (3(x)(x - 10)) * (1/3)
Simplifying further:
15(x - 10) + 15x + 15x = (x)(x - 10)
Expanding and rearranging the equation:
15x - 150 + 15x + 15x = x^2 - 10x
45x - 150 = x^2 - 10x
Rearranging again:
x^2 - 10x - 45x + 150 = 0
x^2 - 55x + 150 = 0
Now, we can use factoring, completing the square, or the quadratic formula to solve for "x".
To solve this problem using rational equations, we can start by setting up an equation based on the given information.
Let's assume that the parent's speed between the house and work is x miles per hour.
Since the parent usually makes the whole trip in 20 minutes, which is equivalent to 1/3 of an hour, we can write the equation as:
5/x + 5/(x - 10) = 1/3
To solve this equation, we can find a common denominator and then simplify:
(5(x - 10) + 5x)/(x(x - 10)) = 1/3
(5x - 50 + 5x)/(x(x - 10)) = 1/3
(10x - 50)/(x(x - 10)) = 1/3
Next, we can cross multiply:
3(10x - 50) = x(x - 10)
30x - 150 = x^2 - 10x
Rearranging the equation to set it equal to zero:
x^2 - 40x + 150 = 0
Now, we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, a = 1, b = -40, and c = 150:
x = (-(-40) ± √((-40)^2 - 4(1)(150))) / (2(1))
x = (40 ± √(1600 - 600)) / 2
x = (40 ± √1000) / 2
x = (40 ± 10√10) / 2
x = 20 ± 5√10
Therefore, the speed between the house and work can be expressed as either 20 + 5√10 mph or 20 - 5√10 mph.
As for finding more practice problems using rational equations, there are several online resources and math textbooks that provide practice exercises. You can search for rational equation practice problems online or check out math textbooks that cover algebra or precalculus topics. Additionally, there are websites and apps that offer math practice questions and quizzes, which can help you improve your skills in solving rational equations.