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Thread: [Homework] Help

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    Default [Homework] Help

    We have like 2 worksheets to do on Rate X Time = Distance formula for math lol.

    I need help, i dont get any of it. Any body willing to help me?

    (id give the problems on msn)

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    why not give problems on here?
    I don't even know how many people I've helped with math around here....lol
    Proud owner of "Efferator" my totally boted main account!
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    Quote Originally Posted by Brain View Post
    why not give problems on here?
    I don't even know how many people I've helped with math around here....lol
    You never help. Don't lie.
    Math Nerd <3

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    Ok brain, its gonna rack up the post count, want to follow me to bbq pit?

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    bbq still does post count, general dump if you want to avoid the posts

    you can add my msn if need be, I only got about an hour around here, then an hour gone, then a few more hours
    Proud owner of "Efferator" my totally boted main account!
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    Quote Originally Posted by Brain View Post
    bbq still does post count, general dump if you want to avoid the posts

    you can add my msn if need be, I only got about an hour around here, then an hour gone, then a few more hours
    OH, ill just post a problem, i don't JUST want the answer, i want how to do it with the formula

    sofia drove 290 milesfrom her college to home and used 23.2 gallons of gasoline . her ssister zena drove 225 miles from her college to homeand used 15 galls of gas, whos veichacle has better gas mileag? justify ur answer

    Sorry for spelling, was looking at the paper

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    sofia:

    290miles / 23.2 gallons = 12.5 miles per gallon (m/g)


    zena:

    225miles / 15 gallons = 15 miles per gallon (miles/gallon)


    For every gallon of fuel used by sofia, her car goes 12.5 miles, while zena's car will 15 miles on one gallon of fuel
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    Quote Originally Posted by Brain View Post
    sofia:

    290miles / 23.2 gallons = 12.5 miles per gallon (m/g)


    zena:

    225miles / 15 gallons = 15 miles per gallon (miles/gallon)


    For every gallon of fuel used by sofia, her car goes 12.5 miles, while zena's car will 15 miles on one gallon of fuel
    So the work would be

    rt = d
    r23.2 = 290
    __________
    23.2 = 23.2

    r = 12.5 ?

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    Quote Originally Posted by Camaro' View Post
    OH, ill just post a problem, i don't JUST want the answer, i want how to do it with the formula

    sofia drove 290 milesfrom her college to home and used 23.2 gallons of gasoline . her ssister zena drove 225 miles from her college to homeand used 15 galls of gas, whos veichacle has better gas mileag? justify ur answer

    Sorry for spelling, was looking at the paper
    Basically what you are going to do is find the MPG (Miles per gallon)

    In order to do that you simply divide the Mi (Miles) by the G (Gallons) - Sorry if the notation is wrong, we use liters and I don't know what represents what.

    So for Sofia, we take 290mi and divide by 23.2g to get the MPG.
    This equals 12.5MPG if my calculator is working right.

    For Zena you take 225mi and divide by 15g.
    Equals 15MPG.

    So by looking at these two numbers, you can see that 12.5 Miles per every gallon is less than 15 Miles per every gallon, and I'm sure you'd rather drive further on every gallon, so Zena has a better MPG

    E: Kinda got ninja'd but at least I may have helped.

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    Quote Originally Posted by Camaro' View Post
    So the work would be

    rt = d
    r23.2 = 290
    __________
    23.2 = 23.2

    r = 12.5 ?
    correctomundo!

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    wait...where is work coming from?

    rate*time=distance

    force*distance=work (assuming they act in the same direction)


    edit.......

    OH, like, equation work, yeah, you're on the money!
    My work would work as well, just put this before it

    d = rt
    (solve for r)

    r = d/t
    Last edited by Brain; 10-13-2009 at 11:41 PM.
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    brain, no need to incorporate physics into a simple algebra equation

    EDIT: oh you thought he meant like... physics work... not showing math work
    Last edited by Runaway; 10-13-2009 at 11:46 PM.

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    23.2 X 290 = 6728 and 15 X 225 = 3375. Zena has better gas millage.
    (common multiples)
    In both of those, they would each have had to drive 65250 (290 X 225) miles each, and that is how much gas each of them would have used.

    Ghetto on the spot math formulas ftw! Pretty sure this works.
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    Im lost, so ill just post the next one


    A truck travels 40 miles from point a to point b in excatly 1 hr . when the rtuck is halfway between point a - b , a car starts from point and travels at 50 mph.
    How many miles has the car traveled when the truck reaches point b >?

    the formula for 2 things that meet is this i think rt+rt=d

    do i do this?

    r * 1 + r * 30 = 50 im so lost

    The last one was much easier

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    well logically think through this one, the truck takes a total of one hour to drive from a to b, so half way through, there is still 0.5 hours left of his trip.
    So the car has 0.5 hours to get where ever it's going, we know it's moving at 50miles/hour, so

    d=rt=50mph*0.5hour=25 miles
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    Alright, well you know:

    Trucker is going: 40 MPH
    Car is going: 50 MPH

    - When the truck reaches halfway, he has gone 20 miles (because halfway between points a and b is 20 miles).

    - since the truck is going 40 MPH, it will take him 30 minutes to get to point b from the halfway mark.

    - the car will therefore have 30 minutes to drive.

    - since the car is going 50 MPH, it will go 25 miles in 30 minutes.

    EDIT: ninja'd....

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    Guys, i get it, but dont understand how to use d= rt and d =rt to find the answer, i can find it logically but not in paper.

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    You have to find the d = rt for the trucker, and then use that data to plug into the d = rt for the car:

    since we want to see how far he gets to the halfway point:


    d = 40 miles/hour * .5 hours
    20 = 40 miles/hour * .5 hours
    d = 20

    Since we know he has 20 miles from halfway to point B, we can say this:


    d = rt
    20 miles = 40 miles/hour * t
    20 miles = 40 miles/hour * .5 hours
    t = .5 hours

    so now for the second d = rt equation,
    we plug in .5 hours for t from the first equation,
    and use 50 MPH, (since this is known from the initial problem), for r:

    d = 50 miles/hour * .5 hours
    25 miles = 50 miles/hour * .5 hours
    d = 25 miles

    Therefore, the car traveled 25 miles.
    Last edited by Runaway; 10-14-2009 at 12:11 AM.

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    OK Finally, thanks to runaway cop, Garrett, and Brain, all of there ideas helped me figure it out

    I used bits of brain's that i understood, bits of cops, and bits of garrets to make a puzzle

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    Problem: A train can travel 504 miles in 6 hrs, at the same rate how far can it travel in 9 hrs?

    We are using a primary equation to find a constant which will be used for the second equation:

    Train constants we know:
    - D = 504 miles
    - T = 6 hours

    D = rt
    504 miles = r * 6 hours
    504 miles / 6 hours = r
    R = 84 miles/hour

    Now we can use that rate in the next equation:

    New train constants:
    - R = 84 miles/hour
    - T = 9 hours

    D = rt
    D = 84 miles/hour * 9 hours
    D = 756 miles

    The train can travel 756 miles in 9 hours.
    Last edited by Runaway; 10-14-2009 at 12:27 AM.

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    Problem: Aaron rides his bike 5 miles every 45 minutes. How long will it take Aaron to ride 8 miles at this rate?

    Info we know:
    - D = 5 miles
    - T = 45 minutes OR .75 hours

    D = rt
    5 miles = r * .75 hours
    5 miles / .75 hours = r
    R = 6.66 miles/hour

    New info we know:
    - R = 6.66 miles/hour
    - D = 8 miles

    D = rt
    8 miles = 6.66 miles/hour * t
    8 miles / 6.66 miles/hour = t
    T = 1.2 hours

    It would take Aaron 1.2 hours OR about 1 hour and 12 minutes to go 8 miles.

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    Thanks to brain in the beginning,

    Runaway, and garrett, i not only get my homework, its done!

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