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Math Help - Just need help with these 2 equations

  1. #1
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    Just need help with these 2 equations

    1.Find all solutions of the equation and express them in the form
    solutions:


    2.Two ships leave a harbor at the same time, traveling on courses that have an angle of 130degrees between them. If the first ship travels at 22 miles per hour and the second ship travels at 28 miles per hour, how far apart are the two ships after 2.9 hours?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by badandy328 View Post
    1.Find all solutions of the equation and express them in the form
    solutions:


    2.Two ships leave a harbor at the same time, traveling on courses that have an angle of 130degrees between them. If the first ship travels at 22 miles per hour and the second ship travels at 28 miles per hour, how far apart are the two ships after 2.9 hours?
    1. x^2 + 4x + 5 = 0
    => x = [-4 +/- sqrt(16 - 4(5))]/2 ......by the quadratic formula
    => x = [-4 +/- sqrt(-4)]/2
    => x = (-4 +/- 2i)/2
    => x = -2 + i or -2 - i

    2. I will attach a figure soon. notice that if the first ship travels at 22 mph for 2.9 hrs, it will cover a distance of 63.8 miles. if the second ship travels at 28 mph for 2.9 hours, it will cover a distance of 81.2 miles. so we have a triange with two adjacent sides, one 63.8 miles in length, the other 81.2 miles in length and an angle of 130 degrees between them. we want to find the length of the side opposite to the 130 degree angle. we use the cosine rule (or the law of cosines as some call it).

    by the cosine rule:
    a^2 = b^2 + c^2 - 2bc*cos(A)
    => a^2 = (63.8)^2 + (81.2)^2 - 2(63.8)(81.2)cos(130)
    => a^2 = 10663.88 - 10361.12*cos(130)
    => a^2 = 10663.88 - 10361(-0.64279)
    => a^2 = 17323.904
    => a = sqrt(17323.904)
    => a = 131.62 miles

    therefore the distance between the ships after 2.9 hrs is 131.62 miles
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jhevon View Post
    1. x^2 + 4x + 5 = 0
    => x = [-4 +/- sqrt(16 - 4(5))]/2 ......by the quadratic formula
    => x = [-4 +/- sqrt(-4)]/2
    => x = (-4 +/- 2i)/2
    => x = -2 + i or -2 - i

    2. I will attach a figure soon. notice that if the first ship travels at 22 mph for 2.9 hrs, it will cover a distance of 63.8 miles. if the second ship travels at 28 mph for 2.9 hours, it will cover a distance of 81.2 miles. so we have a triange with two adjacent sides, one 63.8 miles in length, the other 81.2 miles in length and an angle of 130 degrees between them. we want to find the length of the side opposite to the 130 degree angle. we use the cosine rule (or the law of cosines as some call it).

    by the cosine rule:
    a^2 = b^2 + c^2 - 2bc*cos(A)
    => a^2 = (63.8)^2 + (81.2)^2 - 2(63.8)(81.2)cos(130)
    => a^2 = 10663.88 - 10361.12*cos(130)
    => a^2 = 10663.88 - 10361(-0.64279)
    => a^2 = 17323.904
    => a = sqrt(17323.904)
    => a = 131.62 miles

    therefore the distance between the ships after 2.9 hrs is 131.62 miles
    here's the diagram, i labeled the points a,b,c and the angles A,B,C
    Attached Thumbnails Attached Thumbnails Just need help with these 2 equations-tri.gif  
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  4. #4
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    thanks sooo much. Still it says the first one is wrong? Are you sure that's right?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by badandy328 View Post
    thanks sooo much. Still it says the first one is wrong? Are you sure that's right?
    yup. pretty sure that's right. you can check for yourself by using the quadratic formula. but let's try another method to be sure. how about completing the square.

    x^2 + 4x + 5 = 0
    => x^2 + 4x = -5
    => x^2 + 4x + (2)^2 = -5 + (2)^2
    => (x + 2)^2 = -1
    => x + 2 = +/- sqrt(-1)
    => x = -2 +/- sqrt(-1)
    => x = -2 +/- i

    which is what we got doing it by the quadratic formula
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    Quote Originally Posted by qbkr21 View Post
    Can it solve cubics?
    Try it (I do not have a TI because I hate calculators, hate them with a love).

    Furthermore, let me warn you more complicated equations are solved through Newton's method.

    They fall as quicky as did Rome if you are not careful.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    what program/utility is this? o it's a TI calc.
    Last edited by Jhevon; March 11th 2007 at 06:45 PM. Reason: got answer from TPH
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    what program/utility is this? o it's a TI calc.
    It is a TI calculator.

    I know because it makes my blood run cold as did the "Vulture Eye".

    The newer models have some wire you can comment to your computer to upload images.

    (Or qbkr took a digital photo, gave it to the computer, then posted it here. But I do not think qbkr is that nice to do something like that).
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  10. #10
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    Re:

    This is the TI Voyage 200 Voyage™ 200 by Texas Instruments - US and CANADA . Massive screen with lightning speed. I guess it can solve cubics. I am showing you two screens because the result is so long. Also I just connected my calc to my computer through the USB it came with.






    They do not sell these calculators in stores (Best Buy, Walmart, ect.) You must buy them through an authorized dealer. I drove an hour and a half to pick this one up before my Combinatorics exam.
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  11. #11
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    Tell it to solve,
    x^3+x=1

    I want to see how it presents the answer.
    (Because I have done this by hand and It leads to a nested radical).
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  12. #12
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    Re:

    Hey ThePerfectHacker,

    In looking over your signature I just wanted to know how Ken was?
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  13. #13
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    Re:

    This probably isn't what you are looking for. The square come out as an irrational fraction, but for this particular problem I am getting a complex #.




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  14. #14
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    Quote Originally Posted by qbkr21 View Post
    This probably isn't what you are looking for. The square come out as an irrational fraction, but for this particular problem I am getting a complex #.
    It seems it is not programmed to solve by radicals.
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  15. #15
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    Re:

    How do you solve cubics what is the formula?

    x^2+x=1
    x^2+x-1=0

    Then what's the next step?


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