Page 1 of 3 123 LastLast
Results 1 to 15 of 33

Math Help - Mid-Terms Review

  1. #1
    Member
    Joined
    Apr 2006
    Posts
    85

    Mid-Terms Review

    I just have a few questions I don't quite understand. Hope you guys can help me.

    Calculate the distance between each of the following pairs of points.

    1) I(8,2) and J(-2,-3)

    This means find the slope of the two points right? My final answer (finding the slopes) was \frac{1}{2}. Is that right? Is it even slope at all or is it something completely different.

    2) If the line joining A (a,4) and B (2,3) is parallel to the line joining C (-5,3) and D (1,1), find the value of a.

    I am guessing to do this you must find the slope then the y or x intercepts?

    3) The express 6\sqrt{2} + \sqrt{32} is equivalent to...

    It's been so long since I did radicals I have no idea why the teacher would put this on the test... anyways, I'm at a complete blank. From mixed radical to the radical that isn't mixed (Lol, I'm that lost, forgot the name...), it's square the mixed number then add it to the one in the square root right? So 6\sqrt{2} will become \sqrt{38}. Right? Wrong? If right, then thanks I know how to solve the problem. If wrong, please tell me the right way :P


    Thanks for the help guys.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Rocher View Post
    I just have a few questions I don't quite understand. Hope you guys can help me.

    Calculate the distance between each of the following pairs of points.

    1) I(8,2) and J(-2,-3)

    This means find the slope of the two points right? Mr F says: Wrong! It means find the distance.

    My final answer (finding the slopes) was \frac{1}{2}. Is that right? Is it even slope at all or is it something completely different. Mr F says: Does the question mention slope? No. So even though that actually is the correct slope, it's the wrong answer. Does the question mention distance? Yes. The formula is d = \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}.

    2) If the line joining A (a,4) and B (2,3) is parallel to the line joining C (-5,3) and D (1,1), find the value of a.

    I am guessing to do this you must find the slope then the y or x intercepts? Mr F says: Half right. Find the slope of the line passing through C and D. I know you can find slope because you found the correct slope between the points given in question 1 (pity the question wasn't asking for slope). Now find the slope of the line passing through A and B. It will have a in it. Now put the two slopes equal to each other and solve for a.

    3) The express 6\sqrt{2} + \sqrt{32} is equivalent to...

    It's been so long since I did radicals I have no idea why the teacher would put this on the test... anyways, I'm at a complete blank. From mixed radical to the radical that isn't mixed (Lol, I'm that lost, forgot the name...), it's square the mixed number then add it to the one in the square root right? So 6\sqrt{2} will become \sqrt{38}[/tex]. Right? Wrong? Mr F says: Wrong for two reasons: 1. It becomes \sqrt{36 \times 2} = \sqrt{72}. NOT \sqrt{36 + 2} = \sqrt{38}!!! 2. They probably want you to simplify the expression, in which case you need to reduce \sqrt{32}.

    If right, then thanks I know how to solve the problem. If wrong, please tell me the right way :P Mr F says: \sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4 \sqrt{2}. So 6\sqrt{2} + \sqrt{32} is equivalent to 6\sqrt{2} + 4\sqrt{2} = 10 \sqrt{2}.

    Thanks for the help guys. Mr F says: Just hit the thanks button, dude.
    ..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2006
    Posts
    85
    1) My final answer is \sqrt{-15}. Correct?

    2) My final answer is -1. Correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by Rocher View Post
    1) My final answer is \sqrt{-15}. Correct?

    2) My final answer is -1. Correct?
    1) No, think... how could you possibly get a negative result from: a^2 + b^2. You can't... you did not square the differences.

    2) Yes, you correctly matched the slopes.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2006
    Posts
    85
    Quote Originally Posted by colby2152 View Post
    1) No, think... how could you possibly get a negative result from: a^2 + b^2. You can't... you did not square the differences.
    It's I(8,2) and J (-2,-3). There are some negatives in there, and so you can... but anyways, I show what I did...

    <br />
d=\sqrt{(-3-2)+(-2-8)}<br />
 =\sqrt{(-5)+(-10)}<br />
 =\sqrt{-15}<br />
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Dec 2007
    From
    Georgia
    Posts
    85
    Quote Originally Posted by Rocher View Post
    It's I(8,2) and J (-2,-3). There are some negatives in there, and so you can... but anyways, I show what I did...

    <br />
d=\sqrt{(-3-2)^2+(-2-8)^2}<br />
 =\sqrt{(-5)^2+(-10)^2}<br />
 =\sqrt{25 + 100}<br />
 =\sqrt{125}<br />
 =5\sqrt{5}<br />
    I fixed it for ya Checked the above quote, you forgot to square each term. That's why you ended up with negative numbers.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by Rocher View Post
    It's I(8,2) and J (-2,-3). There are some negatives in there, and so you can... but anyways, I show what I did...

    <br />
d=\sqrt{(-3-2)+(-2-8)}<br />
 =\sqrt{(-5)+(-10)}<br />
 =\sqrt{-15}<br />
    You still haven't squared the differences.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Apr 2006
    Posts
    85
    Thanks, but (-5)^2+(-10)^2 equals out to a negative number on my calculator o.O Also, I have two more questions I hope you can help me with...

    1) 32^{\frac{3}{5}}a^{-\frac{2}{3}}b^2 \div (216a^4b^2)^\frac{1}{3}
    How in the world are you supposed to do fraction exponents!?!

    2) (\frac{27x^{-5}}{125y^2})^{-\frac{2}{3}}
    Again, how are you supposed to do fraction exponents?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Rocher View Post
    Thanks, but (-5)^2+(-10)^2 equals out to a negative number on my calculator o.O Also, I have three more questions I hope you can help me with...
    hehe, i guess this scores one for the people who say calculators should be banned from calc classes...

    you entered the numbers wrong or something. (-5)^2 > 0 and (-10)^2 > 0, there is no way their sum is negative

    if your calculator has gone haywire, or you're not sure you're entering the numbers correctly, note that (-5)^2 + (-10)^2 = 5^2 + 10^2 = 25 + 100 = 125
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Rocher View Post
    1) 32^{\frac{3}{5}}a^{-\frac{2}{3}}b^2 \div (216a^4b^2)^\frac{1}{3}
    How in the world are you supposed to do fraction exponents!?!
    note that \frac {x^a}{x^b} = x^{a - b}

    you think you can try it now?

    2) (\frac{27x^{-5}}{125y^2})^{-\frac{2}{3}}
    Again, how are you supposed to do fraction exponents?
    just treat fractional exponents as you would regular numbers

    here you need the rule: \left( x^a \right)^b = x^{ab} in addition to the rule i gave you earlier

    you think you can take it from here?

    you really should post new questions in a new thread

    for the meaning of fractional exponents, see post #3 here. (i don't suppose you have trouble with negative exponents?)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Jhevon View Post
    hehe, i guess this scores one for the people who say calculators should be banned from calc classes...

    you entered the numbers wrong or something. (-5)^2 > 0 and (-10)^2 > 0, there is no way their sum is negative

    if your calculator has gone haywire, or you're not sure you're entering the numbers correctly, note that (-5)^2 + (-10)^2 = 5^2 + 10^2 = 25 + 100 = 125
    It's a sure bet that the expression was keyed in as -5^2 + -10^2, that is, no brackets. The sad part is the uncritical acceptance of what the calculator spits out ....
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Apr 2006
    Posts
    85
    Quote Originally Posted by Jhevon View Post
    note that \frac {x^a}{x^b} = x^{a - b}

    you think you can try it now?

    just treat fractional exponents as you would regular numbers

    here you need the rule: \left( x^a \right)^b = x^{ab} in addition to the rule i gave you earlier

    you think you can take it from here?

    you really should post new questions in a new thread

    for the meaning of fractional exponents, see post #3 here. (i don't suppose you have trouble with negative exponents?)
    No lol, I still have no idea what to do... There are three exponents on the side, so how are you supposed to add / subtract them?!
    Follow Math Help Forum on Facebook and Google+

  13. #13
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by Rocher View Post
    Thanks, but (-5)^2+(-10)^2 equals out to a negative number on my calculator o.O Also, I have two more questions I hope you can help me with...

    1) 32^{\frac{3}{5}}a^{-\frac{2}{3}}b^2 \div (216a^4b^2)^\frac{1}{3}
    How in the world are you supposed to do fraction exponents!?!

    2) (\frac{27x^{-5}}{125y^2})^{-\frac{2}{3}}
    Again, how are you supposed to do fraction exponents?
    Add exponents of like bases when there bases are multiplied with each other:

    a^{-\frac{2}{3}} \div a^4 = a^{-\frac{2}{3}}a^{-4}\Rightarrow a^{-\frac{2}{3}-\frac{12}{3}}=a^{\frac{-14}{3}}
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Apr 2006
    Posts
    85
    -14? Isn't it supposed to be +10?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    Apr 2006
    Posts
    85
    Agh, I give up... can you guys just show me how to do those two step by step? There are two more questions like them and with the two answers I might be able to do the other ones... I just don't get this stuff.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 3 123 LastLast

Similar Math Help Forum Discussions

  1. Help on Pre-Calculus Review
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 7th 2010, 02:41 PM
  2. Calculus Review
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 6th 2010, 06:52 AM
  3. Need help on Review
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 5th 2009, 10:53 PM
  4. Help with review
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: January 22nd 2008, 03:07 PM
  5. please help me with this exam review. thanks a lot.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 26th 2007, 09:13 PM

Search Tags


/mathhelpforum @mathhelpforum