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Thread: Math Help Please!

  1. #26
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    Alright, well pm me then if you have any other questions

    ooo Number 3....

    Ok, so to find the area of that rectangle, we need the length times height (Area = L * W in your book). You have the height, 5, so we need to find that bottom length, the length.

    Using the pythagorean theorum, sort of backwards, we are going to use the diagonal measurement with the height, to find the bottom length.

    A^2 * B^2 = C^2 - (Where C is the long side - 13, and A is the height - 5. Just remember that A and B are the shorter sides.

    A^2 * B^2 = C^2 (BTW, ^2 = squared, can't figure out a better way to put it)

    5^2 * B^2 = 13^2

    25 * B^2 = 169 (Simplified)

    B^2 = 169 / 25 (Divided both sides by 25. It cancelled on the left side, since there was a times 25)

    B^2 = 6.76

    B = 2.6 (Found square root of both sides)


    Now, to find the area of that rectangle, we use the length that we just found (2.6), times the height that they gave us (5). The area of the rectangle is 13.


    Simple stuff once you get the hang of it. I hope that was clear enough.

    So, to find the area of a similar problem:

    - Plug the two numbers that they gave you into the Pythagorean Theorum, where A and B are the shorter sides, and C is the long side (Diagonal).

    - Simplify the equation and find the square root of both sides.

    - Use the number that you got for the unknown variable in the equation, times the other short side. (If its a rectangle) If its not a rectangle, then use the equation for that polygon.

  2. #27
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    Updated:
    6/03/08!

    added 3 semester exam questions which is tomorow gonna go sleep and come in the morning to see the answer!

    please help me ! thanks btw ill rep!

  3. #28
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    omgz no1 did it :S sh*tz

  4. #29
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    aff just answered an already answered question without pinkly edditing.

    damn you rockman

    u gets mein angry interwebs face

    >;[

  5. #30
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    #3 answer + explanation:
    60 sq units - you divided the rectangel into 2 equal triangles. by using the pathagorean therom (a^2 + b^2 = c^2) you get 5^2 + x^2 = 13^2 from there you simplfy....25 + x^2 = 169 so now subtrct 25 on each side to get: x^2 = 144 takes sq root of 144 and u get the x val of 12. so now you know the side lengths 5 (given) and 12(just figured out). so multiply them and get 60 square units (inches meters or whatever)

  6. #31
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    thanks snaes... but too late semeseter exam done and summer is here! wooot

  7. #32
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    i looked and looked at... im in enriched algebra in 8th grade... i give up i looked at it for like 30 minutes
    <Wizzup> And he's a Christian
    <Wizzup> So he MUST be trusted
    ___________________________________________
    <Wizzup> she sounds like a dumb bitch

  8. #33
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    hey there guys

    4 months or so? well here still no new thread but school started and i forgot algebra 1 lol yes im dumb lol :P got all geometry still in me haha
    well now taking algebra 2:
    and posted new questions please help meeee by tonite
    ill rep you :P

    btw there like 7 questions out of 25 lol so those i dont get much's :P

    thank you!!!


    The Questions:

    AUG 26 08

    #1-----2(3/4x - 5/8)=7/4

    #2-----4(1/3 - 3x/5)=8/60

    #3-----x/5 + x/3= 10

    #4-----x/2 + 7= 3x/4

    #5-----x/4 + 3= x/8 - 1

    #6-----6x/5 * 10/3 DIVIDE 4/7=14/3

    #7-----4/5 DIVIDE 9/10 DIVIDE 8/3x= 1/3

  9. #34
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    here is number 5 for you...did all this stuff 2 years ago and sadly have forgotten most of it.

    1. x/4 + 3 = x/8 - 1
    Simplify this to
    2. x/4 + 4 = x/8
    To get rid of the 8 multiply both sides by 8
    3. 2x + 32 = x
    Subract 32 from both sides
    4. 2x = x - 32
    Subract x from both sides
    5. x = -32

    If you input -32 into the original problem you get

    (-32)/4 + 3 = (-32)/8 -1

    both equal -5

    Hoped I helped...I will edit if i figure out more

    EDIT: Figured out 4...pretty simple here

    1. x/2 + 7 = 3x/4
    Get rid of the 4 by multiplying by 4 on each side
    2. 2x + 28 = 3x
    Get rid of 2x by subtracting by 2x on each side
    3. x = 28

    If you input 28 into the equation you get
    (28)/2 + 7 = 3(28)/4

    both equal 21

    Lets try numero 3

    1. x/5 + x/3 = 10
    First you have to get common denominators, in this case it would be 15
    2. 3x/15 + 5x/15 = 10
    Now you add the two fractions together
    3. 8x/15 = 10
    To get rid of the 15 you multiply both sides by 15
    4. 8x = 150
    Divide 150 by 8
    5. x = 18.75

    If you input 18.75 for x you get

    (18.75)/5 + (18.75)/3 = 10

    This comes to 3.75 + 6.25 = 10 which is true

    Hope i helped there Timer...come on your that good of a programmer and a total noob like me schools you in math...jk, hope you do well on your Tests!

  10. #35
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    THankx broooo!

    so it's -32 right?
    and the -5 was to check the answer ?

    am i correct?

    AUG 26 08

    #1-----2(3/4x - 5/8)=7/4

    #2-----4(1/3 - 3x/5)=8/60

    #3-----x/5 + x/3= 10

    #4-----x/2 + 7= 3x/4

    #5-----x/4 + 3= x/8 - 1 CHECKED OFF

    #6-----6x/5 * 10/3 DIVIDE 4/7=14/3

    #7-----4/5 DIVIDE 9/10 DIVIDE 8/3x= 1/3

  11. #36
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    yessir, you are correct. Its a little trick i learned for Alg. tests....ALWAYS CHECK ANSWERS...it will save your neck on some difficult problems. Trust me, got a solid A in alg 1

  12. #37
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    dannng k thanks hey you got any others like teh last 2 and some others ?

  13. #38
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    ahh to be young again!

    #1-----2(3/4x - 5/8)=7/4

    #2-----4(1/3 - 3x/5)=8/60

    #3-----x/5 + x/3= 10

    #4-----x/2 + 7= 3x/4

    #5-----x/4 + 3= x/8 - 1

    #6-----6x/5 * 10/3 DIVIDE 4/7=14/3

    #7-----4/5 DIVIDE 9/10 DIVIDE 8/3x= 1/3
    #1
    First take the 2 through to get
    6/4x - 10/8 = 7/4
    next simplify
    3/2x - 5/4 = 7/4
    next, multiply by 4.
    12/2x - 20/4 = 7
    simplify again
    6/x - 5 = 7
    add 5 to both sides
    6/x = 2
    times both sides by x
    6 = 2x
    and divide by 2
    x = 3.

    #2
    again, take the 4 through first.
    4/3 - 12x/5 = 8/60
    Find a common denominator to combine the fractions. 15 will do so multiply by the difference to get
    20/15 - 36x/15 = 8/60
    put the fractions together to get
    (20 - 36x)/15 = 8 / 60
    multiply both sides by 15
    20 - 36x = 8 / 4
    simplify
    20 - 36x = 2
    subtract 20 and multiply by -1 to make things neat to get
    36x = 18
    divide by 36
    x = 18 / 36
    and simplify
    x = 1/2

    #3
    first, find the common denominator. Looks like 15 again.
    3x/15 + 5x/15 = 10.
    simplify
    8x/15 = 10
    and multiply both sides by 15
    8x = 150
    and divide by 8 to get
    x = 75/4

    #4
    first, move the x's to the same side by subtracting the x from the left to get
    x/2 - 3x/4 + 7 = 0
    combine the x's. first find a common denominator for them so it looks like
    2x/4 - 3x/4 + 7 = 0 ->
    -x/4 + 7 = 0
    subtract the 7 over
    -x/4 = -7
    multiply by a negitive 4 to get
    x = 28

    #5
    Again, get the x's to the same side
    x/4 - x/8 + 3 = -1
    and combine them..
    2x/8 - x/8 + 3 = -1 ->
    x/8 + 3 = -1
    subtract the 3 over
    x/8 = -4
    and multiply both sides by 8 to get
    x = -32

    #6 (I hate these problems, but they are common)
    First I would flip the 4/7 to the top to get (multiplication is much easier then division.)
    6x/5 * 10/3 * 7/4 = 14/3
    then simplify it
    420x/60 = 14/3
    and simplify it some more
    7x = 14/3
    divide both sides by 7 to get
    x = 2/3

    #7 4/5 DIVIDE 9/10 DIVIDE 8/3x= 1/3
    Ok, the trick to this is going to be to flip it, simplify it, then flip it again so the first step
    ((4/5) / (9/10)) * 3x/8 = 1/3
    then flip again
    (4/5) * (10/9) * (3x/8)
    then multiply through
    120x/360 = 1/3
    and simplify
    x/3 = 1/3
    Times both sides by 3 to get
    x = 1

    I was somewhat tired while doing this so check my work (It aint my homework) But I believe the answers for the most part should be correct. At very least the method to get them is.

    If you go onto higher math (Calculus) this stuff will be like second nature to you. I sometimes slip a negative here and there. but I've pretty much gotten to the point where I can look at any algebra problem and instantly know what steps I want to take. Its like calculus gives you some sort of magic math algebra intuition (Scary I know)

  14. #39
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    Guys i put up 2 questions please help me!!!
    will post more!

  15. #40
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    Quote Originally Posted by ~~Joker~~ View Post
    TOOK OUT OLD QUESTIONS

    NEW QUESTIONS(10/12/08):


    1)At Used Book Sale, 5 paperback books cost $3.75. The total cost, c, of purchasing n paperback books can be found by-
    a) subtracting n from c
    b) dividing n by the cost of 1 book
    c) Multiplying n by c
    d) multiplying n by the cost of 1 book

    2)Randy has $30 to spend at a town fair. the admission price is $6 and each ride costs $2. which inequality can be solved to find how many rides randy can afford?
    NOTE: (> ARE ALL UNDERLINED, like _ and > on top of that underline!!!)
    a) 30>6 + 2r
    b) 30<6 + 2r
    c) 30> 6r + 2
    d) 30< 6r + 2


    WILL RE-EDIT AND POST LIKE 5 MORE SOON!!
    ILL REP YOU FOR SURE AND LOVE YOU
    1) d
    2) b

    1) c := n * $3.75
    c := 5 * 3.75
    c := 18.75.

    2) 30 := 6 + 2r where r = rides
    24 := 2r
    12 := r;
    12 rides can be ridden.
    By reading this signature you agree that mixster is superior to you in each and every way except the bad ways but including the really bad ways.

  16. #41
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    1) d.
    2) b.

    Easy stuff, pre-algebra?

  17. #42
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    yeah, its pre-algebra. I am just starting to take it.

  18. #43
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    Added 4 more QUESTIONS!!!!

    1-2 DONE!

    3-6 please!!!!

    and rep to both for helping on 1-2!!!

    please need by tonight! thank you!

  19. #44
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    psedopost representing help on msn, solving everysingle problem (please no report for spam, I helped him, just not here on msn , be a nice person)

    Join the fastest growing merchanting clan on the the net!

  20. #45
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    lyer, l0l jp thanks dude REP to you and derek

    -.- derek post here...
    l0l

  21. #46
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    I got 3,4,5 done for him. #6 is one of those that I would need the book or an example problem to be able to solve. One of those types or equations I don't possess off the top of my head. Like the other 3, I use them constantly in every math class you go through...


    3)
    ERA = 9 * ER / IP
    So this leads to plugging in numbers…
    2.25 = 9 * ER / 212
    Multiply by 212 to get rid of fractions - > 447 = 9 * ER
    Divide by 9 to solve for ER (Earned Runs) -> 53 = ER


    4)
    Area of circle – > A = Pie * r^2
    So Area of the circle is -> A = 3.14 * 2^2 - > A = 3.14 * 4 -> A = 12.56
    So it wants what was shaven off, the excess.
    The area of the whole piece is 4 * 4 (area of a square) so 16
    16 – 12.56 = 3.44 – Answer ( A )


    5 )
    Pythagorean Theorem – A^2 + B^2 = C^2
    A & B are sides/legs of the triangle, and C is the hypotenuse(the red line/the diagnol)
    So C = Sqrt(50^2 + 94^2) – Answer ( D )


    P.S. – anything that is ^ means and exponent. So x^2 means x squared…

    "Failure is the opportunity to begin again more intelligently" (Henry Ford)


  22. #47
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    ADDED new 6 questions 10/16/08 please help me here guys i get everything else except these questions!!!!

    please need by 2nite! please explain! not just the answer.

  23. #48
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    I did the first three. Maybe I'll do the others later, but not right now.

    1. m = -1/2
      Plug in -1 for x and 5 for y to get your y-intercept, which is the last value you'll need to write it in slope-intercept form.
      y - 5 = -1/2(x - (-1))
      y - 5 = -1/2x - 1/2
      y = -1/2x + 9/2
    2. You need to find the slope first.
      (5 - 3) / (-4 - 2) (rise / run or change in y / change in x)
      m = 2 / -6 or -1/3
      Then, same thing as the last problem - plug in a point.
      y - 3 = -1/3(x - 2)
      y - 3 = -1/3x + 2/3
      y = -1/3x + 11/3
    3. y-intercept of 6 means b = 6. Therefore, we know the equation will look something like y = mx + 6. Then, we can find the slope using the rise / run formula.
      (6 - 0) / (-4 - 0)
      m = 6 / -4 or -3 / 2
      y = -3/2x + 6
    4. This is actually #5.
      Anyways, since the line y = -3 is horizontal, you'll need a vertical line to be perpendicular to it. That means x = something. The line that goes through (-4, -7) is the line x=-4.
    :-)

  24. #49
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    added new question, quite easy hopefully i just hate factoring lol.

    Thanks ill rep

  25. #50
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    Look in your book. I'm positive there's at least a section or some mention of factoring. Not to mention the entire internet is your resource.

    The answer is (2x - 1)(x - 3) by the way.
    :-)

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