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Thread: Need help Interpreting a Graph

  1. #1 Need help Interpreting a Graph 
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    Can someone please help me understand the graph posted in the image. So I have answered the first two question but need help with the rest.
    18. The temperature of the asthenosphere is 1000 Celsius.
    19. The approximate depth of the earthquakes that occur in this region is 300 to 400km.
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    What are you ideas?

    For 20. Look at where dots, which represent earthquake location, lay with respect to the subduction zone near the surface. Do you see a relationship such as "when X happen, Y also happens?"
    21. Look at the isotherms (lines of constant temperature) on both sides and around the subduction zone.


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  4. #3  
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    give it up, go find some girls, beer, weed, whatever your fancy. NONE of you will ever prove anything because nthere are very few things that have ever been proved as fact. For all your study all your thinking all your phillisosphying is a waste of time for you know about as much as the idiots who are "trying" to run the show, which adds up to absolutely nothing and furthermore , to pretend you have somehow superior knowledge to any one else as to what happened in the past puts you in the same category as these religious fanatics who also know nothing apart from what they have been taught by people who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught by peoplle who know nothing apart from what they have been taught and so on. Sure there is nothing wrong with you having a theory about what happened in the past or what will happen in the future but let me tell you this for fact, what has been has been and what will happen will happen and you will still be a poor pathetic human being with no real idea what might have happened or what might happen just like the rest of us.
     

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    funny you should mention that did you know that the equation for the unit circle x^2+y^2=1 is incorrect if x or y is assigned to 0.75 the awnser you get when solving for x or y is wrong
     

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    Genius Duck Moderator Dywyddyr's Avatar
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    Quote Originally Posted by fiveworlds View Post
    funny you should mention that did you know that the equation for the unit circle x^2+y^2=1 is incorrect if x or y is assigned to 0.75 the awnser you get when solving for x or y is wrong
    Don't talk crap.
    Last edited by Dywyddyr; February 23rd, 2013 at 03:47 PM.
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    i always thought it was the pythagoran theorem side1^2+side2^2=hyp^2 though its all a bit confusing maybe smoking weed would help
     

  8. #7  
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    Quote Originally Posted by fiveworlds View Post
    i always thought it was the pythagoran theorem side1^2+side2^2=hyp^2
    It is.
    And the radius (1 in this case) is the hypotenuse.



    though its all a bit confusing maybe smoking weed would help
    Or getting some brains.
    "[Dywyddyr] makes a grumpy bastard like me seem like a happy go lucky scamp" - PhDemon
     

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    reminds me to watch resident evil .... zombies
    0.75^2=0.5625
    1-0.5625=0.4375
    sqareroot(0.4375)=0.661437
    greater than half y axis therefore wrong
    Last edited by fiveworlds; February 23rd, 2013 at 06:06 PM.
     

  10. #9  
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    Quote Originally Posted by fiveworlds View Post
    reminds me to watch resident evil .... zombies
    0.75^2=0.5625
    1-0.5625=0.4375
    sqareroot(0.4375)=0.661437
    greater than half y axis therefore wrong
    Try using correct mathematics.

    x = 0.75. 0.752 = 0.5625
    y = 0.661438. 0.6614382 = 0.437500227844
    0.5625+0.437500227844 = 1
    10.5 = 1

    I have no idea where "half y axis" comes in, the final value you have is the y value.
    The diagram clearly shows that the y value is 0.661438 (okay, you have 7 as the final figure, but that's AutoCAD's rounding for you).
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    half y axis comes in when you have a point on the x axis greater than the midpoint of the circle then the y point should be smaller than 0.5 here however y=0.661438 which is greater than 0.5 and therefore your point is not on your circle its somewhere outside the circle
     

  12. #11  
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    Unadulterated crap.
    Please provide some evidence that "half y axis" is valid.
    YOU gave the equation (no mention of a proviso based on "half y axis").
    I gave a numerical solution AND a graphical one.
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    i just your maths sigh
     

  14. #13  
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    And now we discover that not only are you mathematically ignorant but you can't construct a comprehensible sentence in English either.
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    funny how you can draw a triangle of hyp=1 in your unit circle triangle using the formula without it going anywhere near a single point on your circle. draw your hyp as your diameter and construct both sides
     

  16. #15  
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    Quote Originally Posted by fiveworlds View Post
    funny how you can draw a triangle of hyp=1 in your unit circle triangle using the formula without it going anywhere near a single point on your circle
    Ah, understood. You're blind too.
    The hypotenuse is radial - it ends ON THE CIRCLE.
    You know, where it meets the top (end point) of the vertical line.

    Just for clarification (since you appear to be having difficulty) look at the triangle - that's the thing made up of three straight lines [I assume you can count to 3 and that you know what a straight line is?], of the lines one is horizontal [it goes across the page], one is vertical [it goes UP the page] and the third is sloped at an angle. The end points of that line are, from left to right, at the centre of the circle and on the circumference.
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    your diameter is two place the start of your triangle at 0.235 it will have a radius of 1 therefore 1.235 however both points on your circle are at 0,2 along the line this triangle goes near neither
     

  18. #17  
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    More incomprehensible drivel.
    The equation doesn't use diameter.

    Did you ever have an education? Of any sort?
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    to be honest i prefer using diameter one point and saying that the rate of change of the length of the sides is proportional to the rate of change between the angle of one side and the diameter maintaining an angle of 90 degrees
     

  20. #19  
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    More inane drivel.
    And a complete failure to address any of my points.
    Plus the continued failure to explain where this "half y axis" comes in.
    And not forgetting that your claim of "funny how you can draw a triangle of hyp=1 in your unit circle triangle using the formula without it going anywhere near a single point on your circle" is totally erroneous.

    Why do you bother posting?
    You could spend the time actually learning something.
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    triangle of fixed angle 90 degrees. A triangle maintains 180 degrees therefore a change of one in an angle would mean a proportional change in another angle 59,61 the length of both sides changes proportionally to the rate of change of angle but you aparently are arguing against this
     

  22. #21  
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    Quote Originally Posted by fiveworlds View Post
    triangle of fixed angle 90 degrees. A triangle maintains 180 degrees therefore a change of one in an angle would mean a proportional change in another angle 59,61 the length of both sides changes proportionally to the rate of change of angle but you aparently are arguing against this
    Until you specify exactly how you apply this then it's rather hard to argue for or against.1
    Please try to post in comprehensible manner.
    Is English your first language?


    1 But, at first glance, I'd say you can't.
    Last edited by Dywyddyr; February 24th, 2013 at 06:28 AM.
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    uses well
    cos(@)=adj/d
    sec(@)=d/adj
    sin(@)=opp/d
    csc(@)=d/opp
     

  24. #23  
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    Yes, another essentially meaningless post.
    How does that explain exactly (or even roughly) how it's applied?
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    also you have
    sin(@1)=opp/d
    sin(@2)=adj/d
     

  26. #25  
    Genius Duck Moderator Dywyddyr's Avatar
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    I'm not sure if you have some sort of mental aberration or you truly don't understand the subject.
    Does (continued) posting of non-sequiturs somehow, in your mind, explain what your point (such as it may be 1) is?

    1 I say this because, on the face of it, you don't appear to have one. Other than claiming to "prefer" a method that, if it actually does work, involves far more work than the equation already demonstrated.
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    sin@1+sin@2=(opp+adj)/hyp
    @1+@2=90
    sin@1+sin(90-@1)=(opp+adj)/hyp
    (sin@1+sin90-@1)/(opp+adj)=1/diameter
     

  28. #27  
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    Quote Originally Posted by fiveworlds View Post
    sin@1+sin@2=(opp+adj)/hyp
    @1+@2=90
    sin@1+sin(90-@1)=(opp+adj)/hyp
    (sin@1+sin90-@1)/(opp+adj)=1/hyp
    Got it.
    You don't understand the subject and consequently choose to resort to non-explanations in the hopes that I'll somehow think you have some grasp of geometry.

    Unless you can explain 1 don't bother replying.

    1 Make (an idea, situation, or problem) clear to someone by describing it in more detail or revealing relevant facts or ideas.
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    im looking for alternate ways of arriving at a solution if you cannot use the conventional ways are there other ways to solve for an awnser
     

  30. #29  
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    Quote Originally Posted by fiveworlds View Post
    im looking for alternate ways of arriving at a solution if you cannot use the conventional ways are there other ways to solve for an awnser
    Still a fail.
    You've claimed (post #18) that you prefer using a different method. How can you prefer that method if you don't know whether works or not?
    And, still, you've done nothing whatsoever to explain how that method should work.
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    it isnt a different method just a manipulation of it currently trying to learn fourier transforms an manipulations of sin, cos and tan
     

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    Of course it's a different f*cking method!
    The original (workable) version doesn't, for example, use this "half y axis" crap.
    The original (workable) version doesn't use the diameter.
    The original (workable) version is algebraic, not one one that uses Fourier transforms.
    The original (workable) method WORKS. Unlike the crap you're (not quite) posting.

    I'm still waiting for you to provide some sort of explanation on how the diameter is used.
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    (sigh) Closed.
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