# Need help Interpreting a Graph

• February 14th, 2013, 12:37 PM
chrisman10
Need help Interpreting a Graph
Attachment 1820
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.
• February 20th, 2013, 10:33 AM
Lynx_Fox
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.
• February 23rd, 2013, 09:57 AM
unofall
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.
• February 23rd, 2013, 03:58 PM
fiveworlds
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
• February 23rd, 2013, 04:28 PM
Dywyddyr
Quote:

Originally Posted by fiveworlds
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.
• February 23rd, 2013, 04:49 PM
fiveworlds
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 :)
• February 23rd, 2013, 04:53 PM
Dywyddyr
Quote:

Originally Posted by fiveworlds
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.

http://i23.photobucket.com/albums/b3...pse3262751.png

Quote:

though its all a bit confusing maybe smoking weed would help :)
Or getting some brains.
• February 23rd, 2013, 05:01 PM
fiveworlds
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
• February 23rd, 2013, 07:47 PM
Dywyddyr
Quote:

Originally Posted by fiveworlds
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).
• February 23rd, 2013, 08:11 PM
fiveworlds
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
• February 24th, 2013, 05:38 AM
Dywyddyr
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.
• February 24th, 2013, 05:50 AM
fiveworlds
• February 24th, 2013, 05:54 AM
Dywyddyr
And now we discover that not only are you mathematically ignorant but you can't construct a comprehensible sentence in English either.
• February 24th, 2013, 06:03 AM
fiveworlds
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
• February 24th, 2013, 06:09 AM
Dywyddyr
Quote:

Originally Posted by fiveworlds
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.
• February 24th, 2013, 06:22 AM
fiveworlds
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
• February 24th, 2013, 06:24 AM
Dywyddyr
More incomprehensible drivel.
The equation doesn't use diameter.

Did you ever have an education? Of any sort?
• February 24th, 2013, 06:48 AM
fiveworlds
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
• February 24th, 2013, 06:58 AM
Dywyddyr
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.
• February 24th, 2013, 07:05 AM
fiveworlds
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
• February 24th, 2013, 07:10 AM
Dywyddyr
Quote:

Originally Posted by fiveworlds
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.

1 But, at first glance, I'd say you can't.
• February 24th, 2013, 07:37 AM
fiveworlds
uses well
sin(@)=opp/d
csc(@)=d/opp
• February 24th, 2013, 07:42 AM
Dywyddyr
Yes, another essentially meaningless post.
How does that explain exactly (or even roughly) how it's applied?
• February 24th, 2013, 08:28 AM
fiveworlds
also you have
sin(@1)=opp/d
• February 24th, 2013, 08:48 AM
Dywyddyr
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.
• February 24th, 2013, 10:29 AM
fiveworlds
@1+@2=90
• February 24th, 2013, 10:35 AM
Dywyddyr
Quote:

Originally Posted by fiveworlds
@1+@2=90

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.
• February 24th, 2013, 11:18 AM
fiveworlds
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
• February 24th, 2013, 11:23 AM
Dywyddyr
Quote:

Originally Posted by fiveworlds
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.
• February 24th, 2013, 11:46 AM
fiveworlds
it isnt a different method just a manipulation of it currently trying to learn fourier transforms an manipulations of sin, cos and tan
• February 24th, 2013, 11:58 AM
Dywyddyr
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.
• February 24th, 2013, 05:47 PM
Lynx_Fox
(sigh) Closed.