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☆ENGLISH ONLY☆コミュのDESPERATE HELP FOR G.R.E MATH PROBLEM

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pls somebady help me!!
A line in the xy-coordinate place has x-intercept A and y-intercept b, where a and b are nonezero. The slope of this line is
?1/ab ?−b/a ?−a/b ?b/a ?a/b

The answer is ?. I need explanation...

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-b/a

try drawing the graph. cant do it from a keitai, sorry.
hey there,

Is this a linear function, or a quadratic etc? looking at さぃ's answer the slope is a linear function....
I'm not too bright, but for everyone to help you easier you might want to define your operators better ;

do you mean 1/ab - b/a - a/b + b/a + a/b ?

remember the basic equation for linear functions ;

y = mx + c

so what you typed was the 'gradient' or 'm', no?

It's a long time since I did any mathematics but I get

m = (1-2b^2)/ba where b^2 means b squared, or b×b

so if you put A in as 'c' in the equation you should get y = b

i.e

((1-2b^2)/ba) + A = b

the problem here is your 'a' and 'A'. Are they the same number? If not you can't get an explicit solution. As さぃ said you could try plotting a graph....but without A you are stuck (unless you plot the axis in terms of A, a, or whatever)

Anyway sorry I probably got this totally wrong....just give your operators and nomenclature a quick check. Then reduce your 'm' down to it's simplest form....if all goes well when you put x=a you should get y=b

(Sorry for all the math words)
hey...what text encoding are you using? I think there's some info I can't see on screen.....well....whatever....
from b to a, value-x increased by a and value-y decreased by b so the answer is -b/a. the formula for the slope is Y1-Y2/X1-X2 or Y2-Y1/X2-X1 (distance between two points)
Good Luck=)
The slope of the line connected by two points, A (Xa, Ya) and B (Xb, Yb), is determined by (Yb - Ya) / (Xb - Xa). In this case, point A is on (a, 0), and point B is on (0, b).

Therefore, the slope is yielded by (Yb - Ya) / (Xb - Xa) = (b - 0) / (0 - a) = b / -a = - b / a.

QED. :)
Probably meaningless, but the slope of a line between two points is as だいはど and 佳代 put it for straight lines only. Be careful with other similar math questions that may involve higher functions of 'x', exponentials, logarithms etc - you won't be able to conveniently divide out for the gradient between two points.

Was this a University entrance exam question!!?
Thank you so much everybady!! I really appreicate your consideration!! I would say GRE sucks!! :P but I could finally get this. it's nice to have smart ppl!! :)
Hi "..."

>> for straight lines only

Mathematically speaking, lines are always straight, aren't they? ;) ;)
ermm, not sure....I don't think so. If you were to draw a two dimensional grid with finite increments of x and y. Then say draw a circle, perfect or not - doesn't matter. Then we zoom in extremely close, it would seem from one finite point to another the circle is 'straight'. But this is a resolution problem. At the 'zoomed in' level if we were to resize our veiwing grid to much smaller increments the line would appear curved again, no?
i.e The solution is just approaching linearity at inifinietly small increments of x or y. I think any function with x^1 is linear - higher powers of x are not 'linear' or 'straight'. Looking at the equation of this circular line it can not be straight, as if we move a little in the x direction the y value would change in a non linear way, although at certain resolutions it would appear so...
I think this was the way it works no?

Anyway I should have defined my words better, by "straight" I mean 'linear', and by linear I mean functions that can be described by y = mx + c....

Are you saying that at extremely high resolutions even severe curves like an exponential say can be approximated to y = mx + c? I agree! This is one of the first numerical methods mathematicians use to interpolate over small increments or approximate non-linear functions. For anyone thinking that's a load of useless info, every curved line on your computer screen right now has been described by this process of approximation!

Anyway I think I used the non mathematical word "line" not the mathematical term "line". That's what sounded funny maybe - I wasn't sure how much mathematical wording stinky would understand, so I used a couple of words in lamens terms. I think you might be referring to "curves" and "lines" right? If so "yes!" lines are always 'straight'. But the everyday English word "line" could be 'curved' - like in "I like her lines♪" kinda thing

Maths is fun!
I was wish I was bright enough to know more!

Anyway here's my defintion of lines etc put more concisely ;

1-dimensional = dot/point
2-dimensional = line
3-dimensional = shape/solid
4-dimensional = shape/solid that changes shape dependant on time

so any "line" for me is something described in xy space, could be straight, could be curved. "Line" just means it's 2-D

That's why I wanted to let stinky know the formula you guys gave for his gradient is only precise for y = mx + c or put another way 'straight lines'....any higher powers of x and he'll need differentitation.

p/s the best thing about mathematics ;
you can put an non real number, for example the sqaure root of minus one (sometimes called "j"), into a hyberbolic function - say a hyperbolic sine, and you get a real number result! A real number that can be described by 1,2,3,4 etc from an abstract concept - "j" !!!

It's the only real magic there is on earth I think....
Thank you... Fujimura sensei!!
yeah prospective students often obsessively persist on GRE/TOEFL scores. In my case, I have aweful GPA... so must make it up at GRE exam. anyways I really appriciate your consideration. :)
Here's the simplest explanation I can think of:
1. x-intercept a means (a,0) is on the line.
2. y-intercept b means (0,b) is on the line.
3. Given the two points, you can determine the slope:
m = (b-0)/(0-a) = -b/a

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