SOLUTION: http://172.85.76.57/Library/00001230_84_624_A1.png Can you please go to the link and post the answer here? I'm bad at this kind of thing.

Question 999630: http://172.85.76.57/Library/00001230_84_624_A1.png Can you please go to the link and post the answer here? I'm bad at this kind of thing.
Found 2 solutions by MathLover1, Theo:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!

you just need the coordinates of two points that lie one line to find the equation of that line

use y-intercept which is at (,)
so, we already know :

to find a slope, use (,) and the point (,) which also lie on a line

then, your equation is:

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
here's the picture of your graph.

from what i can see, the graph crosses the y-axis at (0,1) and it crosses the x-axis at (1.5,0)

the slope is the change in y divided by the change in x.

the formula for that is (y2-y1) / (x2-x1).

if you let (x1,y1) = 0,1) and you let (x2,y2) = (1.5,0), then:

(y2-y1) / (x2-x1) becomes (0-1) / (1.5-0) which becomes -1 / 1.5 which becomes -2 / 3 if you need to have a simplified version where both numerator and denominator are integers.

-2/3 was derived by multiplying -1/1.5 * 2/2.

the y-intercept is the value of y when x is equal to 0.

since one of your points is at (0,1) which means the value of y when x = 0, your y-intercept is already shown at y = 1.

the slope intercept form of the equation for a straight line is y = mx + b.

m is the slope and b is the y-intercept.

the equation of your line is therefore y = -2/3 * x + 1

when x = 0, the value of y is -2/3 * 0 + 1 which is equal to 1.

when x = 1.5, the value of y is -2/3 * 1.5 + 1 which is equal to -3/3 + 1 which is equal to -1 + 1 which is equal to 0.

these two points of (0,1) and (1.3,0) are the same two points you show on the graph you supplied, as best i can determine.

here's a graph of that equation.

if i got the points correctly, the equation is y = -2/3 * x + 1

RELATED QUESTIONS
http://172.85.76.57/Library/00001230_84_619_A1.png Go to the link, and post the answer... (answered by Fombitz,MathLover1)
http://media261.acellus.com/Library/00001230_161_1401_A1.png Copy and Paste Link to... (answered by Theo)
Can you please help me explin the process in coming up with this equation as I need to... (answered by stanbon)
Hi. Please help me find the angle in radians of this circle. open pic on the link below... (answered by Alan3354)
``Hi. Please help me find the angle in radians of this circle. open pic on the link below (answered by Edwin McCravy)
http://media261.acellus.com/Library/00001230_92_733_A1.png Copy and paste to see the... (answered by Fombitz)
http://172.85.76.59/Library/00001211_48_2070_A1.png (Link to question above,... (answered by josgarithmetic)
I couldn't quite figure out which section to post this in. Here is a direct screenshot... (answered by rothauserc)
http://i8.photobucket.com/albums/a38/darael14/polynomial_4.jpg this is a link to a... (answered by oscargut,Earlsdon)