yeah no
dam the square root function on this is touchy
yeah no
dam the square root function on this is touchy
but the slope is unknown also......
How can i have 2 variables? Why would the X value even fucking matter, and what x value?? 9, 5 is the value, what other x value is there?
Well just use my formula and extrapolate backwards, X in my formula is slope, so plug into all the x's the formula using the x value that is used to find the slope.
the distance is the square root of x^2 - 18x +106
ok right now my best guess is:
L(x) = SQRT(x^2 + (5x/(x-9))^2)
it works at least when the slope is -1
and by distance i meant LENGTH SWNJDN FUCK YOWMEP
take the point specified in the question (9, 5) and take the point where the line meets the x axis (x, 0)
use the distance formula.
sqrt((x2-x1)^2 + (y2-y1)^2)
simplify.
you get my answer.
take the point specified in the question (9, 5) and take the point where the line meets the x axis (x, 0)
use the distance formula.
sqrt((x2-x1)^2 + (y2-y1)^2)
simplify.
you get my answer.
God please explain to me what x value the L is being based on, the legth on X as in the base of the triangle? I swear to god my slope answer is 100% right, i don't understand what other x value there is...
and i specifically don't get how you can get the distance without knowing the slope!
hypotenuse = sqrt(x^2 + y^2)
so we know L(x) = the same. we want to find y.
draw the triagle and think about x varying and what happens to y, knowing that (9,5) is the 'pivot' so to speak. as x tends to infinity, y tends to 5. as x tends 9, y tends to inifity. 5x/(x-9) is a function consistent with those limits, so it's your y. plug it in and there you go.
god: high school drop out getting everyones college homework right this semester
can we delete this thread and/or pretend it never existed
I still am sort of confused on how you get y.
Your an absolutely wonderful man, any girl that is yours is lucky and I know that from personal experience. ~KMT
why was this moved out of the archives?