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im assuming a,b,and c are degrees. then b+c=90
Well, no they are all side measurements
and also, i am figuring for y.
is a the hypotenuse?
no, c is.
then c^2 - b^2 = a^2
do that and the quadratic formula to solve for y
Yes, the pythagorean theorem... and to tell the truth i have absolutely no idea what that means
what does "that" refer to?
"that" refers to do that and the quadratic formula to solve for y
just plug in everything you know into the pythagorean theorem. then rearrange what you get so that you can use the quadratic formula.
(3y-10)^2 - (y+40)^2 = 90^2
well.... i did the same thing except i used a squared plus b squared equals c squared
that works too
now just rearrange it all so that everything is on one side of the equation
okay but i dont come up with y being a perfect square.
i tried this: a=90,b=y+40,c=3y-10 a2+b2=c2 squared=2 so 902+(y+40)2=(3y-10)2 8100+y2+1600=3y2+100 9700+y2=3y2+100 9600+y2=3y2=2y2+100 9600=2y2 4800=y2
you squared incorrectly. (y+40)^2 is y^2+80y+1600. same with (3y-10)^2, that was squared incorrectly too
how do you get 80y?
use the foil method.
hmmm. write out (y+40)x(y+40). then distribute. so it becomes y*y + y*40 + 40*y + 40*40
ooooohhhhh!!!! okay genius.... (in a good way) thank you so much!
lol no problem. by the way, the answer isn't a whole number.
I'm in seventh grade honors and lets just say my teacher isnt exactly very to the point. and oh are you serious? so its not a perfect square?
ohhh jeez. i was having a hard time the whole time because i thought it HAD to be a perfect square... :) whoops.