HELP!!!!!!!!!
@undeadknight26 @jim_thompson5910
Know what the pythagoreom theory is?
a^2 + b^2 = c^2? That's what you'd do. A and B = the known sides.
If I remember correctly, that is. Maybe Jim can shed some light on the matter?
It could be B and C are the known sides and you solve for A.
@jim_thompson5910
\[height= \sqrt{113^{2}-15^{2}}\]
hint: see attached
what do i have to solve?
http://assets.openstudy.com/updates/attachments/57354835e4b0dabdb023469a-jim_thompson5910-1463110317279-hint1.jpg \[\Large {\color{red}{x}}^2+{\color{green}{y}}^2={\color{blue}{z}}^2\] \[\Large {\color{red}{x}}^2+{\color{green}{15}}^2={\color{blue}{113}}^2\] \[\Large x^2+225=12769\] \[\Large x^2+225-225=12769-225\] \[\Large x^2=12,544\] \[\Large \sqrt{x^2}=\sqrt{12,544}\] \[\Large x=112\]
Ah so I was wrong. It was B and C and you solve for A. Makes sense.
Excellent work as usual Jim.
Thanks
yes great work! next! The length of the hypotenuse of a right triangle is 14 inches. If the length of one leg is 6 inches, what is the approximate length of the other leg? 11.5 inches 12.6 inches 15.2 inches 20.0 inches
Give this one a try |dw:1463110831671:dw|
the steps will be the same as before, just with different numbers this time
i still don't get it
which step are you stuck on?
all :( especially the beginning..
i sort of get it more at the end..
do you see how I set x, y and z to be the side lengths?
x is unknown y is 15 z is 113
but why the letters y and z?
just any 3 letters really. x,y,z makes the most sense for me
it could be a,b,c or p,q,r or any 3 letters you want
oh ok...
the letters are just placeholders for a number example: y is a placeholder for 15
ok yea i get that.. moving on..
do you understand what I mean when I write \(\Large x^2\) ?
no
it means `x squared` squaring any number means you multiply that number by itself examples \(\Large 7^2\) = 7 squared = 7*7 = 49 \(\Large 9^2\) = 9 squared = 9*9 = 81 \(\Large 14^2\) = 14 squared = 14*14 = 196
oh yea i get that.. ok next..
As practice, tell me what \(\Large 16^2\) is equal to
256
good
so hopefully you can see how I turned \(\Large 15^2\) into \(\Large 225\) ?
yes...
and you're able to see how I turned \(\Large 113^2\) into \(\Large 12769\) ?
yes...
so you can see how I solved that last equation for x? you understand all the steps?
I'm pretty sure
ok you'll use those same steps (just with different numbers) on this next problem
|dw:1463111890201:dw| \[\Large x^2+y^2 = z^2\] \[\Large x^2+6^2 = 14^2\] solve for x
Join our real-time social learning platform and learn together with your friends!