The entire thing is a contradiction.y = 0, y = 3, x + y = 0, x + y = 2, 2x + y +3 z = 0 and 2x + y +3 z = 4.
This is a Hitskin.com skin preview
Install the skin • Return to the skin page
This is a Hitskin.com skin preview
Install the skin • Return to the skin page
The entire thing is a contradiction.y = 0, y = 3, x + y = 0, x + y = 2, 2x + y +3 z = 0 and 2x + y +3 z = 4.
Gangstar wrote:I don't mean those theorems. I mean theorems that you use to prove proofs.
You do limits in only the first day of cal class...honest. That's not even considered basic...
Redwan wrote:okay screw calc. This is actual hw.
tell me if what i did is correct.
question: given identity: sin^2(x)+cos^2(x)=1, solve for x in "3*sin^2(x) = 1+ cos(x)"
change identity to sin^2(x)= 1- cos^2(x)
3*sin^2(x) = 1+ cos(x)
3(cos^2(x)) = 1+ cos(x)
3- 3* cos^2(x) = 1+ cos(x)
3cos^2(x) + cos (x) -2 = 0
(3cos(x)-2)(cos(x)+1)=0
If d= cos(x) then (3d-2)(d+1)=0 d=-1, 2/3
cos(x)= -1, 2/3
cos (x)= -1
arccos(-1)=x
cos(x)= 2/3
arccos(2/3)= x
thnx.
|
|