Recent content by kaydis

no I think your right, I swapped them around

how would you do it? sorry I'm super confused :oldconfused:

for Vx=Vcosθ I got 2.262347939 if I do Vx=θcos(V) i got -4.242640687

so if: horizontal: Vx = Vcos verticle : Vy = Vsin this means that: Vx = Vcos = (3π/4)cos6 = 2.262347939 Vy = Vsin = (3π/4)sin6 = -0.658357257 is this correct?

i know that they're the parts of a vector and that you use trigonometry to find the componants?

sorry, i just double checked and its 3π/4

I'm not too sure where to start, but 6 is the magnitude and (2pi/4) is the angle? So if I were to plot this on a graph I could get the value of the point on the horizontal and vertical axis. I've tried to plot is using graphing software but it wasn't working so I'm not sure if i was doing it...

awesome, thank you so much for all the help! :smile:

I think i did it! -c=0 -2a-2c=-1 = -2a-2(0)=-1 = -2a=-1 = -a=-1/2 or -0.5 SO a=0.5 -a-3b+0=1 = -a-3b=1 = -0.5-3b=1 = -3b=1.5 = -b=1/2 or 0.5 SO b=-0.5 when i plug them back in i get the following: a+b=0 -------->...

could i add them? 2b + -2b = canceled out and -2c + c = -c and -1 + 1 = 0 so -c=0

a+b=0 so.. a= -b -2a-2c = -1 = -2(-b)-2c = -1 = 2b-2c=-1 -a-3b+c= 1 = -(-b)-3b+c= 1 = b-3b+c=1 = -2b+c=1 i think this is right but i don't know where to go from here

So I've tried and I'm stuck... a+b=0 -2a-2c = -1 -a-3b+c= 1 1 1 0 ¦ 0 -2 0 -2 ¦ -1 (R2 - 2*R1) -1 -3 1 ¦ 1 (R3 + R1) 1 1 0 ¦ 0 0 2 2 ¦ -1 (R2 - 1/2) 0 4 1 ¦ 1 1 1 0 ¦ 0 0 1 1 ¦ -1/2 0 4 1 ¦ 1 (R3 - 4R2) 1 1 0 ¦ 0 0 1 1 ¦ -1/2 0...

Yes! awesome, what is the easiest way to work this out? or is it just trial and error?

-2a-2c = -1 -a-3b+c= 1

equate the powers? i.e. a+b = 0