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Logarius
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#1
Oct 21 2017 10:56pm
Paying 200 fg to whoever can solve these and show at least a little work. I dont have time right now and this shit is due soon
JoyStix
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#2
Oct 21 2017 10:58pm
Shako
BMar5452
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#3
Oct 21 2017 11:00pm
I learned this in 8th grade
School is usless who rerembers anything
AD2Syndicate
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#4
Oct 21 2017 11:01pm
Mathway.com my man. You’re welcome
Logarius
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#5
Oct 21 2017 11:02pm
Quote (AD2Syndicate @ Oct 21 2017 10:01pm)
Mathway.com my man. You’re welcome
I use mathway bro I just dont know how to input triangle values so the program understands it
laughinggas
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#6
Oct 21 2017 11:07pm
x = 1.598, y=2.176
sin 53.7 = 0.806
cos 53.7 = 0.592
tan 53.7 = 1.361
AJ = 2
((3^(1/2)^2) + 1^2 = Z^2 --> Z = 2
AL is also 2 lol.
This post was edited by laughinggas on Oct 21 2017 11:12pm
Logarius
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#7
Oct 21 2017 11:12pm
Quote (laughinggas @ Oct 21 2017 10:07pm)
x = 1.598, y=2.176
sin 53.7 = 0.806
cos 53.7 = 0.592
tan 53.7 = 1.361
I need some work shown for the x,y coordinates
laughinggas
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#8
Oct 21 2017 11:13pm
Quote (Logarius @ Oct 22 2017 12:12am)
I need some work shown for the x,y coordinates
you paying tho right? lol
sin (53.7) = y/2.7 --> solve for y = 2.176
cos (53.7) = x/2.7 --> solve for x = 1.598
laughinggas
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#9
Oct 21 2017 11:16pm
aj = use hypotenus formula.
3^(1/2) = x
1 = y
x^2 + y^2 = z^2
Plug in x and y and solve for z, which gives you 2, which is Length of AJ.
Same idea for AL.
This post was edited by laughinggas on Oct 21 2017 11:16pm
Logarius
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#10
Oct 21 2017 11:16pm
Quote (laughinggas @ Oct 21 2017 10:13pm)
you paying tho right? lol
sin (53.7) = y/2.7 --> solve for y = 2.176
cos (53.7) = x/2.7 --> solve for x = 1.598
Yes I am paying even though you can find sin, cos and tan in like 2 seconds on the calc, I just need some work to show
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