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here is what i have... idk if it is right though... what do u think? is it correct? :/ |dw:1361912525340:dw| is that right??? :/
From there you can rationalize the denominator |dw:1361912794736:dw|
or is it supposed to be (-2/sq.rt.(3))/3 ??? :/
but what you have is more or less correct
that's the same thing...just a bit more convoluted
I would just go with |dw:1361913060716:dw|
okay, so my steps are all right?? but the answer is just -2 sw.rt. 3 / 3 instead of -2/sq.rt. 3??
yes the final answer is |dw:1361941352439:dw| and your steps are correct
okay!! awesome :) thanks a bunch!!! :D
Consider the right triangle ABC that is attached... a. What is the length of the hypotenuse of triangle ABC? b. What is the length of the shorter leg of triangle ABC?
where are the letters A, B, C?
it doesn't say :/ thats what i was thinking also... cuz aren't they usually labeled? :/
there are 3 triangles so I have no idea which triangle they're referring to
is there any way you can assume where A, B, C are labeled?? cuz its a 30-60-90 triangle right?? one of the special triangles? :/ ***tho i don't quite remember the rule it has... :/
yeah but where do they go and which point is left out?
are you sure it's not labeled?
yeah its not labeled.. :/ its kinda weird... then can we do this then?? ill ask my teacher tomorrow about it and then i will reply back over here ?? would that work? :)
**reply back here after i find out the labels?
sure that will work
okay hahaa :) thanks!! i will be back sometime tomorrow then!!! **if i can get online tomorrow night!!!*** hehe thankssss :)
okay... :) so i just asked my teacher and she said that the entire big triangle is ABC and she gave me a hint also!! :) she said that you will be finding the hypotenuse and the length of the leg by the angles given.. :/ but yeah! The HUGE triangle is triangle ABC :)
anddd idk how to do it tho... I'm still a bit confused on this whole process :/ :(
do u get it?? :/ cuz iheartfood is mucho confused :(
sin(30) = opp/hyp sin(30) = 15/x 1/2 = 15/x 1*x = 2*15 x = 30
sin(60) = opp/hyp sin(60) = 15/y sqrt(3)/2 = 15/y sqrt(3)*y = 2*15 sqrt(3)*y = 30 y = 30/sqrt(3) y = 30*sqrt(3)/3 y = 10*sqrt(3)
a^2 + b^2 = c^2 30^2 + (10*sqrt(3))^2 = z^2 900 + 100*3 = z^2 900 + 300 = z^2 keep going to find z
that's the approximate answer, the exact answer is z = 20*sqrt(3)
so what should i write it out as??
and i show all the work that u drew /wrote above right??? those are the steps?
yeah that's one way to get the hypotenuse
okay so that is the first part of the question right? what about the second part?
the shorter leg is 10*sqrt(3) units and I showed how to get that above as well
okay... so i for a the answer is z = 20*sqrt(3) which = 34.64 ?? and b is 10*sqrt(3) which equals 17.32 ?? is that right??? and it is all based on the steps u wrote above??
yes both are correct
awesome!! okay :) thanks a bunch!!!! :)