Please help... so confused HOLD ON! im going to draw a picture!

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Please help... so confused HOLD ON! im going to draw a picture!

Mathematics
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|dw:1328575316576:dw| solve for x
use \[a^2+b^2=c^2\]
well i think that would work but we're supposed to make it a proportion

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usually it involves the small triangle vs the large triangle
ya but i cant do that... at least i dont think i can
do you know about the angle on top? Is is split in the middle? Is it 45 degrees since the larger part is 90 degrees? I can't tell by your drawing. I know it is hard to draw
well the whole triangle is 90 degrees but it doesnt tell you if its split through the middle
|dw:1328575806467:dw| if so we can assume the side that is in between them is x...
but how can you do that since x is the length of the side on the bottom?
Is the drawing from a book or a worksheet? the problem with geometry is that you can not assume anything
its in the book and ya thats why i dont like it!
|dw:1328575902416:dw| We can assume that 90 degrees, the other one we can assume is the 45 degrees because that line that splits the triangle has a special name but I can't recall it at the moment.
You can scan the picture and attach the file. but I think I might have it figured out
well its online so give me a sec
\[x^2+x^2=(x+3)^2\] \[2x^2=x^2+6x+9\] \[x^2-6x-9=0\] not that does not work !!!!!
I meant that does not work!!! Sorry can't seem to type as well
there is a picture of the problem!
1 Attachment
ok upload it, I will be back in about 5 mins
ok you got my attention.
ok x+3 is the leg of the big triangle the hypotenuse is 12-x for the big triangle
well the problem never tells you anything about the angles or anything and as you can tell the line doesnt go right through the right angle symbol
ok x is the leg of the small triangle and x+3 is the hypotenuse of the small triangle there is our proportion
|dw:1328576462186:dw|
but how did you get 12-x since 12 is the whole LARGE triangle's hypotenuse?
im sorry im just super confused!
|dw:1328576638458:dw|
so (x+3)(x+3)=12 x
No, I think we are ok cross multiply \[12x=(x+3)^2\]
yes now solve it
x^2+3x+3x+9=12x x^2+6x+9=12x x^2+9=6x right? but i dont know what do to next because we havent done anything like this in a while!
x=3 I used the quadratic formula
\[x^2-6x+9=0\]
ok but how did you get x=3?
|dw:1328576877634:dw|
use the quadratic formula
see all of that confuses me and i dont think we've done that yet!
well thanks for your help but i dont think im going to get it.... but thank you for your trying... :)
Is this a challenge question?
i dont think so.... but online they sometimes leave off something like a picture or a number so that might be what happened... thanks for you patients.
You will understand this eventually, your teacher has to teach quadratics. One day it will make sense.
haha ok thanks :)
Oh well another reason why computers will not replace teachers. I am sure that it is an oversight of the program. Hang in there :)
thank you and again thanks for your help1

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