anonymous
  • anonymous
Please help... so confused HOLD ON! im going to draw a picture!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1328575316576:dw| solve for x
precal
  • precal
use \[a^2+b^2=c^2\]
anonymous
  • anonymous
well i think that would work but we're supposed to make it a proportion

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precal
  • precal
usually it involves the small triangle vs the large triangle
anonymous
  • anonymous
ya but i cant do that... at least i dont think i can
precal
  • precal
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
anonymous
  • anonymous
well the whole triangle is 90 degrees but it doesnt tell you if its split through the middle
precal
  • precal
|dw:1328575806467:dw| if so we can assume the side that is in between them is x...
anonymous
  • anonymous
but how can you do that since x is the length of the side on the bottom?
precal
  • precal
Is the drawing from a book or a worksheet? the problem with geometry is that you can not assume anything
anonymous
  • anonymous
its in the book and ya thats why i dont like it!
precal
  • precal
|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.
precal
  • precal
You can scan the picture and attach the file. but I think I might have it figured out
anonymous
  • anonymous
well its online so give me a sec
precal
  • precal
\[x^2+x^2=(x+3)^2\] \[2x^2=x^2+6x+9\] \[x^2-6x-9=0\] not that does not work !!!!!
precal
  • precal
I meant that does not work!!! Sorry can't seem to type as well
anonymous
  • anonymous
there is a picture of the problem!
1 Attachment
precal
  • precal
ok upload it, I will be back in about 5 mins
precal
  • precal
ok you got my attention.
precal
  • precal
ok x+3 is the leg of the big triangle the hypotenuse is 12-x for the big triangle
anonymous
  • anonymous
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
precal
  • precal
ok x is the leg of the small triangle and x+3 is the hypotenuse of the small triangle there is our proportion
precal
  • precal
|dw:1328576462186:dw|
anonymous
  • anonymous
but how did you get 12-x since 12 is the whole LARGE triangle's hypotenuse?
anonymous
  • anonymous
im sorry im just super confused!
precal
  • precal
|dw:1328576638458:dw|
anonymous
  • anonymous
so (x+3)(x+3)=12 x
precal
  • precal
No, I think we are ok cross multiply \[12x=(x+3)^2\]
precal
  • precal
yes now solve it
anonymous
  • anonymous
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!
precal
  • precal
x=3 I used the quadratic formula
precal
  • precal
\[x^2-6x+9=0\]
anonymous
  • anonymous
ok but how did you get x=3?
precal
  • precal
|dw:1328576877634:dw|
precal
  • precal
use the quadratic formula
anonymous
  • anonymous
see all of that confuses me and i dont think we've done that yet!
anonymous
  • anonymous
well thanks for your help but i dont think im going to get it.... but thank you for your trying... :)
precal
  • precal
Is this a challenge question?
anonymous
  • anonymous
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.
precal
  • precal
You will understand this eventually, your teacher has to teach quadratics. One day it will make sense.
anonymous
  • anonymous
haha ok thanks :)
precal
  • precal
Oh well another reason why computers will not replace teachers. I am sure that it is an oversight of the program. Hang in there :)
anonymous
  • anonymous
thank you and again thanks for your help1

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