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in a 45 45 90 triangle i need to find what x is

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would it be 6 square root of 2

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Other answers:

Use Pythagoras theorem \[x^2+x^2=12^2\]
you cant solve this with pythagoras theorem you have to use trig I'm pretty sure
\[2x^2=12^2\]\[2x^2=144\]\[x^2=\frac{144}{2}\]\[x^2=72\]\[x= \sqrt{72}\]\[x= \sqrt{36*2}\]\[x=6 \sqrt{2}\]
Yes you can you have two sides that are equal
Hey @sarver1995 Been a while :) 45-45-90 triangles: |dw:1342917784578:dw|
oh yeah just noticed :)
\[x\sqrt{2} = 12\]\[x = \frac{12}{\sqrt{2}}\]\[x = \frac{12}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\]\[x = \frac{12\sqrt{2}}{2}\]\[x = 6\sqrt{2}\]There you go :)
SOH CAH TOA Sin = opposite/hypotenuse cos = adjacent/hypotenuse tan = opposite/adjacent cos(45) = x/12 cos(45)(12) = x
from this I noticed that you can simply get one of the sides by multiplying the hypotenuse by \[\frac{1}{\sqrt{2}}\]
Only if the triangle is a 45 45 90
Same thing...use the ratio above... |dw:1342918025299:dw|
same thing like I said \[\frac{1}{\sqrt{2}}*x=10\]
solve for x
to calcmathlete yeah i am now reviewing and oh course still confused i only have a few more lessons and i am finished with 11th grade. its been really rough
|dw:1342918127585:dw| would x be 10 square root of 2
Yes :) \[x = 10\sqrt{2}\]
Good job mate!
|dw:1342918257888:dw| okay not sure how to do this one
it cant be 1/2 square root of 2
Does it tell you if it's a right triangle?
45 45 90
It tells you that?
Ok. The ratio again. x, x, x√2 x = 1.2 x√2 = ?
i will try to draw it again maybe i ddin't do it right
By the way, if the x's get confusing because they are using x as a variable, I'll just use y.
Ok. Same situation. The legs are always congruent in a 45-45-90 triangle. The hypotenuse is always in this ratio: \[Leg \times \sqrt{2}\]Can you figure it out since you have the leg?
1.2 x square root of 2
not sure on this one
Yup. which can be rewritten as \(1.2\sqrt{2}\)
okay i have one more like these can i try it with us
Alright shoot away :)
\[x = \frac{\sqrt{24}}{\sqrt{2}} = \sqrt{\frac{24}{2}} = \sqrt{12} = ?\]
2 square root of 3
yup :)
okay i have a different one i just need u to help me understand the problem it will take a little while to draw
|dw:1342918993022:dw| Refer to the figure to complete the following item. Given: If m arc vu =80° and marc st =40°, then 1 =
Oh gosh...a proof?
yeah lol
Do you have to like fill in a blank or is it like write an entire proof?
its just asking what angle 1 would equal
just an answer
I think that it's 50...
can you tell me how you got that because i have three more problems like this one
\[m\angle1 = \frac 12(arcVT - arcST)\]
If arc vu m = 70° and m arc st = 30°, then 2 = would it be the same formula for this one
I kind of have to go...sorry :/ If it maintains the same shape, it should be the same sort of formula...
thanks for your help its really appreciated and God bless you

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