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anonymous
 4 years ago
in a 45 45 90 triangle i need to find what x is
anonymous
 4 years ago
in a 45 45 90 triangle i need to find what x is

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342917395925:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342917641735:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would it be 6 square root of 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Use Pythagoras theorem \[x^2+x^2=12^2\]

Australopithecus
 4 years ago
Best ResponseYou've already chosen the best response.0you cant solve this with pythagoras theorem you have to use trig I'm pretty sure

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2x^2=12^2\]\[2x^2=144\]\[x^2=\frac{144}{2}\]\[x^2=72\]\[x= \sqrt{72}\]\[x= \sqrt{36*2}\]\[x=6 \sqrt{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes you can you have two sides that are equal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hey @sarver1995 Been a while :) 454590 triangles: dw:1342917784578:dw

Australopithecus
 4 years ago
Best ResponseYou've already chosen the best response.0oh yeah just noticed :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x\sqrt{2} = 12\]\[x = \frac{12}{\sqrt{2}}\]\[x = \frac{12}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\]\[x = \frac{12\sqrt{2}}{2}\]\[x = 6\sqrt{2}\]There you go :)

Australopithecus
 4 years ago
Best ResponseYou've already chosen the best response.0SOH CAH TOA Sin = opposite/hypotenuse cos = adjacent/hypotenuse tan = opposite/adjacent cos(45) = x/12 cos(45)(12) = x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0from this I noticed that you can simply get one of the sides by multiplying the hypotenuse by \[\frac{1}{\sqrt{2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Only if the triangle is a 45 45 90

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342917788335:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Same thing...use the ratio above... dw:1342918025299:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0same thing like I said \[\frac{1}{\sqrt{2}}*x=10\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342918127585:dw would x be 10 square root of 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes :) \[x = 10\sqrt{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342918257888:dw okay not sure how to do this one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it cant be 1/2 square root of 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does it tell you if it's a right triangle?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. The ratio again. x, x, x√2 x = 1.2 x√2 = ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i will try to draw it again maybe i ddin't do it right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0By the way, if the x's get confusing because they are using x as a variable, I'll just use y.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342918455394:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. Same situation. The legs are always congruent in a 454590 triangle. The hypotenuse is always in this ratio: \[Leg \times \sqrt{2}\]Can you figure it out since you have the leg?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01.2 x square root of 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yup. which can be rewritten as \(1.2\sqrt{2}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay i have one more like these can i try it with us

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alright shoot away :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342918731847:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x = \frac{\sqrt{24}}{\sqrt{2}} = \sqrt{\frac{24}{2}} = \sqrt{12} = ?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay i have a different one i just need u to help me understand the problem it will take a little while to draw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342918993022:dw Refer to the figure to complete the following item. Given: If m arc vu =80° and marc st =40°, then 1 =

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you have to like fill in a blank or is it like write an entire proof?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its just asking what angle 1 would equal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think that it's 50...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you tell me how you got that because i have three more problems like this one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[m\angle1 = \frac 12(arcVT  arcST)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If arc vu m = 70° and m arc st = 30°, then 2 = would it be the same formula for this one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I kind of have to go...sorry :/ If it maintains the same shape, it should be the same sort of formula...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks for your help its really appreciated and God bless you
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