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Mathematics 19 Online
OpenStudy (natalie1234):

PLEASE HELP WILL GIVE MEDAL AND FAN!!<3 What is the length x?

OpenStudy (natalie1234):

OpenStudy (misty1212):

HI!!

OpenStudy (natalie1234):

Hello!! (:

OpenStudy (misty1212):

this looks like a big drag, but i bet we can do it we have to find a couple lengths first

OpenStudy (natalie1234):

haha ok

OpenStudy (misty1212):

my draw tool is not working you have one triangle on the bottom , the right triangle with both sides one you see the one i mean?

OpenStudy (natalie1234):

yeah i see it

OpenStudy (misty1212):

|dw:1421988445493:dw|

OpenStudy (misty1212):

finally ok by pythagoras the length of the hypotenuse is \(\sqrt2\)

OpenStudy (misty1212):

now we move on to the next triangle

OpenStudy (natalie1234):

alright

OpenStudy (natalie1234):

the one in the middle right?

OpenStudy (misty1212):

|dw:1421988553618:dw|yea that one

OpenStudy (misty1212):

one leg is \(\sqrt2\) the other is 1 so pythagoras again for the hypotenuse \[\sqrt{\sqrt2^2+1^2}=\sqrt{2+1}=\sqrt3\]

OpenStudy (misty1212):

one more to go!

OpenStudy (natalie1234):

okay haha(:

OpenStudy (misty1212):

but now we don't bother drawing it one leg is \(\sqrt3\) the other is 1, pythagoras for the last time gives \[\sqrt{\sqrt3^2+1^2}=\sqrt{3+1}=\sqrt4\]

OpenStudy (misty1212):

better known as \(\huge \color\magenta2\)

OpenStudy (natalie1234):

so is that the answer? or you just did the square root of 4

OpenStudy (misty1212):

lol "nutellia"

OpenStudy (misty1212):

yeah final answer is 2, since the square root of four is 2

OpenStudy (natalie1234):

haha yup thats what they call me /.\ lol and thank you so much! you were really nice and helpful(:

OpenStudy (misty1212):

your welcome darling \(\color\magenta\heartsuit\)

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