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Mathematics 17 Online
OpenStudy (brucebaner):

help me with this math plz help me

OpenStudy (brucebaner):

OpenStudy (brucebaner):

@sammixboo

OpenStudy (brucebaner):

@EclipsedStar

OpenStudy (brucebaner):

@ikram002p

OpenStudy (brucebaner):

@ybarrap

OpenStudy (brucebaner):

@ybarrap

OpenStudy (brucebaner):

@Michele_Laino

OpenStudy (godlovesme):

i got u, use the pythagorean theorem

OpenStudy (godlovesme):

ever heard of it? :P

OpenStudy (brucebaner):

no

OpenStudy (brucebaner):

im new to this

OpenStudy (godlovesme):

ohhh okay heres the formula \[a^2+b^2=c^2\]

OpenStudy (brucebaner):

kk

OpenStudy (shamim):

Do u know base , perpendicular, hypotenuse of a right angle triangle?

OpenStudy (godlovesme):

|dw:1434399693304:dw|

OpenStudy (godlovesme):

u can mix up the legs ( a and b in the formula) it doesnt matter

OpenStudy (brucebaner):

yes i think?

OpenStudy (godlovesme):

|dw:1434399839501:dw| so in this case we are solving for X, which is the hypotenuse

OpenStudy (godlovesme):

so our a can be either 7 or 4 and we're looking for c \[4^2 + 7^2 = c^2\] ^^ that's after we substitute out givens

OpenStudy (godlovesme):

make sense? c:

OpenStudy (brucebaner):

yes

OpenStudy (godlovesme):

\[4^2 + 7^2 = c^2 \] \[16 + 49 = c^2\] \[65 = c^2\]

OpenStudy (godlovesme):

try to do the last step c;

OpenStudy (godlovesme):

*hint* c is squared right? but we're looking for c not c^2

OpenStudy (godlovesme):

take of the square root! :D

OpenStudy (godlovesme):

\[\sqrt{65} = \sqrt{c}\] that should give u x :)

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