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Mathematics 7 Online
OpenStudy (anonymous):

PLEASE HELP !!) Find the length of AH for which MATH is a parallelogram. A. 3 B. 4 C. 6 D. 10

OpenStudy (anonymous):

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OpenStudy (anonymous):

5f-17=90 ?

OpenStudy (anonymous):

where did 90 come from?

OpenStudy (anonymous):

idk. nvm. so how is it done?

OpenStudy (anonymous):

5f - 17 = 2f-5

OpenStudy (anonymous):

solve for f

OpenStudy (mayankdevnani):

it is because the diagonals of parallelogram bisect each other

OpenStudy (anonymous):

yes

OpenStudy (mayankdevnani):

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OpenStudy (anonymous):

i have about 4 minutes to solve this.. could anyone help?

OpenStudy (anonymous):

yes?

OpenStudy (anonymous):

5f - 17 = 2f-5 solve for f

OpenStudy (anonymous):

f=4!

OpenStudy (anonymous):

plug that in to 5f - 17

OpenStudy (mayankdevnani):

you are correct!! @ashleyvee

OpenStudy (anonymous):

5 * (4) - 17

OpenStudy (anonymous):

it said i got that answer wrong.. so it isnt 4??

OpenStudy (anonymous):

no.. listen to what i am saying

OpenStudy (anonymous):

it was 3

OpenStudy (mayankdevnani):

hey!! its f=4 not length of AH

OpenStudy (anonymous):

4 is f... plug f into 5f - 17 you get 3

OpenStudy (anonymous):

3 is the length of PH... which is 1/2 of AH

OpenStudy (anonymous):

so AH is 6

OpenStudy (anonymous):

it's easier if you try to understand it than only caring about answers in my experience at least

OpenStudy (mayankdevnani):

now add: \[\huge 5f-17+2f-5\] and plug f=4 \[\huge 5(4)-17+2(4)-5\] can you solve it???

OpenStudy (mayankdevnani):

@ashleyvee

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