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Mathematics 8 Online
OpenStudy (anonymous):

Find the value of x.

OpenStudy (anonymous):

Find the value of X. http://scde.mrooms.org/file.php/2247/Images/Ch10Test/10_6.JPG

OpenStudy (anonymous):

can someone please help me?

OpenStudy (anonymous):

can someone please help me?

OpenStudy (anonymous):

the link asks for a log in

OpenStudy (jdoe0001):

url requires a login so... ahemm, take a quick screenshot of the material, crop it and post it here

OpenStudy (anonymous):

OpenStudy (anonymous):

can you see it now?

OpenStudy (anonymous):

yes, is it multiple choice ?

OpenStudy (anonymous):

no it just says to find the value of x

OpenStudy (anonymous):

if the arc lengths of a circle are equal, then these chords of the same circle are also equal

OpenStudy (anonymous):

chords making those arc lengths.

OpenStudy (jdoe0001):

yeap... notice the "tickmarks" |dw:1437177463813:dw|

OpenStudy (anonymous):

so set them equal to each other. right ?

OpenStudy (anonymous):

yes

OpenStudy (jdoe0001):

no, those lines don't mean the circle is growing hair are tickmarks pointing out the EQUALity of the arcs

OpenStudy (anonymous):

so would x=28?

OpenStudy (jdoe0001):

well... |dw:1437177639960:dw|

OpenStudy (jdoe0001):

hmm shoot.. should be a minus there |dw:1437177712846:dw|

OpenStudy (anonymous):

@gatey so do you only want the answer or do you actually want to understand the problem?

OpenStudy (anonymous):

can i have both please and thank you

OpenStudy (anonymous):

12x-8=5x+13 that is the first step you need to do set them equal to eachother. then you need to subtract 5x from 12x getting the equation 7x-8=13 then you add the 8 to the 13 and get the equation 7x=21 divide by 7x making x=3

OpenStudy (anonymous):

got it ?

OpenStudy (anonymous):

YES!! i made a mistake by plugging 3 into the 12x-8 equation when i shouldn't have. thank you

OpenStudy (anonymous):

but i have another problem i need your help on, i was hoping you could help me?

OpenStudy (anonymous):

maybe i can

OpenStudy (anonymous):

but post it as a new question

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