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Mathematics 8 Online
OpenStudy (anonymous):

if x varies inversely as y and directly as z^(2), then which one is correct?

OpenStudy (anonymous):

A. \[\frac{ x }{ yz ^{2} }\] is a constant. B. \[\frac{ xy }{ z ^{2} }\] is a constant. C. \[\frac{ xz ^{2} }{ y }\] is a constant. D. \[\frac{ z ^{2} }{ y }\] is a constant. E. \[\frac{ 1 }{ y } +z ^{2}\] is a constant.

OpenStudy (anonymous):

hei

OpenStudy (anonymous):

@kryton1212

OpenStudy (anonymous):

?

OpenStudy (shubhamsrg):

x is directly proportional to z^2 i.e. x= kz^2 x is inversely proportional to y i.e. x=c/y^2 combine both.

OpenStudy (shubhamsrg):

correction: that'll be x= c/y

OpenStudy (anonymous):

\[x=\frac{ k _{1} }{ y } +k _{2}z ^{2}\])

OpenStudy (shubhamsrg):

nopes, combination here means to multiply. final ans would be x is directly proportional to z^2 /y

OpenStudy (anonymous):

D?

OpenStudy (shubhamsrg):

the questions asks you which among them is constant .. x is directly proportional to z^2 /y that means x = c z^2/y or c= xy/z^2 i.e. xy/z^2 is constant.

OpenStudy (anonymous):

ummm..

OpenStudy (shubhamsrg):

??

OpenStudy (anonymous):

what do you mean?

OpenStudy (shubhamsrg):

leme ask you a question, what if I ask you, what do you understand when we say x is directly proportional to z^2 ? and also when x is inversely proportional to y ?

OpenStudy (anonymous):

x=kz^(2) x=1/y

OpenStudy (shubhamsrg):

make that x= c/y some other constant "c"

OpenStudy (anonymous):

yup

OpenStudy (shubhamsrg):

now, k = x/z^2 and c= xy combining them would imply x = K z^2/ y right ? for some new constant "K"

OpenStudy (anonymous):

xz^(2)/y ?

OpenStudy (shubhamsrg):

@msumner no.. x = K z^2 /y hence xy/z^2 must be constant .

OpenStudy (shubhamsrg):

@kryton1212 problems?

OpenStudy (shubhamsrg):

how about a 3rd opinion ? @Callisto

OpenStudy (anonymous):

got it

OpenStudy (shubhamsrg):

why the hell you keep on deleting your comments ?! -_- @msumner

OpenStudy (anonymous):

haha

OpenStudy (anonymous):

so the answer is D? I am just 14 and learning this new stuff

OpenStudy (anonymous):

@shubhamsrg said the answer is B

OpenStudy (anonymous):

@hartnn would you kindly help me?

OpenStudy (callisto):

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