How do I find for x when given y? [closed]












0












$begingroup$


Given y how do I find for x when c=1?



$1-sqrt{1-x^2/c^2}$ = y



e.g.



$1-sqrt{1-.886^2/1^2}$ = y = 0.5363147619






exponential-function







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closed as off-topic by mrtaurho, Shailesh, José Carlos Santos, Henrik, A. Pongrácz Jan 15 at 10:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Shailesh, José Carlos Santos, Henrik, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Do you know how to find the inverse of a function?
    $endgroup$
    – Tyler6
    Jan 15 at 5:19










  • $begingroup$
    I tried but I got stuck
    $endgroup$
    – Agla
    Jan 15 at 5:21
















0












$begingroup$


Given y how do I find for x when c=1?



$1-sqrt{1-x^2/c^2}$ = y



e.g.



$1-sqrt{1-.886^2/1^2}$ = y = 0.5363147619






exponential-function







share|cite|improve this question









$endgroup$



closed as off-topic by mrtaurho, Shailesh, José Carlos Santos, Henrik, A. Pongrácz Jan 15 at 10:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Shailesh, José Carlos Santos, Henrik, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Do you know how to find the inverse of a function?
    $endgroup$
    – Tyler6
    Jan 15 at 5:19










  • $begingroup$
    I tried but I got stuck
    $endgroup$
    – Agla
    Jan 15 at 5:21














0












0








0





$begingroup$


Given y how do I find for x when c=1?



$1-sqrt{1-x^2/c^2}$ = y



e.g.



$1-sqrt{1-.886^2/1^2}$ = y = 0.5363147619






exponential-function







share|cite|improve this question









$endgroup$




Given y how do I find for x when c=1?



$1-sqrt{1-x^2/c^2}$ = y



e.g.



$1-sqrt{1-.886^2/1^2}$ = y = 0.5363147619






exponential-function




exponential-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 15 at 5:14









AglaAgla

32




32




closed as off-topic by mrtaurho, Shailesh, José Carlos Santos, Henrik, A. Pongrácz Jan 15 at 10:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Shailesh, José Carlos Santos, Henrik, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by mrtaurho, Shailesh, José Carlos Santos, Henrik, A. Pongrácz Jan 15 at 10:21


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Shailesh, José Carlos Santos, Henrik, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Do you know how to find the inverse of a function?
    $endgroup$
    – Tyler6
    Jan 15 at 5:19










  • $begingroup$
    I tried but I got stuck
    $endgroup$
    – Agla
    Jan 15 at 5:21


















  • $begingroup$
    Do you know how to find the inverse of a function?
    $endgroup$
    – Tyler6
    Jan 15 at 5:19










  • $begingroup$
    I tried but I got stuck
    $endgroup$
    – Agla
    Jan 15 at 5:21
















$begingroup$
Do you know how to find the inverse of a function?
$endgroup$
– Tyler6
Jan 15 at 5:19




$begingroup$
Do you know how to find the inverse of a function?
$endgroup$
– Tyler6
Jan 15 at 5:19












$begingroup$
I tried but I got stuck
$endgroup$
– Agla
Jan 15 at 5:21




$begingroup$
I tried but I got stuck
$endgroup$
– Agla
Jan 15 at 5:21










1 Answer
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1












$begingroup$

$1-sqrt {1-dfrac {x^2}{c^2}}=y$

As $c=1$,
$1-sqrt {1-x^2}=y$
$sqrt {1-x^2}=1-y$
$1-x^2=(y-1)^2$
$x^2=1-(y-1)^2$
$x=sqrt {1-(y-1)^2}$
$x=sqrt {1-y^2+2y-1}$
$x=sqrt {2y-y^2}$






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $1-sqrt {1-dfrac {x^2}{c^2}}=y$

    As $c=1$,
    $1-sqrt {1-x^2}=y$
    $sqrt {1-x^2}=1-y$
    $1-x^2=(y-1)^2$
    $x^2=1-(y-1)^2$
    $x=sqrt {1-(y-1)^2}$
    $x=sqrt {1-y^2+2y-1}$
    $x=sqrt {2y-y^2}$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $1-sqrt {1-dfrac {x^2}{c^2}}=y$

      As $c=1$,
      $1-sqrt {1-x^2}=y$
      $sqrt {1-x^2}=1-y$
      $1-x^2=(y-1)^2$
      $x^2=1-(y-1)^2$
      $x=sqrt {1-(y-1)^2}$
      $x=sqrt {1-y^2+2y-1}$
      $x=sqrt {2y-y^2}$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $1-sqrt {1-dfrac {x^2}{c^2}}=y$

        As $c=1$,
        $1-sqrt {1-x^2}=y$
        $sqrt {1-x^2}=1-y$
        $1-x^2=(y-1)^2$
        $x^2=1-(y-1)^2$
        $x=sqrt {1-(y-1)^2}$
        $x=sqrt {1-y^2+2y-1}$
        $x=sqrt {2y-y^2}$






        share|cite|improve this answer









        $endgroup$



        $1-sqrt {1-dfrac {x^2}{c^2}}=y$

        As $c=1$,
        $1-sqrt {1-x^2}=y$
        $sqrt {1-x^2}=1-y$
        $1-x^2=(y-1)^2$
        $x^2=1-(y-1)^2$
        $x=sqrt {1-(y-1)^2}$
        $x=sqrt {1-y^2+2y-1}$
        $x=sqrt {2y-y^2}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 5:23









        Mohammad Zuhair KhanMohammad Zuhair Khan

        1,5852625




        1,5852625















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