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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asked Jan 15 at 5:14

AglaAgla

32

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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
  

  
  
  
  

add a comment | 

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

asked Jan 15 at 5:14

AglaAgla

32

$endgroup$

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
  

  
  
  
  

add a comment | 

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

asked Jan 15 at 5:14

AglaAgla

32

$endgroup$

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

asked Jan 15 at 5:14

AglaAgla

32

share|cite|improve this question

asked Jan 15 at 5:14

AglaAgla

32

share|cite|improve this question

share|cite|improve this question

asked Jan 15 at 5:14

AglaAgla

32

asked Jan 15 at 5:14

AglaAgla

32

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

add a comment | 

1 Answer
1

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oldest

votes

1

$begingroup$

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

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

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answered Jan 15 at 5:23

Mohammad Zuhair KhanMohammad Zuhair Khan

1,5852625

$endgroup$

add a comment | 

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

answered Jan 15 at 5:23

Mohammad Zuhair KhanMohammad Zuhair Khan

1,5852625

$endgroup$

add a comment | 

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

answered Jan 15 at 5:23

Mohammad Zuhair KhanMohammad Zuhair Khan

1,5852625

$endgroup$

add a comment | 

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

answered Jan 15 at 5:23

Mohammad Zuhair KhanMohammad Zuhair Khan

1,5852625

$endgroup$

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

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

answered Jan 15 at 5:23

Mohammad Zuhair KhanMohammad Zuhair Khan

1,5852625

share|cite|improve this answer

share|cite|improve this answer

answered Jan 15 at 5:23

Mohammad Zuhair KhanMohammad Zuhair Khan

1,5852625

answered Jan 15 at 5:23

Mohammad Zuhair KhanMohammad Zuhair Khan

1,5852625

answered Jan 15 at 5:23

Mohammad Zuhair KhanMohammad Zuhair Khan

1,5852625

1,5852625

add a comment | 

add a comment | 

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