Proof that if $Y=2X+3$ then $f_Y(y)=frac12f_X(frac{y-3}2)$












0












$begingroup$


I want to prove that if probability density function (PDF) of a random variable $X$ is $f_X(x)$ and another random variable $Y$ is defined as $Y=2X+3$, then the PDF of $Y$ is $f_Y(y)=frac{1}{2} f_X(frac{y-3}{2})$.



My Approach:

We Know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$

similarly,$$P(Y leq y)= int_{-infty}^y f_Y(y) , dy$$
Now,$, Y=2X+3$
$$implies P(Y leq y)= P((2X+3) leq y)$$
$$=P(X leq (frac{y-3}{2}))$$
$$implies P(Y leq y)=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy$$
$$implies int_{-infty}^y f_Y(y) , dy=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy $$
Now how can I proceed further to find a relationship between $f_Y$ and $f_X $?






probability-theory probability-distributions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Use the change of variable $yto2y+3$ in the very last integral.
    $endgroup$
    – Did
    Jan 8 at 9:56












  • $begingroup$
    if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
    $endgroup$
    – Suresh
    Jan 8 at 10:12










  • $begingroup$
    No we do not get that.
    $endgroup$
    – Did
    Jan 8 at 14:30






  • 1




    $begingroup$
    Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
    $endgroup$
    – Did
    Jan 9 at 6:51






  • 1




    $begingroup$
    You are doing the change of variable, backwards.
    $endgroup$
    – Did
    Jan 10 at 10:38
















0












$begingroup$


I want to prove that if probability density function (PDF) of a random variable $X$ is $f_X(x)$ and another random variable $Y$ is defined as $Y=2X+3$, then the PDF of $Y$ is $f_Y(y)=frac{1}{2} f_X(frac{y-3}{2})$.



My Approach:

We Know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$

similarly,$$P(Y leq y)= int_{-infty}^y f_Y(y) , dy$$
Now,$, Y=2X+3$
$$implies P(Y leq y)= P((2X+3) leq y)$$
$$=P(X leq (frac{y-3}{2}))$$
$$implies P(Y leq y)=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy$$
$$implies int_{-infty}^y f_Y(y) , dy=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy $$
Now how can I proceed further to find a relationship between $f_Y$ and $f_X $?






probability-theory probability-distributions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Use the change of variable $yto2y+3$ in the very last integral.
    $endgroup$
    – Did
    Jan 8 at 9:56












  • $begingroup$
    if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
    $endgroup$
    – Suresh
    Jan 8 at 10:12










  • $begingroup$
    No we do not get that.
    $endgroup$
    – Did
    Jan 8 at 14:30






  • 1




    $begingroup$
    Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
    $endgroup$
    – Did
    Jan 9 at 6:51






  • 1




    $begingroup$
    You are doing the change of variable, backwards.
    $endgroup$
    – Did
    Jan 10 at 10:38














0












0








0





$begingroup$


I want to prove that if probability density function (PDF) of a random variable $X$ is $f_X(x)$ and another random variable $Y$ is defined as $Y=2X+3$, then the PDF of $Y$ is $f_Y(y)=frac{1}{2} f_X(frac{y-3}{2})$.



My Approach:

We Know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$

similarly,$$P(Y leq y)= int_{-infty}^y f_Y(y) , dy$$
Now,$, Y=2X+3$
$$implies P(Y leq y)= P((2X+3) leq y)$$
$$=P(X leq (frac{y-3}{2}))$$
$$implies P(Y leq y)=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy$$
$$implies int_{-infty}^y f_Y(y) , dy=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy $$
Now how can I proceed further to find a relationship between $f_Y$ and $f_X $?






probability-theory probability-distributions







share|cite|improve this question











$endgroup$




I want to prove that if probability density function (PDF) of a random variable $X$ is $f_X(x)$ and another random variable $Y$ is defined as $Y=2X+3$, then the PDF of $Y$ is $f_Y(y)=frac{1}{2} f_X(frac{y-3}{2})$.



My Approach:

We Know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$

similarly,$$P(Y leq y)= int_{-infty}^y f_Y(y) , dy$$
Now,$, Y=2X+3$
$$implies P(Y leq y)= P((2X+3) leq y)$$
$$=P(X leq (frac{y-3}{2}))$$
$$implies P(Y leq y)=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy$$
$$implies int_{-infty}^y f_Y(y) , dy=int_{-infty}^{frac{y-3}{2}} f_X(y) , dy $$
Now how can I proceed further to find a relationship between $f_Y$ and $f_X $?






probability-theory probability-distributions




probability-theory probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 9:58









Did

249k23228467




249k23228467










asked Jan 8 at 9:49









SureshSuresh

371110




371110








  • 1




    $begingroup$
    Use the change of variable $yto2y+3$ in the very last integral.
    $endgroup$
    – Did
    Jan 8 at 9:56












  • $begingroup$
    if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
    $endgroup$
    – Suresh
    Jan 8 at 10:12










  • $begingroup$
    No we do not get that.
    $endgroup$
    – Did
    Jan 8 at 14:30






  • 1




    $begingroup$
    Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
    $endgroup$
    – Did
    Jan 9 at 6:51






  • 1




    $begingroup$
    You are doing the change of variable, backwards.
    $endgroup$
    – Did
    Jan 10 at 10:38














  • 1




    $begingroup$
    Use the change of variable $yto2y+3$ in the very last integral.
    $endgroup$
    – Did
    Jan 8 at 9:56












  • $begingroup$
    if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
    $endgroup$
    – Suresh
    Jan 8 at 10:12










  • $begingroup$
    No we do not get that.
    $endgroup$
    – Did
    Jan 8 at 14:30






  • 1




    $begingroup$
    Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
    $endgroup$
    – Did
    Jan 9 at 6:51






  • 1




    $begingroup$
    You are doing the change of variable, backwards.
    $endgroup$
    – Did
    Jan 10 at 10:38








1




1




$begingroup$
Use the change of variable $yto2y+3$ in the very last integral.
$endgroup$
– Did
Jan 8 at 9:56






$begingroup$
Use the change of variable $yto2y+3$ in the very last integral.
$endgroup$
– Did
Jan 8 at 9:56














$begingroup$
if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
$endgroup$
– Suresh
Jan 8 at 10:12




$begingroup$
if we put $y to 2y +3 $ , then we will get $f_Y(y)=2 f_X(2y +3)$,now how to proceed?
$endgroup$
– Suresh
Jan 8 at 10:12












$begingroup$
No we do not get that.
$endgroup$
– Did
Jan 8 at 14:30




$begingroup$
No we do not get that.
$endgroup$
– Did
Jan 8 at 14:30




1




1




$begingroup$
Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
$endgroup$
– Did
Jan 9 at 6:51




$begingroup$
Wrong change of variable, possibly at least partly due to the horrendous practice of using the same symbol for the bound of the integral and for the variable of integration. Somehow, you are trying to get mistakes...
$endgroup$
– Did
Jan 9 at 6:51




1




1




$begingroup$
You are doing the change of variable, backwards.
$endgroup$
– Did
Jan 10 at 10:38




$begingroup$
You are doing the change of variable, backwards.
$endgroup$
– Did
Jan 10 at 10:38










2 Answers
2






active

oldest

votes


















1












$begingroup$


"We know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$... "




Observe that on RHS the $x$ is fixed and variable at the same time. That is confusing notation. You should write something like:$$P(X leq x)= int_{-infty}^x f_X(u) , du$$




You found: $$P(Yleq y)=int_{-infty}^{frac{y-3}2}f(u)du=int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$$



(Notice that I introduce $u$ here in order to avoid the confusing notation).



Now substitute $u=frac{v-3}2$ to get $$cdots=int_{-infty}^{infty}mathbf1_{left(-infty,frac{y-3}2right]}left(frac{v-3}2right)fleft(frac{v-3}2right)dleft(frac{v-3}2right)=int^y_{-infty}frac12 fleft(frac{v-3}2right)dv$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
    $endgroup$
    – Suresh
    Jan 8 at 10:22








  • 1




    $begingroup$
    If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
    $endgroup$
    – drhab
    Jan 8 at 10:25





















2












$begingroup$

$$F_Y(y) = F_Xleft(frac{y-3}{2} right)$$



Now, let's differentiate with respect to $y$ by chain rule



$$frac{d}{dy}F_Y(y) = frac{d}{dy}F_Xleft(frac{y-3}{2} right)=f_Xleft(frac{y-3}{2} right)frac{d}{dy}left(frac{y-3}{2} right)$$



that is $$f_Y(y)=frac12f_Xleft(frac{y-3}{2} right)$$






share|cite|improve this answer









$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065982%2fproof-that-if-y-2x3-then-f-yy-frac12f-x-fracy-32%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$


    "We know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$... "




    Observe that on RHS the $x$ is fixed and variable at the same time. That is confusing notation. You should write something like:$$P(X leq x)= int_{-infty}^x f_X(u) , du$$




    You found: $$P(Yleq y)=int_{-infty}^{frac{y-3}2}f(u)du=int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$$



    (Notice that I introduce $u$ here in order to avoid the confusing notation).



    Now substitute $u=frac{v-3}2$ to get $$cdots=int_{-infty}^{infty}mathbf1_{left(-infty,frac{y-3}2right]}left(frac{v-3}2right)fleft(frac{v-3}2right)dleft(frac{v-3}2right)=int^y_{-infty}frac12 fleft(frac{v-3}2right)dv$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
      $endgroup$
      – Suresh
      Jan 8 at 10:22








    • 1




      $begingroup$
      If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
      $endgroup$
      – drhab
      Jan 8 at 10:25


















    1












    $begingroup$


    "We know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$... "




    Observe that on RHS the $x$ is fixed and variable at the same time. That is confusing notation. You should write something like:$$P(X leq x)= int_{-infty}^x f_X(u) , du$$




    You found: $$P(Yleq y)=int_{-infty}^{frac{y-3}2}f(u)du=int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$$



    (Notice that I introduce $u$ here in order to avoid the confusing notation).



    Now substitute $u=frac{v-3}2$ to get $$cdots=int_{-infty}^{infty}mathbf1_{left(-infty,frac{y-3}2right]}left(frac{v-3}2right)fleft(frac{v-3}2right)dleft(frac{v-3}2right)=int^y_{-infty}frac12 fleft(frac{v-3}2right)dv$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
      $endgroup$
      – Suresh
      Jan 8 at 10:22








    • 1




      $begingroup$
      If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
      $endgroup$
      – drhab
      Jan 8 at 10:25
















    1












    1








    1





    $begingroup$


    "We know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$... "




    Observe that on RHS the $x$ is fixed and variable at the same time. That is confusing notation. You should write something like:$$P(X leq x)= int_{-infty}^x f_X(u) , du$$




    You found: $$P(Yleq y)=int_{-infty}^{frac{y-3}2}f(u)du=int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$$



    (Notice that I introduce $u$ here in order to avoid the confusing notation).



    Now substitute $u=frac{v-3}2$ to get $$cdots=int_{-infty}^{infty}mathbf1_{left(-infty,frac{y-3}2right]}left(frac{v-3}2right)fleft(frac{v-3}2right)dleft(frac{v-3}2right)=int^y_{-infty}frac12 fleft(frac{v-3}2right)dv$$






    share|cite|improve this answer









    $endgroup$




    "We know: $P(X leq x)= int_{-infty}^x f_X(x) , dx$... "




    Observe that on RHS the $x$ is fixed and variable at the same time. That is confusing notation. You should write something like:$$P(X leq x)= int_{-infty}^x f_X(u) , du$$




    You found: $$P(Yleq y)=int_{-infty}^{frac{y-3}2}f(u)du=int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$$



    (Notice that I introduce $u$ here in order to avoid the confusing notation).



    Now substitute $u=frac{v-3}2$ to get $$cdots=int_{-infty}^{infty}mathbf1_{left(-infty,frac{y-3}2right]}left(frac{v-3}2right)fleft(frac{v-3}2right)dleft(frac{v-3}2right)=int^y_{-infty}frac12 fleft(frac{v-3}2right)dv$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 10:14









    drhabdrhab

    104k545136




    104k545136












    • $begingroup$
      Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
      $endgroup$
      – Suresh
      Jan 8 at 10:22








    • 1




      $begingroup$
      If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
      $endgroup$
      – drhab
      Jan 8 at 10:25




















    • $begingroup$
      Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
      $endgroup$
      – Suresh
      Jan 8 at 10:22








    • 1




      $begingroup$
      If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
      $endgroup$
      – drhab
      Jan 8 at 10:25


















    $begingroup$
    Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
    $endgroup$
    – Suresh
    Jan 8 at 10:22






    $begingroup$
    Sorry , i am not able to understand how you are getting "$int_{-infty}^{+infty}mathbf1_{left(-infty,frac{y-3}2right]}(u)f(u)du$ " from "$int_{-infty}^{frac{y-3}2}f(u)du=$";could you please explain it...
    $endgroup$
    – Suresh
    Jan 8 at 10:22






    1




    1




    $begingroup$
    If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
    $endgroup$
    – drhab
    Jan 8 at 10:25






    $begingroup$
    If $Asubseteqmathbb R$ is measurable then $int_Af(u);du$ is nothing else but a notation for $int_{-infty}^{infty}mathbf1_A(u)f(u)du$. If $A=(a,b]$ then $int_a^bf(u)du$ is a notation for $int_{(a,b]}f(u)du=int_{-infty}^{infty}mathbf1_{(a,b]}(u)f(u)du$. You can take $a=-infty$ here.
    $endgroup$
    – drhab
    Jan 8 at 10:25













    2












    $begingroup$

    $$F_Y(y) = F_Xleft(frac{y-3}{2} right)$$



    Now, let's differentiate with respect to $y$ by chain rule



    $$frac{d}{dy}F_Y(y) = frac{d}{dy}F_Xleft(frac{y-3}{2} right)=f_Xleft(frac{y-3}{2} right)frac{d}{dy}left(frac{y-3}{2} right)$$



    that is $$f_Y(y)=frac12f_Xleft(frac{y-3}{2} right)$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      $$F_Y(y) = F_Xleft(frac{y-3}{2} right)$$



      Now, let's differentiate with respect to $y$ by chain rule



      $$frac{d}{dy}F_Y(y) = frac{d}{dy}F_Xleft(frac{y-3}{2} right)=f_Xleft(frac{y-3}{2} right)frac{d}{dy}left(frac{y-3}{2} right)$$



      that is $$f_Y(y)=frac12f_Xleft(frac{y-3}{2} right)$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        $$F_Y(y) = F_Xleft(frac{y-3}{2} right)$$



        Now, let's differentiate with respect to $y$ by chain rule



        $$frac{d}{dy}F_Y(y) = frac{d}{dy}F_Xleft(frac{y-3}{2} right)=f_Xleft(frac{y-3}{2} right)frac{d}{dy}left(frac{y-3}{2} right)$$



        that is $$f_Y(y)=frac12f_Xleft(frac{y-3}{2} right)$$






        share|cite|improve this answer









        $endgroup$



        $$F_Y(y) = F_Xleft(frac{y-3}{2} right)$$



        Now, let's differentiate with respect to $y$ by chain rule



        $$frac{d}{dy}F_Y(y) = frac{d}{dy}F_Xleft(frac{y-3}{2} right)=f_Xleft(frac{y-3}{2} right)frac{d}{dy}left(frac{y-3}{2} right)$$



        that is $$f_Y(y)=frac12f_Xleft(frac{y-3}{2} right)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 9:57









        Siong Thye GohSiong Thye Goh

        104k1468120




        104k1468120






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065982%2fproof-that-if-y-2x3-then-f-yy-frac12f-x-fracy-32%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more