Simplifying expressions with radical exponents












0












$begingroup$


This is the expression that needs to be simplified:



$frac{(9-x^2)^{1/2} + x^2(9-x^2)^{-1/2}}{9-x^2}$



My first step is to move the negative exponent to the denominator:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)(9-x^2)^{1/2}}$



Then multiply the denominator and add the exponents:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)^{3/2}}$



I don't know the next steps to reach the given answer of:



$frac{9}{(9-x^2)^{3/2}}$






algebra-precalculus radicals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
    $endgroup$
    – MPW
    Dec 29 '18 at 16:23










  • $begingroup$
    $frac{9}{left(9-x^2right)^{3/2}}$
    $endgroup$
    – David G. Stork
    Dec 29 '18 at 16:53
















0












$begingroup$


This is the expression that needs to be simplified:



$frac{(9-x^2)^{1/2} + x^2(9-x^2)^{-1/2}}{9-x^2}$



My first step is to move the negative exponent to the denominator:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)(9-x^2)^{1/2}}$



Then multiply the denominator and add the exponents:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)^{3/2}}$



I don't know the next steps to reach the given answer of:



$frac{9}{(9-x^2)^{3/2}}$






algebra-precalculus radicals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
    $endgroup$
    – MPW
    Dec 29 '18 at 16:23










  • $begingroup$
    $frac{9}{left(9-x^2right)^{3/2}}$
    $endgroup$
    – David G. Stork
    Dec 29 '18 at 16:53














0












0








0





$begingroup$


This is the expression that needs to be simplified:



$frac{(9-x^2)^{1/2} + x^2(9-x^2)^{-1/2}}{9-x^2}$



My first step is to move the negative exponent to the denominator:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)(9-x^2)^{1/2}}$



Then multiply the denominator and add the exponents:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)^{3/2}}$



I don't know the next steps to reach the given answer of:



$frac{9}{(9-x^2)^{3/2}}$






algebra-precalculus radicals







share|cite|improve this question









$endgroup$




This is the expression that needs to be simplified:



$frac{(9-x^2)^{1/2} + x^2(9-x^2)^{-1/2}}{9-x^2}$



My first step is to move the negative exponent to the denominator:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)(9-x^2)^{1/2}}$



Then multiply the denominator and add the exponents:



$frac{(9-x^2)^{1/2} + x^2}{(9-x^2)^{3/2}}$



I don't know the next steps to reach the given answer of:



$frac{9}{(9-x^2)^{3/2}}$






algebra-precalculus radicals




algebra-precalculus radicals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 16:20









AntoninusAntoninus

33




33












  • $begingroup$
    Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
    $endgroup$
    – MPW
    Dec 29 '18 at 16:23










  • $begingroup$
    $frac{9}{left(9-x^2right)^{3/2}}$
    $endgroup$
    – David G. Stork
    Dec 29 '18 at 16:53


















  • $begingroup$
    Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
    $endgroup$
    – MPW
    Dec 29 '18 at 16:23










  • $begingroup$
    $frac{9}{left(9-x^2right)^{3/2}}$
    $endgroup$
    – David G. Stork
    Dec 29 '18 at 16:53
















$begingroup$
Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
$endgroup$
– MPW
Dec 29 '18 at 16:23




$begingroup$
Your first step is not valid. That can only be done with factors of the whole numerator, not just factors of a term in the numerator.
$endgroup$
– MPW
Dec 29 '18 at 16:23












$begingroup$
$frac{9}{left(9-x^2right)^{3/2}}$
$endgroup$
– David G. Stork
Dec 29 '18 at 16:53




$begingroup$
$frac{9}{left(9-x^2right)^{3/2}}$
$endgroup$
– David G. Stork
Dec 29 '18 at 16:53










3 Answers
3






active

oldest

votes


















0












$begingroup$

Hint: Write $$frac{(9-x^2)^{1/2}}{9-x^2}+frac{x^2(9-x^2)^{-1/2}}{9-x^2}$$ and this is
$$frac{1}{sqrt{9-x^2}}+frac{x^2}{(9-x^2)^{3/2}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
    $endgroup$
    – Antoninus
    Dec 29 '18 at 16:48










  • $begingroup$
    $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
    $endgroup$
    – KM101
    Dec 29 '18 at 16:51










  • $begingroup$
    Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 29 '18 at 16:52










  • $begingroup$
    Thank you, I've got it now!
    $endgroup$
    – Antoninus
    Dec 29 '18 at 17:11





















0












$begingroup$

Your first step is incorrect.



$$frac{a+b^{-n}}{c} color{red}{neq frac{a}{b^n+c}}$$



You can only bring the negative exponents down when the term is a factor of the whole numerator, not a factor of a single term in the numerator, as in



$$frac{ab^{-n}}{c} = frac{a}{b^nc}$$



Instead, you can make the substitution $u = 9-x^2$:



$$frac{u^{frac{1}{2}} + x^2(u)^{-frac{1}{2}}}{u} = frac{u^{frac{1}{2}}(1+x^2u^{-1})}{u} = frac{1+x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2}{ucdot u^{frac{1}{2}}}$$



Plugging in $u = 9-x^2$ and simplifying gives



$$frac{1}{(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9-x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9}{(9-x^2)^{frac{3}{2}}}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Consider the substitution $z = sqrt{9 - x^2}$ and solve for $x$ and use the $z$ variable.






    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%2f3056000%2fsimplifying-expressions-with-radical-exponents%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Hint: Write $$frac{(9-x^2)^{1/2}}{9-x^2}+frac{x^2(9-x^2)^{-1/2}}{9-x^2}$$ and this is
      $$frac{1}{sqrt{9-x^2}}+frac{x^2}{(9-x^2)^{3/2}}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
        $endgroup$
        – Antoninus
        Dec 29 '18 at 16:48










      • $begingroup$
        $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
        $endgroup$
        – KM101
        Dec 29 '18 at 16:51










      • $begingroup$
        Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 29 '18 at 16:52










      • $begingroup$
        Thank you, I've got it now!
        $endgroup$
        – Antoninus
        Dec 29 '18 at 17:11


















      0












      $begingroup$

      Hint: Write $$frac{(9-x^2)^{1/2}}{9-x^2}+frac{x^2(9-x^2)^{-1/2}}{9-x^2}$$ and this is
      $$frac{1}{sqrt{9-x^2}}+frac{x^2}{(9-x^2)^{3/2}}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
        $endgroup$
        – Antoninus
        Dec 29 '18 at 16:48










      • $begingroup$
        $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
        $endgroup$
        – KM101
        Dec 29 '18 at 16:51










      • $begingroup$
        Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 29 '18 at 16:52










      • $begingroup$
        Thank you, I've got it now!
        $endgroup$
        – Antoninus
        Dec 29 '18 at 17:11
















      0












      0








      0





      $begingroup$

      Hint: Write $$frac{(9-x^2)^{1/2}}{9-x^2}+frac{x^2(9-x^2)^{-1/2}}{9-x^2}$$ and this is
      $$frac{1}{sqrt{9-x^2}}+frac{x^2}{(9-x^2)^{3/2}}$$






      share|cite|improve this answer









      $endgroup$



      Hint: Write $$frac{(9-x^2)^{1/2}}{9-x^2}+frac{x^2(9-x^2)^{-1/2}}{9-x^2}$$ and this is
      $$frac{1}{sqrt{9-x^2}}+frac{x^2}{(9-x^2)^{3/2}}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 29 '18 at 16:30









      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

      79.5k42867




      79.5k42867












      • $begingroup$
        Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
        $endgroup$
        – Antoninus
        Dec 29 '18 at 16:48










      • $begingroup$
        $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
        $endgroup$
        – KM101
        Dec 29 '18 at 16:51










      • $begingroup$
        Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 29 '18 at 16:52










      • $begingroup$
        Thank you, I've got it now!
        $endgroup$
        – Antoninus
        Dec 29 '18 at 17:11




















      • $begingroup$
        Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
        $endgroup$
        – Antoninus
        Dec 29 '18 at 16:48










      • $begingroup$
        $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
        $endgroup$
        – KM101
        Dec 29 '18 at 16:51










      • $begingroup$
        Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
        $endgroup$
        – Dr. Sonnhard Graubner
        Dec 29 '18 at 16:52










      • $begingroup$
        Thank you, I've got it now!
        $endgroup$
        – Antoninus
        Dec 29 '18 at 17:11


















      $begingroup$
      Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
      $endgroup$
      – Antoninus
      Dec 29 '18 at 16:48




      $begingroup$
      Thank you for the feedback. I understand the right side of this, but I don't understand the left side. You are taking the square-root of the numerator and denominator, but how does the square-root of the numerator simplify to 1?
      $endgroup$
      – Antoninus
      Dec 29 '18 at 16:48












      $begingroup$
      $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
      $endgroup$
      – KM101
      Dec 29 '18 at 16:51




      $begingroup$
      $$frac{sqrt n}{n} = frac{1}{sqrt n}$$
      $endgroup$
      – KM101
      Dec 29 '18 at 16:51












      $begingroup$
      Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 29 '18 at 16:52




      $begingroup$
      Write $$frac{(9-x^2)^{1/2}}{(9-x^2)^{2/2}}$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 29 '18 at 16:52












      $begingroup$
      Thank you, I've got it now!
      $endgroup$
      – Antoninus
      Dec 29 '18 at 17:11






      $begingroup$
      Thank you, I've got it now!
      $endgroup$
      – Antoninus
      Dec 29 '18 at 17:11













      0












      $begingroup$

      Your first step is incorrect.



      $$frac{a+b^{-n}}{c} color{red}{neq frac{a}{b^n+c}}$$



      You can only bring the negative exponents down when the term is a factor of the whole numerator, not a factor of a single term in the numerator, as in



      $$frac{ab^{-n}}{c} = frac{a}{b^nc}$$



      Instead, you can make the substitution $u = 9-x^2$:



      $$frac{u^{frac{1}{2}} + x^2(u)^{-frac{1}{2}}}{u} = frac{u^{frac{1}{2}}(1+x^2u^{-1})}{u} = frac{1+x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2}{ucdot u^{frac{1}{2}}}$$



      Plugging in $u = 9-x^2$ and simplifying gives



      $$frac{1}{(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9-x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9}{(9-x^2)^{frac{3}{2}}}$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Your first step is incorrect.



        $$frac{a+b^{-n}}{c} color{red}{neq frac{a}{b^n+c}}$$



        You can only bring the negative exponents down when the term is a factor of the whole numerator, not a factor of a single term in the numerator, as in



        $$frac{ab^{-n}}{c} = frac{a}{b^nc}$$



        Instead, you can make the substitution $u = 9-x^2$:



        $$frac{u^{frac{1}{2}} + x^2(u)^{-frac{1}{2}}}{u} = frac{u^{frac{1}{2}}(1+x^2u^{-1})}{u} = frac{1+x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2}{ucdot u^{frac{1}{2}}}$$



        Plugging in $u = 9-x^2$ and simplifying gives



        $$frac{1}{(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9-x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9}{(9-x^2)^{frac{3}{2}}}$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Your first step is incorrect.



          $$frac{a+b^{-n}}{c} color{red}{neq frac{a}{b^n+c}}$$



          You can only bring the negative exponents down when the term is a factor of the whole numerator, not a factor of a single term in the numerator, as in



          $$frac{ab^{-n}}{c} = frac{a}{b^nc}$$



          Instead, you can make the substitution $u = 9-x^2$:



          $$frac{u^{frac{1}{2}} + x^2(u)^{-frac{1}{2}}}{u} = frac{u^{frac{1}{2}}(1+x^2u^{-1})}{u} = frac{1+x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2}{ucdot u^{frac{1}{2}}}$$



          Plugging in $u = 9-x^2$ and simplifying gives



          $$frac{1}{(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9-x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9}{(9-x^2)^{frac{3}{2}}}$$






          share|cite|improve this answer









          $endgroup$



          Your first step is incorrect.



          $$frac{a+b^{-n}}{c} color{red}{neq frac{a}{b^n+c}}$$



          You can only bring the negative exponents down when the term is a factor of the whole numerator, not a factor of a single term in the numerator, as in



          $$frac{ab^{-n}}{c} = frac{a}{b^nc}$$



          Instead, you can make the substitution $u = 9-x^2$:



          $$frac{u^{frac{1}{2}} + x^2(u)^{-frac{1}{2}}}{u} = frac{u^{frac{1}{2}}(1+x^2u^{-1})}{u} = frac{1+x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2u^{-1}}{u^{frac{1}{2}}} = frac{1}{u^{frac{1}{2}}}+frac{x^2}{ucdot u^{frac{1}{2}}}$$



          Plugging in $u = 9-x^2$ and simplifying gives



          $$frac{1}{(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9-x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}}+frac{x^2}{(9-x^2)(9-x^2)^{frac{1}{2}}} = frac{9}{(9-x^2)^{frac{3}{2}}}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 29 '18 at 16:49









          KM101KM101

          6,0861525




          6,0861525























              0












              $begingroup$

              Consider the substitution $z = sqrt{9 - x^2}$ and solve for $x$ and use the $z$ variable.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Consider the substitution $z = sqrt{9 - x^2}$ and solve for $x$ and use the $z$ variable.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Consider the substitution $z = sqrt{9 - x^2}$ and solve for $x$ and use the $z$ variable.






                  share|cite|improve this answer









                  $endgroup$



                  Consider the substitution $z = sqrt{9 - x^2}$ and solve for $x$ and use the $z$ variable.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 29 '18 at 19:12









                  Lorenz H MenkeLorenz H Menke

                  9611




                  9611






























                      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%2f3056000%2fsimplifying-expressions-with-radical-exponents%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

                      Le Mesnil-Réaume

                      Image
                      Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more

                      web3.py web3.isConnected() returns false always

                      Image
                      1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
                      Read more