title: jquery .css() not working
body: <p>I don't know why when I use the .css() function of jQuery it returns me this:</p>
  
  <pre><code> Uncaught TypeError: undefined is not a function
  </code></pre>
  
  <p>My code is this:</p>
  
  <p>HTML:</p>
  
  <pre><code>&lt;div class="photo1 sliderPhoto"&gt;d&lt;/div&gt;
  &lt;div class="photo2 sliderPhoto"&gt;d&lt;/div&gt;
  &lt;div class="photo3 sliderPhoto"&gt;d&lt;/div&gt;
  &lt;div class="photo4 sliderPhoto"&gt;d&lt;/div&gt;
  </code></pre>
  
  <p>CSS:</p>
  
  <pre><code>.sliderPhoto{
      position: absolute;
      top: 0px;
      left: 0px;
      background-position: center center;
      background-size: cover;
      width: 100%;
      height: 100%;
  }
  .photo1{
      background-image: url('../photos/ph1.jpg');
      z-index: 6;
  }
  .photo2{
      background-image: url('../photos/ph2.jpg');
      z-index: 6; 
      display: none;
  }
  .photo3{
      background-image: url('../photos/ph3.jpg');
      z-index: 6;
      display: none;
  }
  .photo4{
      background-image: url('../photos/ph4.jpg');
      z-index: 6;
      display: none;
  }
  </code></pre>
  
  <p>JQuery on $(document).ready:</p>
  
  <pre><code>$photos = [$(".photo1")[0], $(".photo2")[0], $(".photo3")[0], $(".photo4")[0]];
  $photos[0].css("z-index");
  </code></pre>
  
  <p>when I type this jQuery code it doesn't work :(
  but this $photos[0] returns the exact div I need.</p>