Update index.html

Changed images

Author: Gmac4247

index.html

@@ -908,7 +908,7 @@ <h3 style="margin:7px" itemprop="name">Volume of a Cube</h3>
 </div>
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 <figure itemprop="image" class="imgbox" itemscope itemtype="http://schema.org/ImageObject">
-    <img class="center-fit" src="cubeMarkup.jpeg" alt="A cube is a 3 dimensional solid shape with 3 equal perpendicular pairs of parallel straight edges. V = edge × edge × edge = edge³">
+    <img class="center-fit" src="cube.jpeg" alt="A cube is a 3 dimensional solid shape with 3 equal perpendicular pairs of parallel straight edges. V = edge × edge × edge = edge³">
     </figure>
 <br>
 <p style="margin:12px" itemprop="usageInfo">The cube extends the square into three dimensions. That is a direct extrapolation from the area of the square, establishing the basis for volumetric relationships. This is the basis of volume calculation.</p>
@@ -1963,7 +1963,7 @@ <h3 itemprop="name" style="margin:7px">Area of a Circle</h3>
 </div>
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 <figure itemprop="image" class="imgbox" itemscope itemtype="http://schema.org/ImageObject">
-    <img class="center-fit" src="areaOfACircle.jpg" alt="The circle is cut into four quadrants, each placed with their origin on the vertices of a square. The arcs of the quadrants of the circle that equals in area to the square intersect at the quarters on its centerlines. The ratio between the radius of the circle and the side of the square is calculable. r = side × √5 / 4 Area = 3.2r²">
+    <img class="center-fit" src="circleArea.png" alt="The circle is cut into four quadrants, each placed with their origin on the vertices of a square. The arcs of the quadrants of the circle that equals in area to the square intersect at the quarters on its centerlines. The ratio between the radius of the circle and the side of the square is calculable. r = side × √5 / 4 Area = 3.2r²">
 </figure>
 <br>
 <p itemprop="description" style="margin:12px">The area of a circle is defined by comparing it to a square since that is the base of area calculation.</p>
@@ -3208,7 +3208,7 @@ <h3 itemprop="name" style="margin:7px">Area of a Circle Segment</h3>
 </div>
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 <figure itemprop="image" class="imgbox" itemscope itemtype="http://schema.org/ImageObject">
-    <img class="center-fit" src="circleSegmentMarkup.jpg" alt="The area of a circle segment can be calculated by subtracting a triangle from a circle slice. Area = Acos(( r - n ) / r ) × r² - sin( Acos(( r - n ) / r ) × ( r - n ) × r">
+    <img class="center-fit" src="circleSegment.jpg" alt="The area of a circle segment can be calculated by subtracting a triangle from a circle slice. Area = Acos(( r - n ) / r ) × r² - sin( Acos(( r - n ) / r ) × ( r - n ) × r">
     </figure>
 <br>
     <p itemprop="abstract" style="margin:12px">The area of a circle segment can be calculated by subtracting a triangle from a circle slice.</p>
@@ -3397,7 +3397,7 @@ <h3 itemprop="name" style="margin:7px">Surface Area of a Cone</h3>
 </div>
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 <figure itemprop="image" class="imgbox" itemscope itemtype="http://schema.org/ImageObject">
-  <img class="center-fit" src="coneMarkup.jpeg" alt="The surface area of a cone is calculated as a circle slice with a radius equal to the slant height. Area = 3.2r × ( r + √( r² + H² )">
+  <img class="center-fit" src="cone.jpeg" alt="The surface area of a cone is calculated as a circle slice with a radius equal to the slant height. Area = 3.2r × ( r + √( r² + H² )">
   </figure>
 <br>
 <p style="margin:12px" itemprop="abstract">The surface area of a cone is calculated as a circle slice with a radius equal to the slant height and the angle given by the ratio between the height and the slant height.</p>
@@ -3493,7 +3493,7 @@ <h3 itemprop="name" style="margin:7px">Volume of a Sphere</h3>
 </div>
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 <figure itemprop="image" class="imgbox" itemscope itemtype="http://schema.org/ImageObject">
-    <img class="center-fit" src="sphereAndCubeMarkup.jpeg" alt="The edge length of the cube, which has the same volume as the sphere, equals the square root of the area of the square that has the same area as the sphere's cross-section. Volume = ( √ ( 3.2 ) × r )³">
+    <img class="center-fit" src="sphere.jpeg" alt="The edge length of the cube, which has the same volume as the sphere, equals the square root of the area of the square that has the same area as the sphere's cross-section. Volume = ( √ ( 3.2 ) × r )³">
     </figure>
 <p itemprop="description" style="margin:12px"><strong>The volume of a sphere is defined by comparing it to a cube, since that is the base of volume calculation.</strong>
 <section style="margin:12px">
@@ -3728,7 +3728,7 @@ <h3 itemprop="name" style="margin:7px">Volume of a Cone</h3>
 </div>
 <br>
 <figure itemprop="image" class="imgbox" itemscope itemtype="http://schema.org/ImageObject">
-    <img class="center-fit" src="coneAndSphereMarkup.jpeg" alt="Cone volume from sphere = base × height / √8">
+    <img class="center-fit" src="coneAndSphere.jpeg" alt="Cone volume from sphere = base × height / √8">
 </figure>
 <section style="margin:12px">
 <details>
@@ -3904,11 +3904,7 @@ <h4 itemprop="abstract">The volume of a cone can be calculated by algebraically
 <br>
 <br>
 <figure itemprop="image" class="imgbox" itemscope itemtype="http://schema.org/ImageObject">
-    <img class="center-fit" src="coneAndSphereComparison.png" alt="The base of the two shapes is a quadrant circle.">
-</figure>
-<br>
-<figure itemprop="image" class="imgbox" itemscope itemtype="http://schema.org/ImageObject">
-    <img class="center-fit" src="sphereAndConeMarkup.jpeg" alt="The volume of a cone can be calculated by algebraically comparing the volume of a vertical quadrant of a cone with equal radius and height to an octant sphere with equal radius, through a quadrant cylinder. Volume = base × height / √8">
+    <img class="center-fit" src="sphereAndCone.jpeg" alt="The volume of a cone can be calculated by algebraically comparing the volume of a vertical quadrant of a cone with equal radius and height to an octant sphere with equal radius, through a quadrant cylinder. Volume = base × height / √8">
 </figure>
 <br>
 <div itemprop="abstract" style="margin:12px">
@@ -4048,7 +4044,7 @@ <h4 itemprop="abstract">The volume of a cone can be calculated by algebraically
 </div>
 <br>
 <div itemprop="mathExpression">
-<math display="block" style="margin:12px" xmlns="http://www.w3.org/1998/Math/MathML" >
+<math display="block" xmlns="http://www.w3.org/1998/Math/MathML" >
 <mrow>
 <msub>
 <mi>V</mi>
@@ -4130,7 +4126,7 @@ <h3 itemprop="name" style="margin:7px">Volume of a Frustum Cone</h3>
 </div>
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 <figure class="imgbox" itemscope itemtype="http://schema.org/ImageObject">
-    <img class="center-fit" src="frustumOfConeMarkup.png" alt="Subtracting the missing tip from a theoretical full cone gives the volume of a frustum cone. Volume = frustumHeight * (4 / 5 * bottomDiameter^2 * (1 / (1 - topDiameter / bottomDiameter)) - 4 / 5 * topDiameter^2 * (1 / (1 - topDiameter / bottomDiameter) - 1)) / √8">
+    <img class="center-fit" src="frustumCone.jpeg" alt="Subtracting the missing tip from a theoretical full cone gives the volume of a frustum cone. Volume = frustumHeight * (4 / 5 * bottomDiameter^2 * (1 / (1 - topDiameter / bottomDiameter)) - 4 / 5 * topDiameter^2 * (1 / (1 - topDiameter / bottomDiameter) - 1)) / √8">
     </figure>
 <br>
 <p itemprop="description" style="margin:12px">Subtracting the missing tip from a theoretical full cone gives the volume of a frustum cone.
@@ -4166,8 +4162,8 @@ <h3 itemprop="name" style="margin:7px">Volume of a Frustum Cone</h3>
 </math>
 </div>
 <br>
-<div display="block" itemprop="mathExpression">
-<math style="margin:12px" xmlns="http://www.w3.org/1998/Math/MathML">
+<div itemprop="mathExpression">
+<math display="block" xmlns="http://www.w3.org/1998/Math/MathML">
      <mrow>
  <mi>V</mi>
       <mo>=</mo>
@@ -4312,7 +4308,7 @@ <h3 itemprop="name" style="margin:7px">Volume of a Pyramid</h3>
       <meta itemprop="eduSolution" content="10.94216">
 </div>
 </div><figure class="imgbox" itemscope itemtype="http://schema.org/ImageObject">
-  <img class="center-fit" src="conePyramidVolumeMarkup.jpeg" alt="The volume of a pyramid can be calculated with the same coefficient as the volume of a cone. Volume = base × height / √8">
+  <img class="center-fit" src="conePyramidVolume.jpeg" alt="The volume of a pyramid can be calculated with the same coefficient as the volume of a cone. Volume = base × height / √8">
   </figure>
 <br>
 <section>
@@ -4513,7 +4509,7 @@ <h3 itemprop="name" style="margin:12px">Volume of a horizontal Frustum Pyramid</
 	</div>
 	<br>
 <figure class="imgbox">
-    <img class="center-fit" src="frustumOfPyramidMarkup.png" alt="Subtracting the missing tip from a theoretical full pyramid gives the volume of a frustum pyramid. Volume = frustumHeight * (bottomArea * (1 / (1 - topArea / bottomArea)) - topArea * (1 / (1 - topArea / bottomArea) - 1)) / √8">
+    <img class="center-fit" src="frustumPyramid.jpeg" alt="Subtracting the missing tip from a theoretical full pyramid gives the volume of a frustum pyramid. Volume = frustumHeight * (bottomArea * (1 / (1 - topArea / bottomArea)) - topArea * (1 / (1 - topArea / bottomArea) - 1)) / √8">
 </figure>
 <br>
 <p itemprop="description" style="margin:12px">Subtracting the missing tip from a theoretical full pyramid gives the volume of a frustum pyramid. 
@@ -4752,7 +4748,7 @@ <h3 itemprop="name" style="margin:7px">Volume of a Tetrahedron</h3>
 	</div>
 	<br>
 <figure class="imgbox" itemprop="image" itemscope itemtype="http://schema.org/ImageObject">
-  <img class="center-fit" src="tetrahedronMarkup.jpeg" alt="The volume of a tetrahedron is calculated as a pyramid with an equilateral triangle base and fixed proportions. Volume = edge³ / 8">
+  <img class="center-fit" src="tetrahedron.jpeg" alt="The volume of a tetrahedron is calculated as a pyramid with an equilateral triangle base and fixed proportions. Volume = edge³ / 8">
   </figure>
 <br>
 <p itemprop="description" style="margin:12px">