document.write( "Question 1048269: A right triangle has one vertex on the graph of y=x^5 , x>0, at (x,y) another at the origin, and a third on the positive y-axis at (0,y). Express the area A of the triangle as a function of x \n" ); document.write( "
Algebra.Com's Answer #663872 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
A right triangle has one vertex on the graph of \"y=x%5E5\" , \"x%3E0\", at (\"x\",\"y\") \r
\n" ); document.write( "\n" ); document.write( "another at the origin, or (\"0\",\"0\") \r
\n" ); document.write( "\n" ); document.write( "and a third on the positive y-axis at (\"0\",\"y\")
\n" ); document.write( "remember, we let \"y=x%5E5\"; so, replace \"\"y\"\" with \"x%5E5\" to get (\"0\",\"x%5E5\")\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "let's draw a picture:\r
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "since the point (\"x\",\"y\") is \"\"x\"\" units to the \"right\" and \"\"y\"\" units \"up\", this means that the \"base\" of the triangle is \"\"x\"\" units long and the \"height\" is \"\"y\"\" units \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So the base is \"\"b=x\"\" and the height is \"h=x%5E5\"\r
\n" ); document.write( "\n" ); document.write( "Now recall (or look up), the area of any triangle is \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"A=%28bh%29%2F2\" ..........substitute \"b\" and \"h\"\r
\n" ); document.write( "\n" ); document.write( "\"A=%28x%2Ax%5E5%29%2F2\"\r
\n" ); document.write( "\n" ); document.write( "\"A=x%5E6%2F2\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );