document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=408057>Question 408057</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi there, <BR>\n" );
document.write( "My question: For this linear equation y=-(2/3) x+30  I have to graph the x and y intercepts, and determine the heigth of the laser beam when the laser is 30 feet from the building...\r<BR>\n" );
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document.write( "For this linear equation y=-(2/3) x+30  \r<BR>\n" );
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document.write( "A1} the Y axis represents the height (above ground) in feet.<BR>\n" );
document.write( "A2} the X axis represents the distance (from the building) measured in feet.  \r<BR>\n" );
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document.write( "This problem has asked me to interpret the results graphically. I have graphed the following values:  <BR>\n" );
document.write( "The x intercept or the point where the value crosses the x axis represents the point where the laser beams touches the ground. The y intercept, the point where the value crosses the y axis which represents the laser starting point or point where man is located. \r<BR>\n" );
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document.write( "A3} in the equation y=-(2/3) x+30    y=mx+b format (slope intercept)<BR>\n" );
document.write( "(b) Represents the height of the window<BR>\n" );
document.write( "(x)Represents the point where the light intercepts the ground. \r<BR>\n" );
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document.write( "A4} I solved this question by first drawing a grid, starting at the window point (0, 30)Then I used my slope of -2/3 to determine the value of 42 feet out from the building when the lights beam intercepts the ground. (42, 0) <BR>\n" );
document.write( "I have found the light to be 42feet out (x value), 30feet high(y value) by using (rise over run) or slope of -2/3. \r<BR>\n" );
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document.write( "A5} Quadrant 2 is the only relevant quadrant for this problem. If we were doing a standard graphed linear equation, we would include quad. 1, 2 and 3 but since we are not including the area under the ground quad 3 is excluded, and since the beam is directional, quad 1 is also excluded. Quad four is also excluded because in either scenario, the plotted points are never on that plane/ quadrant. <BR>\n" );
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document.write( "A graph was a very good representation of the light because it clearly shows the path the light took to the ground, in a way that brings the problem to life. The graph gave a pictorial representation for each value. Y isn’t simply Y its now, the point where the light originated from. And x has a pictorial value as well; X is the place where the beam intersects the ground. \r<BR>\n" );
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document.write( "My work was returned to me because I did not find the x value correctly, I do not know how to go about finding this. I have graphed this problem but was not able to send a copy so that you could see my work, \r<BR>\n" );
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document.write( "x	y=-(2/3)x+30<BR>\n" );
document.write( "-1	30 2/3<BR>\n" );
document.write( "0	30<BR>\n" );
document.write( "1	29 1/3\r<BR>\n" );
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document.write( "I need to know what the height is in feet when the laser beam is 30 feet away from the building...\r<BR>\n" );
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document.write( "Thank you so much for your help!\r<BR>\n" );
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document.write( "Take care <BR>\n" );
document.write( "Maranda\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #287526 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=287526\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "You are way way way over thinking this thing.\r<BR>\n" );
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document.write( "Take your equation and substitute 0 in place of x.  Do the arithmetic.  The value you get for y is the height of the window.  Take the original equation and substitute 0 in place of y.  Do the arithmetic.  The value you get for x is the distance from the building where the beam hits the ground.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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