document.write( "Question 1151508: standard stop sign is 75cm across from one side of the red octagon to the opposite side, with a 2-cm white border, using a scale of 14, the ENTIRE width of an image of the stop sign is _____cm. the width of the INSIDE red portion is ______cm \n" ); document.write( "
Algebra.Com's Answer #773277 by MathLover1(20849)\"\" \"About 
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so you have \"75\" as the width of the red octagon
\n" ); document.write( "since there is a \"2cm\" boarder around,means you will have to add \"2cm\" to the left and \"2cm\" to the right of the red octagon\r
\n" ); document.write( "\n" ); document.write( "so, a total of \"4cm\" is added to \"75\" which gives you a total width of \"79cm\" \r
\n" ); document.write( "\n" ); document.write( "now using a scale of \"14\" we will get proportionally smaller image:\r
\n" ); document.write( "\n" ); document.write( "\"79cm%2F14+=+5.64cm\"\r
\n" ); document.write( "\n" ); document.write( "so.the width of the red octagon, to nearest tenth is \"5.6cm\"\r
\n" ); document.write( "\n" ); document.write( "since given that the inside red portion is \"75cm\"\r
\n" ); document.write( "\n" ); document.write( "\"75cm%2F14=5.357cm\"\r
\n" ); document.write( "\n" ); document.write( "so, to nearest tenth is \"5.4cm\"\r
\n" ); document.write( "\n" ); document.write( "answer : \r
\n" ); document.write( "\n" ); document.write( "the ENTIRE width of an image of the stop sign is __ \"5.6\" ___cm.
\n" ); document.write( "the width of the INSIDE red portion is __\"5.4cm\"____cm\r
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