document.write( "Question 1058954: On wet concrete, the stopping distance s, in feet, of a car traveling v miles per hour is given by s(v) = 0.055v^2 + 1.1v. At what speed could a car be traveling and still stop at a stop sign 44 feet away? \n" ); document.write( "

Algebra.Com's Answer #674049 by ikleyn(52781) $\"\"$ $\"About$

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\n" ); document.write( "On wet concrete, the stopping distance s, in feet, of a car traveling v miles per hour is given by s(v) = 0.055v^2 + 1.1v.
\n" ); document.write( "At what speed could a car be traveling and still stop at a stop sign 44 feet away?
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document.write( "0.055v^2 + 1.1v = 44.\r\n" );
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document.write( "Simplify and solve for v.\r\n" );
document.write( "As a first step, divide both sides by 0.055. You will get\r\n" );
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document.write( "v^2 + 20v - 800 = 0.\r\n" );
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document.write( "Factor\r\n" );
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document.write( "(v-20)*(v+40) = 0.\r\n" );
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document.write( "Take the positive root and disregard the negative.\r\n" );
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document.write( "Answer.  v = 20 mph is the unique solution.\r\n" );
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