document.write( "Question 22483: find a value of (n) to make each statement true:
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Algebra.Com's Answer #11751 by AnlytcPhil(1806)\"\" \"About 
You can put this solution on YOUR website!
1.  find a value of (n) to make each statement true:\r\n" );
document.write( "22times n= the answer between 80-85 show your work?\r\n" );
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document.write( "Write this:\r\n" );
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document.write( "80 < 22n < 85\r\n" );
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document.write( "Divide all three isdes by 22\r\n" );
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document.write( "80/22 < 22n/22 < 85/22\r\n" );
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document.write( "3.63 < n < 3.863\r\n" );
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document.write( "Now pick a value of n between 3.63 and 3.863, say n = 3.7 or \r\n" );
document.write( "n = 3.8.  Either one will do.  Check n = 3.7:  If n = 3.7, \r\n" );
document.write( "then 22n = 22(3.7) = 81.4 which is between 80 and 85.  Check\r\n" );
document.write( "n = 3.8:  If n = 3.8, then 22n = 22(3.8 ) = 83.6 which is also \r\n" );
document.write( "between 80 and 85. \r\n" );
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document.write( "2. n times 85= the answer between 175-250 show your work?\r\n" );
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document.write( "Write this:\r\n" );
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document.write( "175 < 85n < 250\r\n" );
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document.write( "Divide all three sides by 85\r\n" );
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document.write( "175/85 < 85n/85 < 250/85\r\n" );
document.write( "  \r\n" );
document.write( "2.058823529 < n < 2.941176471\r\n" );
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document.write( "Now pick a value of n between 2.058823529 and 2.941176471,\r\n" );
document.write( "say n = 2.1 or n = 2.9.  Either one of these or any of these\r\n" );
document.write( "2.2, 2.3, ... 2.8 will do.  Check n = 2.1:  If n = 2.1, then\r\n" );
document.write( "85n = 85(2.1) = 178.5 which is between 175 and 250.  Check\r\n" );
document.write( "n = 2.9:  If n = 2.9, then 85n = 85(2.9) = 246.5, which is also \r\n" );
document.write( "between 175 and 250\r\n" );
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document.write( "3.   142 times n= the answer is between 150-180 show your work?\r\n" );
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document.write( "4.    n times 77= the answer is between 80-150 show your work?\r\n" );
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document.write( "These are done the same way.\r\n" );
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document.write( "Edwin\r\n" );
document.write( "AnlytcPhil@aol.com
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