document.write( "Question 1100376: calculate 8.81×10^6 times 3.47×10^3 by using scientific notation and the product rule. express your answer in scientific notation with the proper number of significant figures. \n" ); document.write( "
Algebra.Com's Answer #714867 by Edwin McCravy(20060)\"\" \"About 
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document.write( "8.81×106 times 3.47×103\r\n" );
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document.write( "Each of those is rounded to three significant digits.\r\n" );
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document.write( "(8.81)(3.47)×106+3\r\n" );
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document.write( "30.5707×109\r\n" );
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document.write( "So we round 30.5707 to three significant digits also:\r\n" );
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document.write( "30.6×109  \r\n" );
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document.write( "But that's not scientific notation for the first\r\n" );
document.write( "number must be less than 10.\r\n" );
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document.write( "Write 30.6 as 3.06×101 \r\n" );
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document.write( "3.06×101×109\r\n" );
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document.write( "3.06×109+1 \r\n" );
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document.write( "3.06×1010    <--final answer\r\n" );
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document.write( "Let's analyze what happens here:\r\n" );
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document.write( "Both those numbers to multiply were rounded off\r\n" );
document.write( "themselves, so:\r\n" );
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document.write( "The 8.81 could have been as large as 8.8149 and\r\n" );
document.write( "the 3.47 could have been as large as 3.4749 and\r\n" );
document.write( "the product as large as 30.63089601.\r\n" );
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document.write( "OR\r\n" );
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document.write( "The 8.81 could have been as small as 8.805 and\r\n" );
document.write( "the 3.47 could have been as small as 3.465 and\r\n" );
document.write( "the product as small as 30.509325.\r\n" );
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document.write( "They probably weren't both that large or both that small.\r\n" );
document.write( "So let's average them:\r\n" );
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document.write( "The average of those two answers is 30.57011051, or 30.6.\r\n" );
document.write( "so the only digits we can count as significant are the 3, \r\n" );
document.write( "the 0, and the 6.  The other digits to the right of those \r\n" );
document.write( "would be meaningless, so we drop them.\r\n" );
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document.write( "Edwin
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