document.write( "Question 459816:
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document.write( "Dear yenigelabert@yahoo.com, your question for free tutors has been submitted.\r
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document.write( "Note, if you need an IMMEDIATE answer and can pay a modest amount, you
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document.write( "can go to\r
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document.write( " http://www.algebra.com/cgi-bin/secretoffer.mpl?name=liveperson\r
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document.write( "They have great tutors and cheap rates. Note that algebra.com is completely free.\r
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document.write( "--------------- QUESTION FOLLOWS --------------------------
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document.write( "if 4 -3a=7-2(2a+5) evaluate a^2+7a
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document.write( "Please I NEED HELP
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document.write( "(-2x^3+x-1)-(-x^2+x-3)\r
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document.write( "(2x^2+5x-3)-(3x^3+2x-5)\r
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document.write( "a ship traveling east at 25mph is 10mi from a harbor when another ship leaves the harbor traveling east at 35mph. How long does it take the second ship to catch up.?\r
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Algebra.Com's Answer #315330 by mananth(16946)![]() ![]() You can put this solution on YOUR website! if 4 -3a=7-2(2a+5) evaluate a^2+7a\r \n" ); document.write( "\n" ); document.write( "4-3a=7-4a-10 \n" ); document.write( "-3a+4a=7-10-4 \n" ); document.write( "a=-7\r \n" ); document.write( "\n" ); document.write( "a^2+7a= 49-49 \n" ); document.write( "=0 \n" ); document.write( "..... \n" ); document.write( "(-2x^3+x-1)-(-x^2+x-3) \n" ); document.write( "-2x^3+x-1 +x^2-x+3 \n" ); document.write( "-2x^3+x^2+3\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "(2x^2+5x-3)-(3x^3+2x-5) \n" ); document.write( "-3x^3+2x^2+3x+5\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "a ship traveling east at 25mph is 10mi from a harbor when another ship leaves the harbor traveling east at 35mph. How long does it take the second ship to catch up.?\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "catch up distance = 10 miles \n" ); document.write( "catchup speed = 35 -25 = 10 mph \n" ); document.write( "catchup time = catchup distance / catch up speed \n" ); document.write( "10/10 \n" ); document.write( "=1 hour to catch up\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |