document.write( "Question 1197359: The temperature recorded in a certain region at 7:00 a.m. it was -4°C, at 9:00 a.m. rose 3°C, and then at 10:00 a.m. dropped 2°C. From now until 2:00 p.m. increased 2°C per hour. What was the temperature at 1:00 p.m.?
\n" ); document.write( "A) -4
\n" ); document.write( "B) -3
\n" ); document.write( "C) -1
\n" ); document.write( "D) 3
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Algebra.Com's Answer #830585 by MathLover1(20849)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "at \"7%3A00\"a.m. it was \"-4\"°C\r
\n" ); document.write( "\n" ); document.write( " at \"9%3A00+\"a.m. rose \"3\"°C=>\"-4\"°C+\"3\"°C=\"-1\"°C\r
\n" ); document.write( "\n" ); document.write( " at \"10%3A00\" a.m. dropped \"2\"°C=>\"-1\"°C-\"2\"°C=\"-3\"°C\r
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\n" ); document.write( "\n" ); document.write( "From now until \"2%3A00+\"p.m. increased \"2\"°C per hour:\r
\n" ); document.write( "\n" ); document.write( "from \"10%3A00\" a.m to \"1%3A00\" p.m is \"3\" hours, temperature increased \"3%2A2\"°C =\"6\"°C\r
\n" ); document.write( "\n" ); document.write( "since at \"10%3A00\" a.m was \"-3\"°C, the temperature at \"+1%3A00\" p.m is: \"-3\"°C +\"6\"°C= \"3\"°C\r
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\n" ); document.write( "\n" ); document.write( "answer: D) \"3\"\r
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