document.write( "Question 68251: I need help to some questions and I need help badly can anyone help me please the questions are as followed:if a=0.65 and b=1.28 and c=-5over2 evaluate - a(-b)+c but the answer I get out of it is three different ones every time its between -4.43,-1.87,and -0.57 I'm confused.The next one is solve the equation for q:6+[Q-2]=-1.The next one is if p=0.75 and q=-1.6,evaluate:p+q-(-0.7).And the last one is solve the equation for b:3over5=9over10b.Thank you for whoever can help thanks \n" ); document.write( "
Algebra.Com's Answer #48570 by ankor@dixie-net.com(22740)\"\" \"About 
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need help to some questions and I need help badly can anyone help me please the questions are as followed:
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\n" ); document.write( "if a=0.65 and b=1.28 and c=-5 over 2 evaluate - a(-b)+c
\n" ); document.write( "Since the answer is in decimals change -5/2 to -2.5 (c)
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\n" ); document.write( "-0.65(-1.28) + (-2.5)
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\n" ); document.write( "+.832 - 2.5; minus times a minus is a plus, plus a minus is a minus
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\n" ); document.write( "-1.668
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\n" ); document.write( " solve the equation for q:
\n" ); document.write( "6 + [Q-2] = -1
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\n" ); document.write( "Get rid of the brackets:
\n" ); document.write( "6 + Q - 2 = -1
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\n" ); document.write( "Q + 4 = -1; did the math on the left
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\n" ); document.write( "Subtract 4 from both sides:
\n" ); document.write( "Q = - 1 - 4
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\n" ); document.write( "Q = -5
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\n" ); document.write( "you can check that in the original equation:
\n" ); document.write( "6 + [Q-2] = -1
\n" ); document.write( "6 + [-5-2] = -1
\n" ); document.write( "6 + (-7) = -1
\n" ); document.write( "6 - 7 = -1
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\n" ); document.write( "if p=0.75 and q=-1.6, evaluate:
\n" ); document.write( "p + q -(-0.7)
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\n" ); document.write( "Remember a minus a minus is a plus, remove the brackets and you have:
\n" ); document.write( "p + q + 0.7
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\n" ); document.write( "Substitute for p & q
\n" ); document.write( ".75 + (-1.6) + .7
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\n" ); document.write( ".75 - 1.6 + .7 = -.15
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\n" ); document.write( "solve the equation for b: 3 over 5= 9 over 10b
\n" ); document.write( "\"3%2F5\" = \"9%2F%2810b%29\"
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\n" ); document.write( "We want to get rid of the denominators, mult both sides by 10b
\n" ); document.write( "\"10b%283%29%2F5\" = \"10b%289%29%2F%2810b%29\"
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\n" ); document.write( "After canceling we have:
\n" ); document.write( "2b(3) = 9
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\n" ); document.write( "6b = 9
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\n" ); document.write( "b = 9/6
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\n" ); document.write( "b = 3/2 or 1.5
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\n" ); document.write( "Check, substitute 1.5 for b:
\n" ); document.write( "\"3%2F5\" = \"9%2F%2810%281.5%29%29\"
\n" ); document.write( "\"3%2F5\" = \"9%2F%2815%29\" proves that b = 1.5
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\n" ); document.write( "Hope this helps!
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