Alg_Complete

# In fact this equation is factorable so the solution

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Unformatted text preview: ours) so we need to work rates of each machine. Here are those computations. 1 Job = ( Work Rate of Machine A ) ´ ( t ) 1 Job = ( Work Rate of Machine B ) ´ ( t + 1) Þ Þ 1 t 1 Machine B = t +1 Machine A = Note that it’s okay that the work rates contain t. In fact they will need to so we can solve for it! Plugging into the word equation gives, © 2007 Paul Dawkins 109 http://tutorial.math.lamar.edu/terms.aspx College Algebra æ1ö æ1ö ç ÷ ( 2) + ç ÷ ( 2) = 1 èt ø è t +1 ø 2 2 + =1 t t +1 So, to solve we’ll first need to clear denominators and get the equation in standard form. 2ö æ2 ç+ ÷ ( t )( t + 1) = (1)( t )( t + 1) è t t +1ø 2 ( t + 1) + 2t = t 2 + t 4t + 2 = t 2 + t 0 = t 2 - 3t - 2 Using the quadratic formula gives, t= 3 ± 17 2 Converting to decimals gives, t= 3 + 17 = 3.5616 2 and t= 3 - 17 = -0.5616 2 Again, the negative doesn’t make any sense and so Machine A will work for 3.5616 hours to stuff a batch of envelopes by itself. Machine B will need 4.5616 hours to stuff a batch of envelopes by itself. Again, unlike the first example, note that the time for Machine B was NOT the second solution from...
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## This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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