Stitz-Zeager_College_Algebra_e-book

92 summation notation 921 571 exercises 1 find the

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Unformatted text preview: so let’s call this quantity R, with units garden hour . Using the fact that rates are additive, we have: rate working together = rate of Taylor working + rate of Carl working 1 garden 3 hour so that R = 1 garden 12 hour . = 1 garden 4 hour + R garden hour Substituting this into our ‘work-rate-time’ equation for Carl, we get: 1 garden = (rate of Carl working) · (t hours) 1 garden = Solving 1 = 1 12 t, 1 garden 12 hour · (t hours) we get t = 12, so it takes Carl 12 hours to weed the garden on his own.5 Even though a system of equations wasn’t formally used in Example 8.7.4, the notion of using two variables and two unknowns is there, albeit it more subtly than in Example 8.7.3. As is common with ‘word problems’ like Examples 8.7.3 and 8.7.4, there is no ‘short-cut’ to the answer. We encourage the reader to carefully think through and apply the basic principles of rate to each (potentially different!) situation. It is time well spent. We also encourage keeping track of units, especially in th...
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