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Solutions to Equation Worksheet: Draining Fluid, Distance Traveled, and Walking Speeds - P, Assignments of Algebra

Solutions to various equation-based problems. The first problem deals with draining and replacing fluid in a radiator, the second problem calculates the total distance traveled at different speeds, and the third problem uses the pythagorean theorem to determine walking speeds. Students can use this document as study notes or summaries to help understand the concepts and solve similar problems.

Typology: Assignments

Pre 2010

Uploaded on 07/23/2009

koofers-user-vo0
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Download Solutions to Equation Worksheet: Draining Fluid, Distance Traveled, and Walking Speeds - P and more Assignments Algebra in PDF only on Docsity! Solutions to Equation Application Worksheet 1. This is very similar to Example 5 on page 94 (which you should have read). If x is the amount of fluid we drain and replace with pure antifreeze, we have: (25% of (10โˆ’ x) quarts) + (100% of x quarts) = (40% of 10 quarts) . .25(10โˆ’ x) + x = .4(10) 2.5โˆ’ .25x + x = 4 .75x = 1.5 x = 2. (1) So, Linnea should drain 2 quarts from the radiator and pour in 2 quarts of antifreeze. 2. This is very similar to Example 6 on p. 95. If r is the speed during the first direction, and t is the time spent going the first direction, then using distance = rateร— time, we have: (rate 1st dir.) (time 1st dir.) + (rate 2nd dir.) (time 2nd dir.) = total distance 160 = rt + (r โˆ’ 10)(3.6โˆ’ t) 160 = rtโˆ’ rt + 3.6r + 10tโˆ’ 36 160 = 3.6r + 10tโˆ’ 36 196โˆ’ 3.6r = 10t 19.6โˆ’ .36r = t (2) We also know that rt = 80, so we have: 80 = rt 80 = (19.6โˆ’ .36r)r 80 = 19.6r โˆ’ .36r2 .36r2 โˆ’ 19.6r + 80 = 0 36r2 โˆ’ 1960r + 8000 = 0 (3) Applying the quadratic formula, we get r = 50 or r = 40/9. However, the second solution makes no sense, as it would require the Sarah to drive 40/9โˆ’10 < 0 mph the second direction. So, we have r = 50 mph, and t = 19.6 โˆ’ .36r = 1.6 hours. So, Sarah drives to Portland at 50 mph for 1.6 hours, and back at 40 mph for 2 hours. You can substitute this back into the original equation to check that it is correct. 3. If r is the rate of the person walking down 11th, using the Pythagorean Theorem, we have: 502 = (20r)2 + (20(r โˆ’ .5))2 2500 = 400r2 + 400(r2 โˆ’ r + .25) 2500 = 400r2 + 400r2 โˆ’ 400r + 100 0 = 800r2 โˆ’ 400r โˆ’ 2400 (4) Applying the quadratic formula, we have r = โˆ’1.5 or r = 2. The only answer that makes sense is r = 2. We can substitute back into the original equation to check. So, one person walks down 11th at 2 m/s while the other walks down Willamette at 1.5 m/s. 1
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