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9 Cards in this Set
- Front
- Back
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Using a chain of Conversions
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1 day X (24 hr/1 day) X (60 min/1 hr) X (60s/1 min)= 86,400 s
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Feet ---> Yards
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120 ft^2 X (1 yd^2/ 9 ft^2)= 120/9 yd^2
=13.3 yd^2 |
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Price Conversion
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1 euro= $1.272
1 euro/ $1.272=1 45 euro X ($1.272/1 euro)= $57.24 |
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Buying Currency
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$1= 10.95 pesos
(10.95 peso/$1)=1 $100 X (10.95 peso/$1)= 1095 peso |
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Distance, Time, and Speed
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A car is traveling 25 miles every half-hour. How is it going?
25 mi. / .5 hr = 25 mi. X (2/1 hr) = 50 mi/hr |
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Jill and Jack's Race
- Jill and Jack ran a 100m. race. Jill won by 5m; Jack had run only 95m when Jill finished. They decide to race again, but this time Jill starts 5m. behind the starting line. Assuming that both runners run at the same pace as before, who will win? |
Jill: 105m./(5m/s)= 105 X (1/5)= 21 s.
Jack: 100m/(4.75m/s)= 100 X (1/4.75) = 21.05s. JILL IS THE WINNER |
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The Cars and the Canary
-Two cars, 120 miles apart, begin driving toward each other on a long straight highway. One car travels 20 mph and the other 40 mph. At the same time, a canary, starting on one car, flies back and forth between the two cars as they approach each other. If the canary always flies 150 mph and turns around instantly at each car, how far has it flown when the cars collide? |
120mi / (60mi/hr)= 120X(1/60)=2hr
-The canary is flying at a speed of 150 mph, so in 2 hrs it will fly 2 hr X (150mi/hr)= 300 mi. |
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Party Decorations
Juan is decorating for a party in a room that has ten large cylindrical posts. The posts are 8 ft. high and have a circumference of 6 ft. His plan is to wrap eight turns of ribbon around each post. How much ribbon does Juan need? |
hypotenuse=
sq.root of (6ft)^2 + (1ft)^2= 6.1 ft -Thus the length of each ribbon segment is 6.1ft, and the total length of the ribbon is 8X6.1ft= 48.8 ft. To decorate 10 posts, Juan needs 10X48.8 ft= 488 ft of ribbon. |
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A Bowed Rail
- Imagine a mile-long bar of metal such as a railroad rail. Suppose that the rail was anchored on both ends (a mile apart) and that, on a hot day, its length expanded by 1 ft. If the added length caused the rail to bow upward in a circular arc, about how high would the center of the rail rise above the ground? |
2 right triangles and the Pythagorean theorem applies. The bases of the two right triangles together give the original rail length of 1 mile, so each base is 1/2mi long. The two hypotenuses together represent the expanded the length of 1mi +1ft, so each hypotenuse is 1/2mi+1/2ft long. Because there are 3280ft in a mile, 1/2mi= 2460ft. The height of the rail off the ground is approx. the height of the triangles:
height of triangle= sq. root of (2640.5ft)^2 - (2640ft)^2= 51.4 ft |
