ALGEBRA 1 Lesson 2-4 Warm-Up. ALGEBRA 1 “Ratios and Proportions” (2-4) What is a “ratio”? What is a “rate”? What is a “unit rate”? ratio: a comparison.

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Presentation transcript:

ALGEBRA 1 Lesson 2-4 Warm-Up

ALGEBRA 1

“Ratios and Proportions” (2-4) What is a “ratio”? What is a “rate”? What is a “unit rate”? ratio: a comparison of two numbers by division – There are three ways to compare two numbers: a to b a : b rate: a ratio that compares two quantities measured in different units. Example: is read as “210 miles ‘per’ or ‘each’ 3 hours” unit rate: the rate in which the denominator is 1 (rate ‘per’ 1 unit of a given quantity) - Unit rate is used in everyday life to determine the best value. It can be found by simplifying the fraction to a denominator of “1” or simply dividing the numerator (top number) by the denominator (bottom number). Example: = or 210  3 = 70 mph The above unit rate is read as “70 miles ‘per’ hour” (70 m/h or 70 mph) a b 210 miles 3 hours 210 miles 3 hours  3 70 miles 1 hours

ALGEBRA 1 A brand of apple juice costs $1.56 for 48 oz. Find the unit rate. The unit rate is 3.25¢ / oz. cost$1.56 ounces48 oz = 1.56  48 = = $0.325 / oz. or 3.25 ¢ per oz. Ratio and Proportion LESSON 2-4 Additional Examples

ALGEBRA 1 Words :Tour de France average speed equals295-km trip average speed In 2000, Lance Armstrong completed the 3630-km Tour de France course in 92.5 hours. Traveling at his average speed, how long would it take Lance Armstrong to ride 295 km? t 3630 t = Write cross products. t 7.5 Simplify. Round to the nearest tenth. Traveling at his average speed, it would take Lance approximately 7.5 hours to cycle 295 km. = Divide each side by Define:Let t = time needed to ride 295 km kilometers hours Equations: = 295 t Ratio and Proportion LESSON 2-4 Additional Examples 3630t 3630 known unknown km. hr. =

ALGEBRA 1 “Ratios and Proportions” (2-4) What is a “unit analysis”? How do you use dimensional analysis to solve conversion problems in which you need to change the units in a rate? unit analysis (also called “dimensional analysis”): the process of changing units using conversion factors (unit rates in the form of fractions that include both the rate in the problem and the desired rate you want to “convert”, or change, to) To use dimensional analysis, start with the units. Multiply the given rate by conversion fractions in which the “old” units cancel out (one is on top of the fraction bar and the other is on the bottom) so that only the “new” units are left (Example: To change the numerator unit, multiply by a conversion factor in which the new units are on top and the old units are on the bottom. To change the denominator unit, multiply by a conversion factor in which the new units are on the bottom and the old units are on the top ). Example: To change hours to minutes or minutes to hours, you can multiply the number of hours by the “conversion fraction” (in red) as in: 3 hours = x = = 180 minutes 300 minutes = x = 5 hours 3 hours 1 60 minutes 1 hour 180 minutes minutes 1 1 hour 60 minutes

ALGEBRA 1 The fastest recorded speed for an eastern gray kangaroo is 40 mi per hour. What is the kangaroo’s speed in feet per second? The kangaroo’s speed is about 58.6 ft/s. 40 mi. 1 h 40 mi 1 h 5280 ft 1 mi 1 h 60 min Ratio and Proportion LESSON 2-4 Additional Examples Use appropriate conversion factors 5280 ft 1 mi 40 mi 1 h 5280 ft 1 mi. 1 h 60 min. 1 min 60 sec = 211,200 ft 3,600 sec = 58.6 ft 1 sec   3,600 make into a unit rate by making the denominator 1

ALGEBRA 1 Solve=. y3y y = 9Simplify. y = 2.25Simplify. 4y44y4 = 9494 Divide each side by 4. y3y3 12 = 3434 Multiply each side by the least common multiple of 3 and 4, which is Ratio and Proportion LESSON 2-4 Additional Examples

ALGEBRA 1 “Ratios and Proportions” (2-4) What is a “proportion”? What is the “extremes of the proportion”? What is the “means of the proportion”? What are “cross products”? Why do “cross products” work? proportion: equal ratios (in other words, equal fractions) Example: extremes of the proportion: the first cross product of a proportion. In the above proportion, the “extremes of the proportion” are a and d. means of the proportion: the second cross product of a proportion. In the above proportion, the “extremes of the proportion” are b and c. cross products: the product of the means equals the product of the extremes (in the above example, ad = bc). Example: Proof That Cross Products Work For Equal Fractions Work: = abab c d for b ≠ 0 and d ≠ 0

ALGEBRA 1 Use cross products to solve the proportion = –. w w 4.5 = – 6565 w(5) = (4.5)(–6)Write cross products. 5w = –27 Simplify. 5w55w5 = –27 5 Divide each side by 5. w = –5.4 Simplify. Ratio and Proportion LESSON 2-4 Additional Examples = –6 5–6 5 Replace w with Check: -1.2 = –1.2  True Statement (Both sides equal one another.)

ALGEBRA 1 Solve the proportion =. z z – 4 6 z z – 4 6 = (z + 3)(6) = 4(z – 4)Write cross products. 6z + 18 = 4z – 16Use the Distributive Property. 2z + 18 = – Subtract 18 from each side. z = –17Simplify. Divide each side by 2. = 2z22z2 –34 2 Ratio and Proportion LESSON 2-4 Additional Examples Check: = Replace w with = –3.5  True Statement (Both sides equal one another.) 2z = z Subtract 4z from each side.

ALGEBRA 1 Solve. 1.Find the unit rate of a 12-oz bottle of orange juice that sells for $ If you are driving 65 mi/h, how many feet per second are you driving? Solve each proportion c6c = y7y = 3 + x = 2 + x x – = 10.75¢/oz. about 95.3 ft/s –17 Ratio and Proportion LESSON 2-4 Lesson Quiz