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BELL-WORK

BELL-WORK. TB pg 371 #12. CW 2.4 Review. TB pg 353 # 13-14,17-18 TB pg 363 # 16 TB pg 371 # 11. HW 2.4(h). Due Monday: Practice Quiz 2.3. HW 2.4(g) Solutions. One; (-1,1) One; (3,1) Infinitely many solutions One; (-3,1) 31.One; (-2,-3) 33.One; (-2,-1). Extra Credit Opportunity.

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BELL-WORK

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  1. BELL-WORK TB pg 371 #12

  2. CW 2.4 Review TB pg 353 # 13-14,17-18 TB pg 363 # 16 TB pg 371 # 11

  3. HW 2.4(h) Due Monday: Practice Quiz 2.3

  4. HW 2.4(g) Solutions • One; (-1,1) • One; (3,1) • Infinitely many solutions • One; (-3,1) 31.One; (-2,-3) 33.One; (-2,-1)

  5. Extra Credit Opportunity We will not discuss the third method of solving systems in class. However, if you read unit 6-3 and correctly complete PW 6-3 #1-9 using elimination, you will earn extra credit points! Due Monday, on a separate sheet of notebook paper.

  6. Guiding question: What are some real-world applications of linear systems?

  7. Real-world application of systems Suppose you plan to start taking aerobics class. One gym charges a $4 membership fee and $2 per class while another gym charges a $5 membership fee and $1 per class. After how many classes will the cost be the same? What is that cost? Solution: Write an expression for the cost at gym 1: y = 4 + 2x Write an expression for the cost at gym 2: y = 5 + x Solve the system! (1,6) Cost will be the same after 1 class with a cost of $6 Complete Got It? 2 on page 361

  8. Real-world application of systems Tyrel and Dalia bought some pens and pencils. Tyrel bought 4 pens and 5 pencils, which cost him $6.71. Dalia bought 5 pens and 3 pencils, which cost her $7.12. What is the price of a pen? What is the price of a pencil? Suppose you have just enough money, in coins, to pay for a loaf of bread priced at $1.95. You have 12 coins, all quarters and dimes. How many quarters do you have? How many dimes do you have?

  9. Applications of linear systems:Break-even problems In the world of business/economics, what is meant by the break-even point? Break-even point = the point at which income equals expenses. Why is the break even point important? Because it tells business people whether or not they will make a profit (if they make above the break-even point) or suffer a loss (if they make below the break-even point) in their business. We can solve a system of linear equations to find the break-even point.

  10. Applications of linear systems:Break-even problems Suppose you have a typing service. You buy a personal computer for $1750 on which to do your typing. You charge $5.50 per page for typing. Expenses are $0.50 per page for ink, paper, electricity, and other expenses. How many pages must you type to break even? Let x = number of pages typed, we can form the system: y = 5.50x y = 1750 + 0.50x Solving the system will give the break even point 5.50x = 1750 + 0.50x 5x = 1750 x = 350 pages

  11. Applications of linear systems:Break-even problems Suppose an antique car club publishes a newsletter. Expenses are $0.35 for printing and mailing each copy, plus $770 total for research and writing. The price of the newsletter is $0.55 per copy. How many copies must the club sell to break even? Let x = number of newsletters y = 0.55x y = 770 + 0.35x At the break even point 0.55x = 770 + 0.35x 0.2x = 770 x = 3850 Complete TB pg 386# 1

  12. Applications of linear systems:Mixture problems A chemist has one solution that is 50% acid. She has another solution that is 25% acid. How many liters of each type of acid solution should she combine to get 10 liters of a 40% acid solution? If x = number of liters of 50% solution y = number of liters of 25% solution then we can write: x + y = 10 0.50x + 0.25y = (0.4)10 and now we simply solve the system: x + y = 10 0.5x + 0.25y = 4 6 liters of 50% and 4 liters of 25%

  13. Applications of linear systems:Mixture problems Suppose you combine ingots of 25% copper alloy and 50% copper alloy to create 40kg of 45% copper alloy. How many kilograms of each do you need? If x = number of kgs of 25% alloy y = number of kgs of 50% alloy then we can write: x + y = 40 0.25x + 0.50y = (0.45)40 and now we simply solve the system: x + y = 40 0.25x + 0.50y = 18 8 kg of the 25% and 32 kg of the 50% Complete TB pg 386 # 2

  14. Applications of linear systems:Resistance problems In various instances of travel, resistance plays a vital role. Examples: flying, going up/down stream, etc. If you fly with a tail wind, Ground speedtailwind = air speed + wind speed G = A + W If you fly into a head wind, Ground speedheadwind = air speed – wind speed G = A – W We can use these equations to write and solve systems!

  15. Applications of linear systems:Resistance problems A traveler flies from Charlotte, NC, to Los Angeles, CA. At the same time, another traveler flies from Los Angeles to Charlotte. The air speed of each plane is the same. The ground speed of the plane to Charlotte is 550 mph and to LA is 495 mph. What is the air speed? What is the wind speed? Which direction is with the tail wind? So we can write the system: 550 = A + W 495 = A – W Solve the system! (Round to the nearest tenth if necessary) A = 522.5 mph W = 27.5 mph TB pg 386 # 3

  16. Who wants to answer the Guiding question? What are some real-world applications of linear systems?

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