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Monday, December 2 nd

Monday, December 2 nd. Warm Up. Review for Final: What is the variable(s) in the expression? What is the constant in the expression? . Grade Check . Grades Left the Semester. 1 more quiz 1 more Warm up(Daily grade) Exponential Test (Test Grade) Semester I Final (Final grade)

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Monday, December 2 nd

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  1. Monday, December 2nd Warm Up Review for Final: What is the variable(s) in the expression? What is the constant in the expression?

  2. Grade Check

  3. Grades Left the Semester • 1 more quiz • 1 more Warm up(Daily grade) • Exponential Test (Test Grade) • Semester I Final (Final grade) • 3 Weekly Reviews-(daily Grade)

  4. Weekly Review • We will have three weekly Reviews • Each will count as a Daily Grade • They are eligible to replace QUIZ grades 

  5. High School GPA A: 4.0 B: 3.0 C. 2.0 D. 1.0 F-Receive no Credit. You will have to retake first semester all over again during semester II.

  6. Tutoring Option 1: Ms. Evans Tuesday and Thursdays Before School 6:40-7:00 After School 2:10-2:30

  7. Tutoring Option 2: Lunch Tuesday and Thursdays Either lunch Go to room 400 FIRST, then to lunch

  8. Tutoring Option 3: Algebra Department *Check Schedule in Back

  9. Discussion Question What’s the difference between exponential growth and exponential decay equations?!

  10. Growth & Decay in graph Decay Growth

  11. Growth & Decay in Equation Decay Growth

  12. Growth and Decay in Table

  13. Reading Math • For compound interest • annually means “once per year” (n = 1). • quarterly means “4 times per year” (n =4). • monthly means “12 times per year” (n = 12).

  14. Example #1 Write a compound interest function to model each situation. Then find the balance after the given number of years. $1200 invested at a rate of 2% compounded quarterly; 3 years. Step 1 Write the compound interest function for this situation. Write the formula. Substitute 1200 for P, 0.02 for r, and 4 for n. = 1200(1.005)4t Simplify.

  15. Step 2 Find the balance after 3 years. A = 1200(1.005)4(3) Substitute 3 for t. = 1200(1.005)12 Use a calculator and round to the nearest hundredth. ≈ 1274.01 The balance after 3 years is $1,274.01.

  16. Example #2 Write a compound interest function to model each situation. Then find the balance after the given number of years. $15,000 invested at a rate of 4.8% compounded monthly; 2 years. Step 1 Write the compound interest function for this situation. Write the formula. Substitute 15,000 for P, 0.048 for r, and 12 for n. = 15,000(1.004)12t Simplify.

  17. Step 2 Find the balance after 2 years. Substitute 2 for t. A = 15,000(1.004)12(2) Use a calculator and round to the nearest hundredth. = 15,000(1.004)24 ≈ 16,508.22 The balance after 2 years is $16,508.22.

  18. Example #3 Write a compound interest function to model each situation. Then find the balance after the given number of years. $1200 invested at a rate of 3.5% compounded quarterly; 4 years Step 1 Write the compound interest function for this situation. Write the formula. Substitute 1,200 for P, 0.035 for r, and 4 for n. = 1,200(1.00875)4t Simplify.

  19. Step 2 Find the balance after 4 years. A = 1200(1.00875)4(4) Substitute 4 for t. = 1200(1.00875)16 Use a calculator and round to the nearest hundredth.  1379.49 The balance after 4 years is $1,379.49.

  20. Lesson Summary Write a compound interest function to model each situation. Then find the balance after the given number of years. 1. The number of employees at a certain company is 1440 and is increasing at a rate of 1.5% per year. Write an exponential growth function to model this situation. Then find the number of employees in the company after 9 years. y = 1440(1.015)t; 1646 2. $12,000 invested at a rate of 6% compounded quarterly; 15 years A = 12,000(1 + .06/4)4t, =$29,318.64

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