Monday december 2 nd
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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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Monday december 2 nd

Monday, December 2nd

Warm Up

Review for Final:

What is the variable(s) in the expression?

What is the constant in the expression?


Monday december 2 nd

Grade

Check


Grades left the semester

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)


Weekly review

Weekly Review

  • We will have three weekly Reviews

  • Each will count as a Daily Grade

  • They are eligible to replace QUIZ grades 


High school gpa

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.


Tutoring

Tutoring

Option 1: Ms. Evans

Tuesday and Thursdays

Before School 6:40-7:00

After School 2:10-2:30


Tutoring1

Tutoring

Option 2: Lunch

Tuesday and Thursdays

Either lunch

Go to room 400 FIRST, then to lunch


Tutoring2

Tutoring

Option 3: Algebra Department

*Check Schedule in Back


Discussion question

Discussion Question

What’s the difference between exponential growth and exponential decay equations?!


Growth decay in graph

Growth & Decay in graph

Decay

Growth


Monday december 2 nd

Growth & Decay in Equation

Decay

Growth


Growth and decay in table

Growth and Decay in Table


Monday december 2 nd

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).


Monday december 2 nd

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.


Monday december 2 nd

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.


Monday december 2 nd

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.


Monday december 2 nd

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.


Monday december 2 nd

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.


Monday december 2 nd

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.


Monday december 2 nd

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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