Thursday, January 25, 2007

LESSON 8.2

Today is January 25, 2007. Our scribes will be discussing lesson 8.2

3 comments:

Unknown said...

8.2

In lesson 8.2, with the given half-life of a radioactive substance, find the amount of a substance that remains after X hours.

0|6 |12| 18 | 24 | 30 | 36
100|50|25|12.5|6.25|3.13|1.56|

TOP:# of hrs. elapsed
BOTTOM: amount of substance remain.

A=BEGINNING(initial)amount

X/# of hrs. in half-life

Y=AB
Where B=1/2

EXAMPLE:
Find the amount of technetium that remains after 75 hrs.

Y=100(1/2)^(75/6)=O.17

PART 2

Continuously compounded interest problems.
OBJECTIVES:use the nuber "e" to solve problems involving continuous growth.

e=2.7182818284904523...

For continuously compounded interest problems, use this formula

Y=Pe^rt
(Y being the amount of money after time t)

(P the principal, beginning amount of money)

(R=interest rate)

(T=time in years)

EXAMPLE:
suppose you invest $100 at an annual interest rate of 4.8% compounded continuously, how much will you have after 3 years?

Y=Pe^rt
Y=100e^(.048(3))
Y=$115.49


Kyle :)

Mrs. S. said...

Kevin Bailey is also the "SCRIBE" for lesson 8.2. Kevin, you may post your BLOG before Monday, January 29, 2007
:)Mrs. S

Kevin Bailey said...

Well it seems to work fine from school. :/ Well without further ado, I present my notes for lesson 8.2

Lesson 8.2
- Properties of Exponential
Functions -

Objective 1
Comparing Graphs

y=ab^x graphs values greater than zero, when a is less than zero y=ab^x is a reflection of y= |a|b^x over the x axis

Objective 2
The Number e

The number e is the asymptote of y=(1+1/x)^. The number e represented in numerical form is e=2.71828...

This unique number is used to calculate compound interest with the "pert" formula.

A = Pe^rt
Where:
A = Ammount
P = Prinicpal
e = The number e
r = rate
t = time