![Picture](/uploads/2/3/5/1/23513330/1389024943.jpg)
Unit One: Real World Problem
Example: The wold coffee consumption from 1990 to 2000 can be modeled by f(x)=-.004x^4+.077x^3 - .38x^2 + .46x+12, where x is the year, x=0 corresponds with 1990, and the consumption is measured in millions of pounds. Find the average rate of change for each time interval. a) 1990-200 b) 1995-2000
a. f(0) = 12 and f(10) = 15.6
m=15.6 -12/ 10-0 .......RATE OF CHANGE
=3.6/10 or .36 million pounds more per year
b. f(5) = 11.925 and f(10) = 15.6
m= 15.6 -11.925/10-5.............RATE OF CHANGE
=3.675/5 or .735 million pounds more per year
Unit Two: Real World Problem
Example: The water level in a bucket sitting on a patio can be modeled by the f(x)= x^3 + 4x^2 - 2x = 7, where f(x) is the height of the water in millimeters and x is the time in days. On what day will the water reach a height of 10 millimeters?
10 = f(x) so...... 10 = x^3 + 4x^2 - 2x +7
get everything on one side which leaves you with.......x^3 + 4x^2 - 2x - 3 = 0
use your p/q formula....so p = plus or minus 1,3
q = plus or minus 1
so your final answer is p/q = +- 1 and 3
Unit Three: Real World Problem
Example: Financial Literacy....200 is being invested. The account has 5% interest rate. If isn't touched what will the balance be in 18 years if the interest is compounded quarterly?
Use the compound interest formula: A= P ( 1 + r/n) ^nt
so: A = 2000(1+ .05/4) ^72
work through the problem normally using the well know order of operations.
you then get the final answer: $4891.84
Unit Four: Real World Problem
Example: A hot air balloon that is moving above a neighborhood has an angle of depression of 20 degrees to one house and 52 degrees to a house down the street. If the height of the balloon is 650 feet, estimate the distance between the two houses.
The first step you can do is sketch yourself a triangle to better understand the problem.
Example: The wold coffee consumption from 1990 to 2000 can be modeled by f(x)=-.004x^4+.077x^3 - .38x^2 + .46x+12, where x is the year, x=0 corresponds with 1990, and the consumption is measured in millions of pounds. Find the average rate of change for each time interval. a) 1990-200 b) 1995-2000
a. f(0) = 12 and f(10) = 15.6
m=15.6 -12/ 10-0 .......RATE OF CHANGE
=3.6/10 or .36 million pounds more per year
b. f(5) = 11.925 and f(10) = 15.6
m= 15.6 -11.925/10-5.............RATE OF CHANGE
=3.675/5 or .735 million pounds more per year
Unit Two: Real World Problem
Example: The water level in a bucket sitting on a patio can be modeled by the f(x)= x^3 + 4x^2 - 2x = 7, where f(x) is the height of the water in millimeters and x is the time in days. On what day will the water reach a height of 10 millimeters?
10 = f(x) so...... 10 = x^3 + 4x^2 - 2x +7
get everything on one side which leaves you with.......x^3 + 4x^2 - 2x - 3 = 0
use your p/q formula....so p = plus or minus 1,3
q = plus or minus 1
so your final answer is p/q = +- 1 and 3
Unit Three: Real World Problem
Example: Financial Literacy....200 is being invested. The account has 5% interest rate. If isn't touched what will the balance be in 18 years if the interest is compounded quarterly?
Use the compound interest formula: A= P ( 1 + r/n) ^nt
so: A = 2000(1+ .05/4) ^72
work through the problem normally using the well know order of operations.
you then get the final answer: $4891.84
Unit Four: Real World Problem
Example: A hot air balloon that is moving above a neighborhood has an angle of depression of 20 degrees to one house and 52 degrees to a house down the street. If the height of the balloon is 650 feet, estimate the distance between the two houses.
The first step you can do is sketch yourself a triangle to better understand the problem.
Your final answer is 714.6 feet apart.
Unit Five: Real World Problems
Example: When Light shines through two narrow slits, a series of light and dark fringes appear. The angle x, in radians, locating the mth fringe can be calculated by sinx= m Y/d, where d is the distance between the tow slits, and Y is the wavelength. Rewrite the formual in terms of cscx.
So this is a simple problem. csc is the opposite of sine so all you have to do is flip the formula.
your final answer is.... d/mY...:)
Unit Five: Real World Problems
Example: When Light shines through two narrow slits, a series of light and dark fringes appear. The angle x, in radians, locating the mth fringe can be calculated by sinx= m Y/d, where d is the distance between the tow slits, and Y is the wavelength. Rewrite the formual in terms of cscx.
So this is a simple problem. csc is the opposite of sine so all you have to do is flip the formula.
your final answer is.... d/mY...:)