Question 39353
Problem 1:

Use the table shown below. Assume the data can be represented as a linear relationship.

a) Using the two points defined by data from 2000 and 2005, develop a linear equation for the data.
Points are (2000,19500) and (2005,31000)
Slope if [31000-19500/[2005-2000]=11500/5=2300
The form of the linear equation is y=mx+b where y=31000,x=2005,m=2300
31000=2300(2005)=b
b=-4580500
EQUATION: y=2300x-4580500

b) Using the equation found in a), predict the tax revenue for 2006.
tax=2300(2006)-4580500=$33,300

c) Use the calculator to determine the "best fit" equation for the data.
y=2204.65x-439074.42

d) Using the equation found in c), predict the tax revenue for 2006.
$32,056

e) Using the equation found in c), estimate tax revenue for the missing 2002 data
$23,237

Year Sales Tax Revenue 
2000 $19500
2001 $21000
2003 $24000
2004 $27300
2005 $31000

Note: data is missing for 2002, the county clerk embezzled the funds for that year.


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Problem #2

A rocket is shot from the top of an oceanside cliff that is 60ft high. The original angle is 45 degrees and the original velocity is 50ft per sec. The formula for the rocket's motion is:

y = f(x) = -0.0128x(squared) + x + 60

The variables x and y are both measured in feet. The variable x represents horizontal displacement (distance from the base of the cliff), and y represents the height of the rocket above the ground.

a) What is the maximum height attained by the rocket? How far has the rocket travelled horizontally at this point?
The equation is a quadratic with a=-0.0128, b=1, c=60
The maximum point occurs at x=-b/2a=-1/2(-0.0128)=39.0625, height =79.53.. ft.

b) At waht point(s) (after launch) is the rocket 70 feet high? (Horizontal displacement)
80=-0.0128x^2+x+60
-0.0128x^2+x-10=0
x=11.774611 y=70; and x=66.350389 y=70

   

c) How far away from the base of the cliff does the rocket land?
0 solutions
Let -0.0128x^2+x+60=0
Solve for "x"
x=117.88749 ft

Cheers
Stan H.