algebra 2 slope: in 2003, y = $1.40; in 2004, y = $1.50.

[kerrielynn]

i need to find the slope of a line representing the change from 2003 to 2004 letting the year represent "x" and price represent "y" the price in 2003 is $1.40 and in 2004 it was 1.50
so far i have m=1.50-1.40/2004-2003 so there for m=.1/1 so how do i take this information and find the slope of the line or is m the slope of the line???

 

[mmm4444bot]

Yes, m is the slope.

m = 0.1 is correct.

 

[Deleted member 4993]

Re: slope

i need to find the slope of a line representing the change from 2003 to 2004 letting the year represent "x" and price represent "y" the price in 2003 is $1.40 and in 2004 it was 1.50
so far i have m=1.50-1.40/2004-2003 so there for m=.1/1 so how do i take this information and find the slope of the line or is m the slope of the line???

Please review definition of slope.

You may want to study:

http://www.purplemath.com/modules/slope.htm

 

[mmm4444bot]

DUPLICATE POST



Please use grouping symbols around numerators and/or denominators that consist of more than one term.

m=1.50-1.40/2004-2003 should be typed as m = (1.50 - 1.40)/(2004 - 2003)

 

[kerrielynn]

so i posted a question last night and this is the second part, and i just need to know how i go about setting this problem up because im not sues if i have enough information. the firs part i had to find the slope representing the change from 2003 to 2004 letting the year represent "x" and the price represent "y".in 2003 the price was $1.40 and in 2004 it was 1.50
i found that the slope equals .1 because:
m = (y2-y1)/(x2-x1)

x1 = 2003
y1 = 1.40

x2 = 2004
y2 = 1.50

m=.1
now the second part of the question is: Now that you know the slope of the line between these years can you write the equation of the line representing the change in population between these years?? i know that the equation is y=mx+b but how do i put the information in to represent the change in population when i am not given and numbers for the population? please help me understand how to set it up so i can solve this problem.

 

[Deleted member 4993]

Re: second part of slope question

so i posted a question last night and this is the second part, and i just need to know how i go about setting this problem up because im not sues if i have enough information. the firs part i had to find the slope representing the change from 2003 to 2004 letting the year represent "x" and the price represent "y".in 2003 the price was $1.40 and in 2004 it was 1.50
i found that the slope equals .1 because:
m = (y2-y1)/(x2-x1)

x1 = 2003
y1 = 1.40

x2 = 2004
y2 = 1.50

m=.1
now the second part of the question is: Now that you know the slope of the line between these years can you write the equation of the line representing the change in population between these years?? i know that the equation is y=mx+b but how do i put the information in to represent the change in population when i am not given and numbers for the population? please help me understand how to set it up so i can solve this problem.

If you know two points on a straight-line - \(\displaystyle (x_1,y_1)\) and \(\displaystyle (x_2,y_2)\) - then the equation of the line is:

\(\displaystyle \frac{y-y_1}{y_2-y_1} \, = \, \frac{x-x_1}{x_2-x_1}\)

or

since you know 'm' - the slope you can use equations discussed here:

http://www.purplemath.com/modules/strtlneq2.htm

 

[soroban]

Re: second part of slope question

Hello, kerrielynn!

It seems that you have never formed the equation of a line, given two points.
You must have done this at least once before encountering a problem like this.


\(\displaystyle \text{The two points are: }\;(2003,1.40)\text{ and }(2004,1.50)\)

\(\displaystyle \text{You found that the slope is: }\;m \:=\:\frac{1.50-1.40}{2004-2003} \:=\:0.1\quad\hdots\quad \text{Good!}\)



Now there are several ways to find the equation . . . Here are two of them.



\(\displaystyle \text{[1] You know the form: }y \:=\:mx+b\)

. . .\(\displaystyle \text{We already know that }m = 0.1\text{, so we have: }\;y \:=\:0.1x + b\)

\(\displaystyle \text{Since }(2003,1.40)\text{ is on the line, we have: }x = 2003,\:y = 1.40\)

. . .\(\displaystyle \text{Substitute: }\;1.40 \:=\:0.1(2003) + b \quad\Rightarrow\quad 1.40 + 20.3 + b \quad\Rightarrow\quad b \:=\:-198.9\)

\(\displaystyle \text{Therefore, the equation is: }\;y \:=\:0.1x - 198.9\)



\(\displaystyle \text{[2] We can use the Point-Slope Formula: }\;y - y_1 \:=\:m(x-x_1)\)

. . .\(\displaystyle \text{and use point }(2003,1.40)\text{ and slope }0.1\)

\(\displaystyle \text{Then we have: }\;y - 1.40 \:=\:0.1(x-2003) \quad\Rightarrow\quad y \:=\:0.1x - 198.9\)