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on the graph I have, there is a point on 3000 $'s and the year 1.
There is another point on 32000 $'s and the year 2. The last point is on 34000 $'s and the year 3.
Does this help?

2007-07-19 03:28:14 · 2 answers · asked by yuleydis77 1 in Science & Mathematics Mathematics

2 answers

Are you trying to find an equation? If the first point is actually $30000, you have an increase of $2000 every year. That means that $2000 per year is the slope. The y-intercept would be the value at year 0, which is not meaningful in the real world, but still makes sense mathematically. If his salary increases $2000 per year, his salary in year 0 would be $2000 less than his salary in year 1, so $28000. That means the equation in slope-intercept form would be y = 2000x + 28000, for integers x > 0.

Looking at your previous question, I see that you also want the salary in 2005. 2005 would be year 5, so just plug in 5 for x in the above equation and solve for y.

2007-07-19 03:33:04 · answer #1 · answered by DavidK93 7 · 0 0

a) List the coordinates of two points on the graph in (x,y) form.
(1,30000)
(2,32000)
(3,34000) this is a thrid point, extra point.

b) Find the slope of this line:
slope m = (32000 - 30000)/(2 - 1) = 2000

c) Find the equation of this line in slope-intercept form.
this is the slope-intercept form: y = mx + b
we know slope m = 2000 (from b), so
y = 2000x + b

Plug in this point (x,y) = (1,30000)
30000 = (2000 * 1) + b
so, b = 28000
y = 2000x + 28000

d) If bob's salary trend continued, what would be salarybe in the year 2005?
the year 2005 means , x= 5
so, y = 2000x + 28000
y = (2000) * 5 + 28000
y = 10000 + 28000 = 38,000
it'll be 38,0000

2007-07-19 12:45:58 · answer #2 · answered by buoisang 4 · 0 0

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