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Question 363502: i need to figure out the answer what I can up with is wrong.
P =7t^2+7000 with a growth rate of 750 each year. For each of these years 2010,2011,2012,2013,2016
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! Your equation and your problem statement don't really track with each other.
The equation is that of a quadratic equation that has an exponential growth rate each year.
I will graph your equation to show you.
In order to do that, I have to make y equal to P, and x equal to t.
Your equation becomes:
y = 7x^2 + 7000
A graph of this equation would look like this:
It looks like a flat line because the x value is increasing by 1 each time and the result is swallowed up by the 7000.
Change the equation to y = 750x^2 + 7000 and you get:
You can see that the graph of the equation is now exponential.
That's because you have x^2 rather than x.
Change the equation to y = 750*x + 7000 and you get:
That last equation is a straight line.
In fact, it's a straight line with an annual increase of 750 per year.
That's because this is a linear equation in the form of:
y = mx + b, where:
m is the slope, and b is the y-intercept.
The slope is 750 which means that for every change in x, y will change by 750.
The y-intercept is 7000 which means that when x = 0, y = 7000.
To show you that this equation increases by 750 per year, you can simply plot the values of x.
When x = 0, y = 750*0 + 7000 = 7000
When x = 1, y = 750*1 + 7000 = 7750
When x = 2, y = 750*2 + 7000 = 8500
When x = 3, y = 750*3 + 7000 = 9250
etc.
I will plot 3 horizontal lines at 7750, 8500, 9250 to show you.
Simply trace a vertical line down from the intersection of those horizontal lines with the graph of the equation and you will see that x = 0,1,2,3 respectively.
Unless I'm missing something, I don't think you have a valid equation to solve the problem given.
If I'm all wet, then send me the original problem statement in its entirety and I'll look at it and give you my opinion.
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