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A student looking at global population figures remarks that between the years 1950 and 2010 the global population can be modelled by the equation 300y= 23t−44 180 (1950≤t≤2010),

Write down the gradient of the straight line represented by the equation 300y= 23t−44 180 (1950≤t≤2010)

What does this represent in the practical situation being modelled?

Respuesta :

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Answer:

The equation is (i guess):

300*y = 23*t  -41,180

This is a linear equation, where y represents the population, and t represents time (in years)

Now, to find the gradient, we must derive the function y with respect to the time.

then we can write:

y = (23/300)*t - (41,180)/300.

now, remember that when we have:

y = a*x^n

we also have:

dy/dx = n*a*x^(n-1)

then in this case, we have:

dy/dt = (23/300) + 0

now we pass aain the 300 to the right:

300*dy/dt = 23.

d(300*y)/dt = 23

what does this mean?

Thie mean that for each unit of that t increases, we have an increase of 23 units in 300*y.

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