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?

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.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

ddd

step-by-step explanation: