Question

Math help

Answer 1

@Zepdrix

Answer 2

Let's use the year 2000 as our starting point to model our exponential function.

Answer 3

So year 2000 is like year 0 for our function.
When x=0, the population=571044

Answer 4

\[\large\rm f(x)=a\cdot b^x\]\[\large\rm f(0)=571044=a\cdot b^0\]\[\large\rm 571044=a\]

So that gives us part of the information we need, our starting coefficient for the exponential.

Answer 5

Errr wait wait wait, is that what they want us to do? :d sec thinking

Answer 6

Ya ya ya, then we'll use another data point to find an appopriate base value for our exponential.

Answer 7

\[\large\rm f(x)=571044\cdot b^x\]
In 2005, think of this as 5 years after our starting point,
x=5, and population=550521\[\large\rm f(5)=550521=571044\cdot b^5\]

Answer 8

Let's divide our exponential's coefficient to the other side,\[\large\rm \frac{550521}{571044}=b^5\]

Answer 9

To undo a 5th power, we'll take a 5th root.\[\large\rm \sqrt[5]{\frac{550521}{571044}}=\sqrt[5]{b^5}\]

Answer 10

Using a calculator gives us a value of abouttttttt ...
0.992706595

Answer 11

So \(\rm b\approx 0.993\)

Answer 12

\[\large\rm f(x)=571044(0.993)^x\]

Answer 13

a ∙ bx = (550,521)(0.992)x


a ∙ bx = (550,521)(0.990)x


a ∙ bx = (550,521)(0.991)x


a ∙ bx = (550,521)(0.993)x

Answer 14

Something like that. Assuming I didn't mess up a number somewhere along the way x'D

Answer 15

Oh you have options, nice

Answer 16

Oh they used 2005 as the starting point, ok that's weird x'D

Answer 17

Wish you had given the options before all that math .... -_-

Answer 18

sorry

Answer 19

I didn't realize you'd need the options for this

Answer 20

Sec thinking :d
They're using 2005 as the starting point,
so all the other values would be negative in relation... hmm seems weird

Answer 21

Using the new numbers I'm coming up with,
b=0.992706495

maybe that's the same value as before <.< i dunno lol

f(x) = 550521 * (0.993)^x

Answer 22

Ya ya I think the last option is the one we want :O