Population growth the population p (in thousands) of oriando, florida from 1980 through 2009 can be modeled by p = 788e0.0248t
where t = 0 corresponds to 1980.
(a) what was the population of phoenix in 2009?
(b) in what year will phoenix have a population of 2,000,000?

(a). 1,617,599

(b). In year 2018.

Step-by-step explanation:

We have been given that population P (in thousands) of Phoenix, Arizona from 1980 through 2009 can be modeled by , where, corresponds to 1980.

(a). To find the population of Phoenix in 2009, we will substitute in our given equation as:

Since our given population is in thousands, so we will multiply 1617.5987731991356 by 1,000 as:

Therefore, the population of Phoenix would be 1,617,599 in 2009.

(b). To find the time, when Phoenix have a population of 2,000,000, we can set an equation as:

We will substitute 2,000 as our given equation represents population in thousands.

Let us solve for t.

Now, we will use natural log to solve for t as:

Now, we need to find 38 years after 1980 that is .

Therefore, in 2018 Phoenix will have a population of 2,000,000.

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

good job you smart son of a gun you are smarter that einstein. 7