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babunello [35]
2 years ago
13

A town's population is 14,280, and it is increasing by 160 people every year. A nearby town has a population of 24,000, and it i

s decreasing by 200 every year. In about how many years will the populations of the towns be equal. (please just write the number)
Mathematics
1 answer:
fgiga [73]2 years ago
7 0

Answer:

27 years

Step-by-step explanation:

Given that :

Town A:

Initial population = 14280

Rate = increase by 160 per year

Let t = time in years

Population is thus :

14280 + 160t - - - (1)

Town B:

Initial population = 24000

Rate = decrease by 200 per year

P = 24000 - 200t - - - (2)

Equate both equations :

14280 + 160t = 24000 - 200t

160t + 200t = 24000 - 14280

360t = 9720

t = 9720 / 360

t = 27

In 27 years

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Troyanec [42]

Answer:

 z= \frac{pd-59}{-2+p}

Step-by-step explanation:

-p(d+z)=-2z+59

-pd-pz=-2z+59

lets bring all z integers on one side

-2z+pz = -pd-59

z(-2+p)=-pd-59

z = (pd-59) / (-2+p)

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For the function defined by f(t)=2-t, 0≤t&lt;1, sketch 3 periods and find:
Oksi-84 [34.3K]
The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
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The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
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Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

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f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

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20? I’m not for sure maybe look in the book? That’s the answer I got and it was right
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Look at the link below I filled in the answers for you. :)

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