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3 years ago

PLEASE HELP

Mathematics

1 answer:

DENIUS [597]3 years ago

6 0

Answer:

Assume that a certain population is P. If the population has decreased 2.8% at the end of one year, this means that the population now is

100%P-2.8%P=97.2%P. Thus a decrease of 2.8% can means a multiplication by 97.2%, which is 97.2/100=0.972.

So, assume the population in 2006 was P. Then the population in 2007 was

0.972P. Similarly, we can see that the population in 2008 was 0.972(0.972P).

The population in 2009 the population was 0.972(0.972(0.972P)), which is .

Finally, in the end of 2010 the population was .

The decline of the population from 2006 to 2010 was 22,000. This means that

. Factorizing P we have:

. We can calculate , thus

We saw that the population after the n'th year after 2006 was .

2040 is the (2040-2006)=34th year after 2006, so the population in 2040 will be

(to calculate : 0.972_*_=*_=*_=*_=*_=, one *_= for each 2)

Thus, the population in 2040 was 0.378*204,842=77,430.

Hope this helps you out . It's alot to read though

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2 years ago

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Answer:

Please check the explanation

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:\left(x-5\right)^3-1$

Given that the output = -3

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now substituting the value y=-3 and solve for x to determine the input 'x'

$\:\:y=\:\left(x-5\right)^3-1$

$-3\:=\:\left(x-5\right)^3-1\:\:\:$

switch sides

$\left(x-5\right)^3-1=-3$

Add 1 to both sides

$\left(x-5\right)^3-1+1=-3+1$

$\left(x-5\right)^3=-2$

$\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

Thus, the input values are:

$x=-\sqrt[3]{2}+5,\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}-i\frac{\sqrt[3]{2}\sqrt{3}}{2},\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}+i\frac{\sqrt[3]{2}\sqrt{3}}{2}$

And the real input is:

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2 years ago