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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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Step-by-step explanation:

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Let's find the length between each points to see which is equal to 6 units, using the distance formula below.

$distance = \sqrt{(y1 - y2)^{2} + {(x1 - x2)}^{2} }$

A. Distance between 2 points

$= \sqrt{ {(10 + 5)}^{2} + ( - 2 + 6)^{2} } \\ = \sqrt{15 ^{2} + {4}^{2}} \\ = \sqrt{241} \\ = 15.524 \: units$

(to 5 s.f.)

Since the distance is greater than 6 units, point A lies outside the circle.

B. Distance between 2 points

$= \sqrt{ {(10 - 4)}^{2} + ( - 2 + 2)^{2} } \\ = \sqrt{ {6}^{2} } \\ = 6 \: units$

Since the distance of B from the center is 6 units, point B lies on Circle O.

Let's check the other 2 points:

C. Distance between 2 points

$= \sqrt{ {(10 - 4)}^{2} + {( - 2 - 6)}^{2} } \\ = \sqrt{ {6}^{2} + ( - 8)^{2} } \\ = \sqrt{100} \\ = 10 \: units$

Since the distance of point C from the center is greater than 6 units, point C lies outside the circle.

D. Distance between 2 points

$= \sqrt{ {(10 - 8)}^{2} + {( - 2 - 8)}^{2} } \\ = \sqrt{2^{2} + ( - 10)^{2} } \\ = \sqrt{104} \\ = 10.198 \: units$

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Point D also lies outside Circle O.

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