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Mars2501 [29]
3 years ago
9

The trees at a national park have been increasing in numbers. There were 1,000 trees in the first year that the park started tra

cking them. Since then, there has been one fifth as many new trees each year. Create the sigma notation showing the infinite growth of the trees and find the sum, if possible.
1 1000
2 200
3 40

Mathematics
1 answer:
ivanzaharov [21]3 years ago
6 0
The sequence forms a Geometric sequence as the rule to obtain the value for the next term is by ratio

Term 1: 1000
Term 2: 200
Term 3: 40

From term 1 to term 2, there's a decrease by \frac{1}{5}
From term 2 to term 3, there's a decrease also by \frac{1}{5}

The rule to find the n^{th} term in a sequence is 
n^{th}=a r^{n-1}, where a is the first term in the sequence and r is the ratio

So, the formula for the sequence in question is
n_{th} term = 1000( \frac{1}{5} ^{n-1} )

The sequence is a divergent one. We can always find the value of the next term by dividing the previous term by 5 and if we do that, the value of the next term will get closer to 'zero' but never actually equal to zero.

We can find a partial sum of the sequence using the formula
S_{∞} = \frac{a}{1-r} for -1<r<1 
Substituting a=1000 and r= \frac{1}{5} we have 
S_{∞} = \frac{1000}{1- \frac{1}{5} } = 1250

Hence, the correct option is option number 1

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A circle is growing so that the radius is increasing at the rate of 3 cm/min. How fast is the area of the circle changing at the
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Answer:

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<em>Notice that this problem requires the use of implicit differentiation in related rates (some some calculus concepts to be understood), and not all middle school students cover such.</em>

We identify that the info given on the increasing rate of the circle's radius is 3 \frac{cm}{min} and we identify such as the following differential rate:

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We now apply the derivative operator with respect to time (\frac{d}{dt}) to this equation, and use chain rule as we find the quadratic form of the radius:

\frac{d}{dt} [A=\pi\,r^2]\\\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}

Now we replace the known values of the rate at which the radius is growing ( \frac{dr}{dt} = 3\,\frac{cm}{min}), and also the value of the radius (r = 12 cm) at which we need to find he specific rate of change for the area :

\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}\\\frac{dA}{dt} =\pi\,*2*(12\,cm)*(3\,\frac{cm}{min}) \\\frac{dA}{dt} =226.19467 \,\frac{cm^2}{min}\\

which we can round to one decimal place as:

\frac{dA}{dt} =226.2 \,\frac{cm^2}{min}

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