Photo AI

Chris bought a bonsai (miniature tree) at a nursery - NSC Mathematics - Question 3 - 2016 - Paper 1

Question icon

Question 3

Chris-bought-a-bonsai-(miniature-tree)-at-a-nursery-NSC Mathematics-Question 3-2016-Paper 1.png

Chris bought a bonsai (miniature tree) at a nursery. When he bought the tree, its height was 130 mm. Thereafter the height of the tree increased, as shown below. | ... show full transcript

Worked Solution & Example Answer:Chris bought a bonsai (miniature tree) at a nursery - NSC Mathematics - Question 3 - 2016 - Paper 1

Step 1

During which year will the height of the tree increase by approximately 11,76 mm?

96%

114 rated

Answer

To determine during which year the height of the tree will increase by approximately 11.76 mm, we first find the common ratio rr of the geometric sequence.

Calculating rr:

r=70100=710 r = \frac{70}{100} = \frac{7}{10}

Next, we use the formula for the nthn^{th} term of a geometric sequence: Tn=arn1T_n = ar^{n-1} where:

  • a=100a = 100 (first term)
  • r=710r = \frac{7}{10}
  • Tn=11.76T_n = 11.76 mm

We set up the equation: 11.76=100(710)n1 11.76 = 100 \left( \frac{7}{10} \right)^{n-1}

Solving for nn:

  1. Divide both sides by 100: (710)n1=11.76100 \left( \frac{7}{10} \right)^{n-1} = \frac{11.76}{100} (710)n1=0.1176 \left( \frac{7}{10} \right)^{n-1} = 0.1176
  2. Take logarithms: (n1)log(710)=log(0.1176) (n-1) \log\left(\frac{7}{10}\right) = \log(0.1176)
  3. Solve for nn: n1=log(0.1176)log(710) n - 1 = \frac{\log(0.1176)}{\log\left(\frac{7}{10}\right)} n=log(0.1176)log(710)+1 n = \frac{\log(0.1176)}{\log\left(\frac{7}{10}\right)} + 1

After performing the calculations, we find that during the 7th7^{th} year, the increase will be approximately 11.76 mm.

Step 2

Determine a formula for $h(n)$

99%

104 rated

Answer

To find the formula for the height of the tree, we start with the height after nn years:

The initial height is h(0)=130mmh(0) = 130 mm, so the height after nn years can be expressed as: h(n)=130+(100+70+49+) for n termsh(n) = 130 + (100 + 70 + 49 + \ldots ) \text{ for } n \text{ terms}

Using the geometric series sum formula: Sn=a1rn1rS_n = a \frac{1 - r^n}{1 - r} where a=100a = 100 and r=710r = \frac{7}{10}, we have: h(n)=130+100+100(1(710)n)1710h(n) = 130 + 100 + \frac{100(1 - (\frac{7}{10})^n)}{1 - \frac{7}{10}}

Simplifying this gives a formula for h(n)h(n): h(n)=130+100+100(1(710)n)0.3h(n) = 130 + 100 + \frac{100(1 - (\frac{7}{10})^n)}{0.3}

Step 3

What height will the tree eventually reach?

96%

101 rated

Answer

The eventual height of the tree can be determined by considering the limit of the height as nn approaches infinity.

As nn approaches infinity, (710)n(\frac{7}{10})^n approaches 0. Therefore, the eventual height is: h()=130+1000.3 h(\infty) = 130 + \frac{100}{0.3} Calculating this gives: h()=130+333.33=463.33 mm h(\infty) = 130 + 333.33 = 463.33 \text{ mm} Thus, the tree will eventually reach a height of approximately 463.33 mm.

Join the NSC students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;