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pochemuha
2 years ago
6

Find (a) arc length and (b) Area of a sector.

Mathematics
1 answer:
barxatty [35]2 years ago
5 0

Answer:

a) 29.45 cm (2 dp)

b) 220.89 cm² (2 dp)

Step-by-step explanation:

<u>Formula</u>

\textsf{Arc length}=r \theta

\textsf{Area of a sector}=\dfrac{1}{2}r^2 \theta

\quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle in radians)}

<u>Calculation</u>

Given:

  • \theta=\dfrac{5 \pi}{8}
  • r = 15 cm

\begin{aligned}\implies \textsf{Arc length} & =r \theta\\& = 15\left(\dfrac{5 \pi}{8}\right)\\& = \dfrac{75}{8} \pi \\& = 29.45\: \sf cm\:(2\:dp)\end{aligned}

\begin{aligned} \implies \textsf{Area of a sector}& =\dfrac{1}{2}r^2 \theta\\\\ & = \dfrac{1}{2}(15^2) \left(\dfrac{5 \pi}{8}\right)\\\\& = \dfrac{225}{2}\left(\dfrac{5 \pi}{8}\right)\\\\ & = \dfrac{1125}{16} \pi \\\\& = 220.89 \: \sf cm^2\:(2\:dp)\end{aligned}

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In 1992, the moose population in a park was measured to be 4260. By 1996, the population was measured again to be 3660. If the population continues to change linearly: A.) Find a formula for the moose population, P , in terms of t , the years since 1990. P ( t ) = B.) What does your model predict the moose population to be in 2008?

<u>Answer:</u>

Model predict the moose population to be in 2008 is 1860

<u>Solution:</u>

Let us consider this problem on graph, taking the x axis to be years since 1990 and y on the graph be the number of moose

We have two points on the graph:  (2, 4260) and (6, 3660)

Using the slope formula we can find the slope :

\mathrm{m}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

m=\frac{(3660-4260)}{(6-2)}

m = -150                                  

Now that we have the slope of the line, we can use the point-slope formula to find the equation for the line as follows:-

\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right)} \\\\ {y-4260=-150(x-2)} \\\\ {y-4260=-150 x+300} \\\\ {y=-150 x+4560}\end{array}

Since the formula that is being sought is supposed to be interms of P and t, we will replace y with P and x with t

P(t) = -150t + 4560

Number of years from 1990 to 2008 is  

2008 - 1990 = 18

So, population in 2008 will be

P(18) = -150 (18) + 4560

P(18) = -2700 + 4560

P(18) = 1860

Thus the model to predict the moose population in 2008 is found

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