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# Mathematics_applications_and_interpretation_paper_2_TZ2_HL

### Question

The diagram shows points in a park viewed from above, at a specific moment in time.
The distance between two trees, at points A and B, is 6.36m.
Odette is playing football in the park and is standing at point O, such that OA = 25.9m and OÂB = $$125^0$$.

(a) Calculate the area of triangle AOB.
Odette’s friend, Khemil, is standing at point K such that he is 12m from A and KÂB = $$45^0$$

(b) Calculate Khemil’s distance from B.

XY is a semicircular path in the park with centre A, such that KÂY = $$45^0$$. Khemil is standing on the path and Odette’s football is at point X. This is shown in the diagram below.

The length KX = 22.2m, KÔX = $$53.8^0$$ and OK̂X = $$51.1^0$$.

(c) Find whether Odette or Khemil is closer to the football.
Khemil runs along the semicircular path to pick up the football.
(d) Calculate the distance that Khemil runs.

(a) attempt to use area of triangle formula
$$\frac{1}{2} \times 25.9 \times 6.36 \times sin (125^0)$$
67.5 $$m^2$$ (67.4700… $$m^2$$)

(b) attempt to use cosine rule
(BK=) $$\sqrt{12^2 + 6.36^2 – 2 \times 12 \times 6.36 \times cos 45^0}$$
8.75 (m) (8.74738… (m))

(c) METHOD 1
attempt to use sine rule with measurements from triangle OKX
$$\frac{OX}{sin 51.1^o} = \frac{22.2}{sin 53.8^o}$$
(OX=) 21.4 (m) (21.4099…)(m)
(21.4 (m) < 22.2 (m))

METHOD 2
sketch of triangle OXK with vertices, angles and lengths

$$51.1^o$$ is smallest angle in triangle OXK
opposite side (OX) is smallest length
therefore Odette is closest

(d) attempt to use length of arc formula
$$\frac{135}{360} \times 2 \pi \times 12$$
28.3 (m) (9 $$\pi$$, 28.2743…) (m)

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Time

Q — 1
11:12
Eng

### Question

A scientist is conducting an experiment on the growth of a certain species of bacteria.
The population of the bacteria, P, can be modelled by the function

P(t) = 1200 × $$k^t$$, t ≥ 0.

where t is the number of hours since the experiment began, and k is a positive constant.
(a) (i) Write down the value of P(0).
(ii) Interpret what this value means in this context.

3 hours after the experiment began, the population of the bacteria is 18 750.

(b) Find the value of k

(c) Find the population of the bacteria 1 hour and 30 minutes after the experiment began.

The scientist conducts a second experiment with a different species of bacteria.
The population of this bacteria, S, can be modelled by the function

S(t) = 5000 × $$1.65^t$$
, t ≥ 0 ,

where t is the number of hours since both experiments began.
(d) Find the value of t when the two populations of bacteria are equal.
It takes 2 hours and m minutes for the number of bacteria in the second experiment to reach 19 000.
(e) Find the value of m, giving your answer as an integer value.

(a) (i) 1200
(ii) the initial population of the bacteria

(b) 1200 × $$k^3$$ = 18750
(k=) 2.5

(c) 1200 × $$2.5^{1.5}$$
4740 (4743.41…)

(d) equating P (t) and S (t) OR equating each function to a common variable
1200 × $$2.5^t$$ = 5000 × $$1.65^t$$ ; 1200 × $$2.5^t$$ = x and 5000 × $$1.65^t$$ = x
t = 3.43 (hours)  (3.43456…)

(e) METHOD 1
5000 × $$1.65^t$$ = 19000
(t=) 2.66586… OR (t – 2 ) 0.66586… (seen)
multiplying by 60 seen to convert to minutes
(m = 39.9521…)
(m=) 40 (minutes) OR 2 hours and 40 minutes
METHOD 2
equating an expression for S (t) to 19000
expressing t as 2 + $$\frac{m}{60}$$
5000 $$\times 1.65^{2 + \frac{m}{60}}$$ = 19000
2 + $$\frac{m}{60}$$ = 2.66586…
(m=) 40 (minutes) OR 2 hours and 40 minutes

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Time

Q — 2
12:14
Eng
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