Algebra: Types of Relationships Extended Response Quiz 9
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Algebra: Types of Relationships Extended Response Quiz 9
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Question 1 of 5
1. Question
1 point(s)Anjali is investigating stopping distances for a car travelling at different speeds. To model this she uses the equation
d = 0.01s^2 + 0.7s,
where d is the stopping distance in metres and s is the car’s speed in km/h.
The graph of this equation is drawn below.
(i) Anjali knows that only part of this curve applies to her model for stopping distances.
In your writing booklet, using a set of axes, sketch the part of this curve that applies for stopping distances. (1 mark)(ii) What is the difference between the stopping distances in a school zone when travelling at a speed of 40 km/h and when travelling at a speed of 70 km/h? (2 marks)
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This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Hint
(i)
(ii) When s=40
d =0.01(40^2)+0.7(40)
=16+28
=44 mWhen s=70
d =0.01(70^2) +0.7(70)
=49+49
=98 m∴ Difference =98 −44
=54 metres -
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Question 2 of 5
2. Question
1 point(s)A movie theatre has 200 seats. Each ticket currently costs $8.
The theatre owners are currently selling all 200 tickets for each session. They decide to increase the price of tickets to see if they can increase the income earned from each movie session.
It is assumed that for each one dollar increase in ticket price, there will be 10 fewer tickets sold.
A graph showing the relationship between an increase in ticket price and the income is shown below.
(i) What ticket price should be charged to maximise the income from a movie session? (1 mark)
(ii) What is the number of tickets sold when the income is maximised? (1 mark)
(iii) The cost to the theatre owners of running each session is $500 plus $2 per ticket sold.
Calculate the profit earned by the theatre owners when the income earned from a session is maximised. (2 marks)-
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Hint
(i) Graph is highest when increase = $6
∴ Ticket price =8+6
=$14(ii) Solution 1
Tickets sold =200−(4×10)
=140Solution 2
Tickets =max income/ticket price
=1960/14
=140(iii) Cost =140×$2+$500
=$780∴ Profit when income is maximised
=1960−780
=$1180 -
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Question 3 of 5
3. Question
1 point(s)The height above the ground, in metres, of a person’s eyes varies directly with the square of the distance, in kilometres, that the person can see to the horizon.
A person whose eyes are 1.6 m above the ground can see 4.5 km out to sea.
How high above the ground, in metres, would a person’s eyes need to be to see an island that is 15 km out to sea? Give your answer correct to one decimal place. (3 marks)
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This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Hint
h∝d^2
h=kd^2
When h=1.6, d=4.5
1.6 =k×4.5^2
∴k =1.6/4.5^2
=0.07901 …Find h when d=15
h =0.07901…×15^2
=17.777…
=17.8 m (to 1 d.p.) -
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Question 4 of 5
4. Question
1 point(s)A golf ball is hit from point A to point B, which is on the ground as shown. Point A is 30 metres above the ground and the horizontal distance from point A to point B is 300 m.
The path of the golf ball is modelled using the equation
h = 30 + 0.2d – 0.001d^2
where
h is the height of the golf ball above the ground in metres, and
d is the horizontal distance of the golf ball from point A in metres.
The graph of this equation is drawn below.
(i) What is the maximum height the ball reaches above the ground? (1 mark)
(ii) There are two occasions when the golf ball is at a height of 35 metres.
What horizontal distance does the ball travel in the period between these two occasions? (1 mark)(iii) What is the height of the ball above the ground when it still has to travel a horizontal distance of 50 metres to hit the ground at point B? (1 mark)
(iv) Only part of the graph applies to this model.
Find all values of d that are not suitable to use with this model, and explain why these values are not suitable. (2 marks)-
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Hint
(i) Max height=40m
(ii) From graph
h=35 when x=30 and x=170
∴ Horizontal distance =170 −30
=140 m
(ii) Ball hits ground at x=300
⇒Need to find y when x=250
From graph, y=17.5m when x=250
∴ Height of ball is 17.5 m at a horizontal
distance of 50m before B.(iv) Values of d not suitable.
If d<0, it assumes the ball is hit away from point B. This is not the case in our example. If d>300, h becomes negative which is
not possible given the ball cannot go
below ground level. -
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Question 5 of 5
5. Question
1 point(s)In 2010, the city of Thagoras modelled the predicted population of the city using the equation
P = A(1.04)^n.
That year, the city introduced a policy to slow its population growth. The new predicted population was modelled using the equation
P = A(b)^n.
In both equations, P is the predicted population and n is the number of years after 2010.
The graph shows the two predicted populations.
(i) Use the graph to find the predicted population of Thagoras in 2030 if the population policy had NOT been introduced. (1 mark)
(ii) In each of the two equations given, the value of A is 3 000 000.
What does A represent? (1 mark)(iii) The guess-and-check method is to be used to find the value of b, in P = A(b)^n.
(1) Explain, with or without calculations, why 1.05 is not a suitable first estimate for b. (1 mark)
(2) With n = 20 and P = 4 460 000, use the guess-and-check method and the equation P = A(b)^n to estimate the value of b to two decimal places. Show at least TWO estimate values for b, including calculations and conclusions. (2 marks)(iv) The city of Thagoras was aiming to have a population under 7 000 000 in 2050. Does the model indicate that the city will achieve this aim?
Justify your answer with suitable calculations. (2 marks)-
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Hint
(i) 2030 occurs at n=20 on the x-axis.
Expected population (no policy)=6 600 000(ii) A represents the population when n=0
which is the population in 2010.(iii)(1) P=A(1.05)n would be steeper and lie above
P=A(1.04)n since 1.05>1.04(iii)(2) Let b=1.03
P =3 000 000×1.03^20
=5 418 000Let b=1.02
P =3 000 000×1.02^20
=4 457 800∴b=1.02
(iv) In 2050, n=40
P =3 000 000×1.02^40
=6 624 119 (nearest whole)Since the population is below 7 million,
the model will achieve the aim. -