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Solve the following equations:
(i) 5(x+3) = 4(x+7). (2 marks)
(ii) (5h + 2)/3 – (h – 5)/4 = 8. (2 marks)
(iii) 2^x + 16 = 144, use ‘guess-and-check’ or any other method and give your answer to two decimal places. (2 marks)
(iv) ∛(2a^2 + 1) = 3 answer to 1 decimal place. (2 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(i) 5x+15=4x+28
x=13
(ii) 4(5h+2)- 3(h-5)= 96
20h+8-3h+15=96
17h+23=96
17h=73
h=4 5/17
(iii) 2^x=128
2^7=128
x=7
(iv) 2a^2+1=27
2a^2=26
a^2=13
a=±3.6
Make r the subject of the formula V = 1/3 πr^2 h (2 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
3V/πh=r^2
r=±√(3V/πh)
The graph below shows the cost of making dresses and the income received from their sale.
(i) How much profit or loss is made when 5 dresses are sold? (1 mark)
(ii) Use the graph to determine the number of dresses which need to be sold to break even. (1 mark)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(i) $100-$90=$10(profit)
(ii) 4 dresses
Make F the subject of the formula C = 5/9 (F – 32) and hence find F when C = 30. (2 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
9C/5+32=F
F=(9×30)/5+32
F=86
Solve for a
(3a + 1)/2 – (2a – 3)/3 = 5. (2 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
3(3a+1)-2(2a-3)=30
9a+3-4a+6=30
5a=21
a=4 1/5
The growth of bacteria in a colony is given by the equation: P = 500(1.07)^t, where P is the number of bacteria in the colony and ‘t’ is the time in months.
(i) Find the number of bacteria in the colony initially. (1 mark)
(ii) Calculate the number of bacteria in the colony after 8 months. (1 mark)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(i) P=500(1.07)^o when t=0
P=500
(ii) P=500(1.07)^8
P=859.1 =859
The formula for the volume of a cone is V = 1/3 π(r^2)h where V = volume, h = height and r = radius.
(i) Show that the radius of the cone is r = √(3V/πh). (1 mark)
(ii) Calculate the radius of the base of a cone that is 20cm high and has a volume of 200 cm^3. (1 mark)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(i) 3v=πr^2 h
r^2=3v/πh
r=√(3v/πh)
(ii) r=√((3×200)/(π×20)) = 3.0901…..
r=3.09 cm
Olivia’s watering can is initially filled with 4 litres of water. However, the watering can has a small hole in the base and is leaking at a rate of 0.1 litres per minute.
(i) Write a linear equation in the form V = mt + c to describe this situation. (2 marks)
(ii) What volume of water remains after 150 seconds? (1 mark)
(iii) How long would it take for all the water to leak out? (1 mark)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(i) V=4-0.1t
(ii) 150 seconds = 2.5 minutes
V=4-0.1×2.5
V=3.75 litres
(iii) 0=4-0.1t
0.1t=4
t=40 minutes
The petrol consumption (Y litres per 100 km) and the speed of a car (s, km/h) are modelled by the formula Y = 0.05s^2 – 3s + 55.
(i) Draw the graph of Y = 0.05s^2 – 3s + 55. Use s from 0 to 70 km/h. (1 mark)
(ii) What is the petrol consumption at 40 km/h? (1 mark)
(iii) What is the speed of the car to achieve the lowest petrol consumption? (1 mark)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(i)
(ii) Y=0.05(40)^2-3×40+55
Y=15(litres per 100km
(iii) 30 km/h
Solve the following equations:
(i) 5(x + 2) = -2(x – 6). (2 marks)
(ii) 2a = 1+ 4a/3. (2 marks)
(iii) √(4b – 8) = 2. (2 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(i) 5x+10=-2x+12
7x=2
x=2/7
(ii) 6a=3+4a
2a=3
a=1.5
(iii) 4b-8=4
4b=12
b=3