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Make y the subject of the equation x = √(yp – 1). (2 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
x = √(yp – 1)
x^2 = yp – 1
yp = x^2 + 1y = (x^2 + 1)/p
Ariana’s parents have given her an interest-free loan of $4800 to buy a car. She will pay them back by paying $x immediately and $y every month until she has repaid the loan in full. After 18 months Ariana has paid back $1510, and after 36 months she has paid back $2770. The information can be represented by the following equations.
x + 18y = 1510
x + 36y = 2770
(i) Solve these equations simultaneously to find the values of x and y. (2 marks)
(ii) How many months will it take to repay the loan in full? (2 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Expand 4x(7x^4 – x^2). (1 mark)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
4x(7x^4 – x^2) = 28x^5 – 4x^3
Solve the equation (5x + 1)/3 – 4 = 5 – 7x. (3 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(5x + 1)/3 – 4 = 5 – 7x
(5x + 1)/3 = 9 – 7x
5x + 1 = 3(9 – 7x)
5x + 1 = 27 – 21x
5x + 21x = 27 – 1
x = 1
26x = 26
Solve these simultaneous equations to find the values of x and y. (3 marks)
y = 2x + 1
x – 2y – 4 = 0
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Solve these equations simultaneously, showing all working. (2 marks)
4x + y = 13
2x – y = 2
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Solve the equation. 5x/4 + 9 = 10x – 12. (3 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
5x/4 + 9 = 10x – 12
4 × 5x/4 + 4 × 9 = 4 × 10x – 4 × 12
5x + 36 = 40x – 48
40x – 5x = 36 + 48
35x = 84
35x/35 = 84/35
x = 2.4
Sarah tried to solve this equation and made a mistake in Line 2.
(W + 4)/3 – (2W – 1)/5 = 1 ……………….. Line 1
5W + 20 – 6W – 3 = 15 ………….……. Line 2
17 – W = 15 ……….……… Line 3
W = 2 ……….……… Line 4
Copy the equation in Line 1 into your writing booklet and continue your solution to solve this equation for W. Show all lines of working. (2 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(W + 4)/3 – (2W – 1)/5 = 1
Multiply through by 15:
15 × (W + 4)/3 – 15 x (2W – 1)/5 = 15 × 1
5(W + 4) – 3(2W – 1) = 15
5W + 20 – 6W + 3 = 15
23 – W = 15
23 – 15 = W
W = 8
Temperature can be measured in degrees Celsius (C) or degrees Fahrenheit (F). The two temperature scales are related by the equation F = 9C/5 + 32.
(i) Calculate the temperature in degrees Fahrenheit when it is –20 degrees Celsius. (1 mark)
(ii) Solve the following equations simultaneously, using either the substitution method or the elimination method. (2 marks)
F = 9C/5 + 32
F = C
(iii) The graphs of F = 9C/5 + 32 and F = C are shown opposite. What does the result from part (ii) mean in the context of the graph? (1 mark)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(i) Substitute C = –20:
F = 9C/5 + 32
= 9(-20)/5 + 32
= –4
the temperature is –4 degrees C.
(ii) F = 9C/5 + 32 ……1
F = C ………………..2
Substitute in :
C = 9C/5 + 32
5C = 9C + 160
4C = 160
C = –40
therefore, F = –40
therefore, C = –40 and F = –40.
(iii) The two lines meet on the extended number plane at (–40, –40).
The weight of an object on the moon varies directly with its weight on Earth. An astronaut who weighs 84 kg on Earth weighs only 14 kg on the moon. A lunar landing craft weighs 2449 kg on the moon. Calculate the weight of this landing craft when on Earth. (2 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
Let E = weight on Earth, m = weight on moon.
Let E= km
Subs E = 84, m = 14:
84 = k(14)
14k = 84
k = 6
therefore, E = 6m
Subs m = 2449:
E = 6(2449)
= 14 694
therefore, the landing craft weighs 14 694 kg.