The graph of y = x\(^2\) and y = x intersect at which of these points?
(0,0), (1,1)
(0,0), (0,1)
(1, 0), (0, 0)
(0, 0) (0, 0)
Correct answer is A
y = x\(^2\) ....(1)
y = x ......(2)
y = y
x\(^2\) - x
x\(^2\) - x = 0
x(x - 1) = 0
x = 0 or x - 1 = 0
x = 0 or x = 1
when x = 0, y = 0\(^2\) = 0
when x = 1, y = 1\(^2\) = 1
Hence; the two graphs interest at (0, 0) and (1, 1)
18
20
30
38
Correct answer is D
Let n(M \(\cup\) N \) = x
Then 20 - x + x + 30
- x = n(M \(\cup\) N)
50 - x = 40
50 - 40 = x
10 = x
x = 10
Hence, n(M \(\cup\N)' = 8 + (20 - 10) + (30 + 10)
= 8 + 10 + 20
= 38
42cm
48cm
52cm
60cm
Correct answer is C
Let the length of a side of the rhombus be n
Then, n\(^2\) = 5\(^2\) + 12\(^2\)
= 25 + 144 = 169
n = \(\sqrt{169}\)
= 13cm
Hence, perimeter of rhombus = 4n = 4 x 13
= 52cm
Given that Y is 20cm on a bearing of 300\(^o\) from x, how far south of y is x?
10cm
15cm
25cm
30cm
Correct answer is A
In \(\bigtriangleup\)YSC, sin 30\(^o\) = \(\frac{YS}{20}\)
|YS| = 20 sin 30\(^o\)
= 20 x 0.5
10m
If tan x = \(\frac{4}{3}\), 0\(^o\) < x < 90\(^o\), find the value of sin x - cos x
\(\frac{1}{10}\)
\(\frac{1}{5}\)
\(\frac{5}{12}\)
1\(\frac{2}{5}\)
Correct answer is B
From the diagram,
h\(^2\) = 4\(^2\) + 3\(^2\) (pythagoras')
h\(^2\) = 16 + 9 = 25
h = \(\sqrt{25}\) = 5
Hence, sin x - cos x
= \(\frac{4}{5} - \frac{3}{5}\)
= \(\frac{1}{5}\)
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