What is the length of a rectangular garden whose perimeter is 32cm and area 39cm2?
25cm
18cm
13cm
9cm
Correct answer is C
perimeter = 2(l + b) = 32
l + b = \(\frac{32}{2}\)
l + b = 16
b = 16 - 1.......(1)
Area = l + b = 39
lb = 39 .....(2)
put (1) into (2)
l(16 - 1) = 39
16l - l2 = 39
l2 = 13l - 3l + 39 = 0
l(l - 13) - 3(1 - 13) = 0
(l - 3)(l + 130 = 0
l - 3 = 0 or l - 13 = 0
l = 3cm or l = 13cm; The length in 13cm
Given that tan x = 1, where 0o \(\geq\) x 90o, evaluate \(\frac{1 - \sin^2 x}{\cos x}\)
2\(\sqrt{2}\)
\(\sqrt{2}\)
\(\frac{\sqrt{2}}{2}\)
\(\frac{1}{2}\)
Correct answer is C
Given tan x = 1
x = tan-1(1)
x = 45o
Now, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\)
= \(\frac{1 - \frac{1}{2}}{\frac{1}{2}}\)
= \(\frac{1}{2} + \frac{1}{\sqrt{2}}\)
= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\)
= \(\frac{\sqrt{2}}{2}\)
A rectangle has length xcm and width (x - 1)cm. If the perimeter is 16cm. Find the value of x
3\(\frac{1}{2}\)cm
4cm
4\(\frac{1}{2}\)cm
5cm
Correct answer is C
l = x; b = x - 1
perimeter = 2(l + b) = 16
l + b = \(\frac{16}{2}\) = 8
l + b = 8
x + x - 1 = 8
2x = 8 + 1
2x = 9
x = \(\frac{9}{2}\)cm
x = 4\(\frac{1}{2}\)
If sin 3y = cos 2y and 0o \(\leq\) 90o, find the value of y
18o
36o
54o
90o
Correct answer is A
sin 3y = cos 2y, but sin \(\theta\) = cos(90 - \(\theta\))
sin 3y = cos(90 - 3y)
cos(90 - 3y) = cos 2y
90 - 3y = 2y
5y = 90
y = \(\frac{90}{5}\)
y = 18o
Simplify 2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)
1
\(\frac{1}{3}\sqrt{3}\)
2\(\sqrt{3} - 5\frac{2}{3}\)
6\(\sqrt{3}\) - 17
Correct answer is B
2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)
= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} + \frac{3}{\sqrt{9 \times 3}}\)
= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} + \frac{3}{3\sqrt{3}}\)
= 2\(\sqrt{3} = 6 \frac{\sqrt{3}}{3} + \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= 2\(\sqrt{3} - 2\sqrt{3} + \frac{\sqrt{3}}{3}\)
= \(\frac{\sqrt{3}}{3} = \frac{1}{3} \sqrt{3}\)