If y = \(\frac{(2\sqrt{x^2 + m})}{3N}\), make x the subject of the formular
\(\frac{\sqrt{9y^2 N^2 - 2m}}{3}\)
\(\frac{\sqrt{9y^2 N^2 - 4m}}{2}\)
\(\frac{\sqrt{9y^2 N^2 - 3m}}{2}\)
\(\frac{\sqrt{9y^2 N - 3m}}{2}\)
Correct answer is B
y = \(\frac{(2\sqrt{x^2 + m})}{3N}\)
3yN = 2(\(\sqrt{x^2 + m})\)
\(\frac{3yN}{2} = \sqrt{x^2 + m}\)
(\(\frac{3yN}{2})^2 = ( \sqrt{x^2 + m})\)
\(\sqrt{\frac{9y^2N^2}{4} - \frac{m}{1}}\)
x = \(\frac{\sqrt{9Y^2N^2 - 4m}}{4}\)
x = \(\frac{\sqrt{9y^2N^2 - 4m}}{2}\)
The sum of the exterior of an n-sided convex polygon is half the sum of its interior angle. find n
6
8
9
12
Correct answer is A
sum of exterior angles = 360o
Sum of interior angle = (n - 2) x 180
360 = \(\frac{1}{2}\) x(n - 2) x 180(90o)
360 = \(\frac{1}{2}\) x(n - 2) x 90o
\(\frac{360}{90}\) = a - 2
4 = n - 2
n = 4 + 2 = 6
Simplify \(\frac{2}{2 + x} + \frac{2}{2 - x}\)
\(\frac{4}{4 - x^3}\)
\(\frac{8}{4 - x^2}\)
\(\frac{4x}{4 - x^2}\)
\(\frac{8 - 4x}{4 - x^2}\)
Correct answer is B
\(\frac{2}{2 + x} + \frac{2}{2 - x}\)
\(\frac{2(2 - x) + 2(2 + x)}{(2 + x)(2 - x)} = \frac{4 - 2x + 4 + 2x}{4 - 2x + 2x - x^2}\)
= \(\frac{8}{4 - x^2}\)
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, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\)
= \(\frac{1 - \frac{1}{2}}{\frac{1}{2}}\)
= \(\frac{1}{2} + \frac{1}{\sqrt{2}}\)
= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\)
= \(\frac{\sqrt{2}}{2}\)