\(\sqrt{2239cm}\)
\(\sqrt{2}\) x 2239cm
\(\frac{\sqrt{3}}{2}\) 2239cm
\(\sqrt{3}\) x 2239cm
4478cm
Correct answer is D
x = \(\sqrt{-2239^2 + 2239^2}\)
= -\(\sqrt{10026242}\)
= 3166.42
y = -\(\sqrt{10026242 + 5013121}\)
= -\(\sqrt{15039363}\)
= 3878
= \(\sqrt{3}\) x 2239
An arc of circle of radius 2cm subtends an angle of 60º at the centre. Find the area of the sector
\(\frac{2 \pi}{3}\)cm2
\(\frac{\pi}{2}\)cm2
\(\frac{\pi}{3}\)cm2
\(\pi\)cm2
Correct answer is A
Area of a sector \(\frac{\theta}{360}\) x \(\pi\)r\(^2\)
= \(\frac{60^o}{360^o}\) x \(\pi\)2\(^2\)
= \(\frac{2 \pi}{3}\)cm2
A cylinder of height h and radius r is open at one end. Its surface area is
2\(\pi\)rh
\(\pi\)r2h
2\(\pi\)rh + \(\pi\)r2
2\(\pi\)rh + 2\(\pi\)r2
Correct answer is C
A cylinder of height h ans radius r is open at one end, its surface area is 2\(\pi\)rh + \(\pi\)r2
Simplify \(\frac{1 - x^2}{x - x^2}\), where x \(\neq\) 0
\(\frac{1}{x}\)
\(\frac{1 - x}{x}\)
\(\frac{1 + x}{x}\)
\(\frac{1}{x - 1}\)
\(\frac{-x - 1}{1}\)
Correct answer is C
\(\frac{1 - x^2}{x - x^2}\), where x = \(\neq\) 0
\(\frac{1^2 - x^2}{x - x^2}\)
= \(\frac{(1 + x)(1 - x)}{x(1 - x)}\)
= \(\frac{1 + x}{x}\)
For the set of numbers 2, 3, 5, 6, 7, 7, 8
The median is greater than the mode
The mean is greater than the mode
The mean is greater than the median
The median is equal to the mean
The mean is less than the median
Correct answer is E
Mean = 5, 42
Median = 6
Mode = 7
The mean is less than the median