N215.30
N134.10
N143.10
N80.00
N50.30
Correct answer is B
\(\bar{x}\) = \(\frac{fx}{N}\)
= \(\frac{6705}{50}\)
= 1341
= N134.10
Simplify f\(\frac{1}{2}\)g2h\(\frac{1}{2}\) \(\div\) f\(\frac{5}{2}\)goh\(\frac{7}{3}\)
(\(\frac{g}{fh}\))2
f2g2h2
\(\frac{5}{4}\)goh\(\frac{7}{9}\)
\(\frac{g^2}{f^5h^7}\)
\(\frac{1}{f^2h^2}\)
Correct answer is A
f\(\frac{1}{2}\)g2h\(\frac{1}{2}\) \(\div\) f\(\frac{5}{2}\)goh\(\frac{7}{3}\) = f\(\frac{1}{2}\) - \(\frac{5}{2}\) g2 - 0 h\(\frac{1}{2}\) - \(\frac{7}{3}\)
f-2 g2 h-2
= \(\frac{g^2}{f^2h^2}\)
= (\(\frac{g}{fh}\))2
(21cm3, 3cm3)
(24cm3, 3cm3)
(24cm3, 21cm3)
(72cm3, 9cm3)
(63cm3, 9cm3)
Correct answer is B
Vol. of the 1st section is side x height
vol. = 3 x 8
= 24cm3
vol. of the second section is 3 x 1 = 3
= 24cm3, 3cm3
58 sq.cm
34 sq.cm
132 sq.cm
77 sq.cm
66 sq.cm
Correct answer is E
Given: length of the arc AOB = 22cm ; L = 6cm
Curved surface area of cone = \(\pi\)rl
Length of an arc = \(\frac{\theta}{360}\) x 2 \(\pi\)L
= 22cm
but length of an arc = circumference of the cone = 2\(\pi\)
where r is the radius of the cone circle
2\(\pi\)r = 22, r = \(\frac{22}{2\pi}\)r
= 11 x \(\frac{7}{22}\)
= \(\frac{7}{2}\)
curved surface area = \(\pi\)rl
= \(\frac{22}{7}\) x \(\frac{7}{2}\) x 6
= 66 sq.cm
The solution to the simultaneous equations 3x + 5y = 4, 4x + 3y = 5 is
(\(\frac{-13}{11}, \frac{1}{11}\))
(\(\frac{13}{11}, \frac{1}{11}\))
(\(\frac{13}{11}, \frac{-1}{11}\))
(\(\frac{11}{13}, \frac{1}{11}\))
(13, 11)
Correct answer is B
3x + 5y = 4, 4x + 3y = 5
3x + 5y = 4 x 4
4x + 3y = 5 x 3
12x + 20y = 16.....(i)
12x + 9y = 15.......(ii)
subtract eqn.(ii) from eqn.(i)
11y = 1
y = \(\frac{1}{11}\)
12x + 20 x \(\frac{1}{11}\) = 16
12x = \(\frac{156}{11}\)
x = \(\frac{13}{11}\)
= \(\frac{13}{11}, \frac{1}{11}\)