The radius of a sphere is 3 cm. Find, in terms of π, its volume.
\(30\pi cm^3\)
\(108\pi cm^3\)
\(27\pi cm^3\)
\(36\pi cm^3\)
Correct answer is D
Given that radius = 3cm.
volume of sphere = \(\frac{4}{3}\times\pi\times r^3\)
= \(\frac{4}{3}\times\pi\times 3^3\)
= \(\frac{4}{3}\times\pi\times 27\)
= \(4\times\pi\times9\)
= \(36\pi cm^3\)
\(1,106.29cm^2\)
\(1,016.29cm^2\)
\(1,106.89cm^2\)
\(1,206.27cm^2\)
Correct answer is A
radius = 8cm , height = 14cm and \(\pi = \frac{22}{7}\)
total surface area of a solid cylinder =\( 2πrh+2πr^2\) = 2πr( h + r )
\( 2 \times \frac{22}{7} \times 8( 8 + 14)\)
\( 2 \times \frac{22}{7} \times 8 \times 22\)
\(\frac{7744}{7}\)
= \(1,106.29cm^2\)
6.29m
7.67m
7.18m
6.65m
Correct answer is A
%error=5%, measured height = 5.98m
let the actual height = y
error=x - 5.98 (since 'y' is more than 5.98)
%error = \(\frac{error}{actual height}\times 100%\)
5% = \(\frac{y - 5.98}{y}\times 100%\)
\(\frac{5}{100} = \frac{y - 5.98}{y}\)
5y = 100(y - 5.98)
5y = 100y - 598
5y - 100y = - 598
-95y = - 598
y = \(\frac{-598}{-95}\)
y = 6.29m( to 2 d.p).
Find the roots of the equations: \(3m^2 - 2m - 65 = 0\)
\(( -5, \frac{-13}{3})\)
\(( 5, \frac{-13}{3})\)
\(( 5, \frac{13}{3})\)
\(( -5, \frac{13}{3})\)
Correct answer is B
Find the roots of the equations: \(3m^2 - 2m - 65 = 0\)
= \( m^2 - 15m + 13m - 65 = 0\)
= 3m(m - 5) + 13( m - 5) = 0
( m - 5)(3m + 13) = 0
m-5 = 0 or 3m + 13 = 0
therefore, m = 5 or \(\frac{-13}{3}\)
therefore the roots of the quadratic equation = ( 5, \(\frac{-13}{3})\)
If \(log_a 3\) = m and \(log_a 5\) = p, find \(log_a 75\)
\(m^2 + p \)
2m + p
m + 2p
\(m + p^2\)
Correct answer is C
Given: \(log_a 3\) = m and \(log_a 5\) = p
\(log_a 75\) = \(log_a (3 × 25)\)
= \(log_a (3 × 5^2)\)
= \(log_a 3 + log_a 5^2\)
= \(log_a 3 + 2log_a 5\)
Since \(log_a 3\) = m and \(log_a 5\) = p
∴ \(log_a 75\) = m + 2p
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