Mathematics Questions and Answers

Given that radius = 3cm.

volume of sphere = \(\frac{4}{3}\times\pi\times r^3\)

= \(\frac{4}{3}\times\pi\times 3^3\)

= \(\frac{4}{3}\times\pi\times 27\)

= \(4\times\pi\times9\)

= \(36\pi cm^3\)

radius = 8cm , height = 14cm  and \(\pi = \frac{22}{7}\)

total surface area of a solid cylinder =\( 2πrh+2πr^2\) = 2πr( h + r )

 \( 2 \times \frac{22}{7} \times 8( 8 + 14)\)

 \( 2 \times \frac{22}{7} \times 8 \times 22\)

\(\frac{7744}{7}\)

= \(1,106.29cm^2\)

%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: \(3m^2 - 2m - 65 = 0\)

= \( m^2 - 15m + 13m - 65 = 0\)

= 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, \(\frac{-13}{3})\)

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