Mathematics Questions and Answers

\(3^{-4}\times 9^2 \times 81^{-1}\\
=log_9 (3^{-4}\times 9^2 \times 81^{-1})\\
=log_9 \left(\frac{1}{3^4}\times 9^2 \times \frac{1}{81}\right)\\
=log_9 \left(\frac{1}{81}\times \frac{81}{1}\times \frac{1}{81}\right)\\
=log_9 \frac{1}{81}\\
=log_9 \frac{1}{9^2}\\
=log_9 9^{-2}\\
=-2log_9 9\\
-2 \times 1\\
=-2\)

m : n = 13 : 11
m\(^2\) - n\(^2\) : (m + n)2
= 13\(^2\) - 11\(^2\) : (13 + 11)\(^2\)
= 169 - 121 : 24\(^2\)
= 48 : 576
= 1 : 12

\(\frac{\frac{7}{9}-\frac{2}{3}}{\frac{1}{3}+\frac{\frac{2}{5}}{\frac{4}{5}}}\\
=\frac{\frac{7-6}{9}}{\frac{1}{3}+\frac{\frac{2}{5}}{\frac{4}{5}}}\\
=\frac{\frac{1}{9}}{\frac{1}{3}+\left(\frac{2}{5}\times \frac{5}{4}\right)}\\
=\frac{\frac{1}{9}}{\frac{1}{3}+\frac{1}{2}}\
=\frac{\frac{1}{9}}{\frac{2+3}{6}}\\
=\frac{\frac{1}{9}}{\frac{5}{6}}\\
=\frac{1}{9}\times\frac{6}{5}\\
=\frac{2}{15}\)

\(\int_{-4}^0(1 - 2x)dx=[x-x^2]_{-4}^0\\= (0 - 0 + C) - (-4(-4)^2 + C)\\
= C - (-4-16+C)\\
= C-(-20+C)\\
= C+20-C\\
= 20\)

y = (cosθ - sinθ)
dy/dx = 2(cosθ - sinθ)(-sinθ - cosθ)
dy/dx = 2(-cosθsinθ - cos2θ + sin2θ)
dy/dx = 2(- cos\(^2\)θ + sin\(^2\)θ)
= -2(cos\(^2\)θ - sin\(^2\)θ)
= -2(1 - 2sin\(^2\)θ)
= -2cos2θ