Mathematics Questions and Answers

Since A(x, y) is the point of equidistance between B and C, then 

AB = AC

(AB)\(^2\) = (AC)\(^2\)

Using the distance formula, 

(x - 0)\(^2\) + (y - 2)\(^2\) = (x - 2)\(^2\) + (y - 1)\(^2\)

x\(^2\) + y\(^2\) - 4y + 4 = x\(^2\) - 4x + 4 + y\(^2\) - 2y + 1

x\(^2\) - x\(^2\) + y\(^2\) - y\(^2\) + 4x - 4y + 2y = 5 - 4

4x - 2y = 1

Let the number of days worked by the assistant = t

∴∴ The bricklayer worked (t + 10) days.

1500(t + 10) + 500(t) = N 95,000

1500t + 15,000 + 500t = N 95,000

2000t = N 95,000 - N 15,000

2000t = N 80,000

t = 40 days

∴∴ The assistant worked for 40 days and received N (500 x 40)

= N 20,000

\(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\)

\((5^2)^{(1 - x)} \times 5^{(x + 2)} \div (5^{-3})^x = (5^4)^{-1}\)

\(5^{2 - 2x} \times 5^{x + 2} \div 5^{-3x} = 5^{-4}\)

\(5^{(2 - 2x) + (x + 2) - (-3x)} = 5^{-4}\)

Equating bases, we have

\(2 - 2x + x + 2 + 3x = -4\)

\(4 + 2x = -4 \implies 2x = -4 - 4\)

\(2x = -8\)

\(x = -4\)

\((0.0439 \div 3.62)\)

= 0.01213

\(\approxeq\) 0.012

= \(\frac{12}{1000}\)

\(V = \frac{\pi h}{3} (R^2 + Rr + r^2)\)

\(V = \frac{\pi R^2 h}{3} + \frac{\pi Rr h}{3} + \frac{\pi r^2 h}{3}\)

\(\frac{\mathrm d V}{\mathrm d R} = \frac{2 \pi R h}{3} + \frac{\pi r h}{3}\)

= \(\frac{\pi}{3} (2R + r)\)