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\(\frac{321_4}{23_4}\) = \(\frac{(3\times4^{2})+(2\times4^{1})+(1\times4^{0})}{(2\times4^{0})+(3\times4^{0})}\) = \(\frac{3\times16+2\times4+1\times1}{2\times4+3\times1}\) = \(\frac{48+8+1}{8+3}\) =\(\frac{57}{11}\) = 5\(\hspace{1mm}remainder\hspace{1mm}2\) ∴ r = 2\(_{10}\) Now\(\hspace{1mm}convert\hspace{1mm}2_{10} \hspace{1mm}to\hspace{1mm}base\hspace{1mm}4\frac{4}{2}\) = 2\(\frac{4}{0}\) = 0\(\hspace{1mm}or\hspace{1mm}2\) ∴ r = 2

3\(\frac{1}{2}\) - (2\(\frac{1}{3}\) x 1\(\frac{1}{4}\)) + \(\frac{3}{5}\)) = \(\frac{7}{2}\) - (\(\frac{7}{3}\) x \(\frac{5}{4}\)) + \(\frac{3}{5}\)) = \(\frac{7}{2}\) - \(\frac{35}{12}\) + \(\frac{3}{5}\) = L.C.M = 60 = \(\frac{(210 - 175 + 36)}{60}\) = \(\frac{71}{60}\) = 1\(\frac{11}{60}\)

To find the interest on \( N250.00 \) for \( 6 \) months at the same rate, we follow these steps: Calculate the Interest Rate Given: P = N150.00, T = 2\(\frac{1}{2}\) years and I = N4.50 Using the formula for simple interest: \(I = P \times r \times T\) Rearranging to find the interest rate \( r \): \(r = \frac{I}{P \times T}\) Substituting the values: \(r = \frac{4.50}{150 \times 2.5}\) Calculating \( P \times T \): \(P \times T = 150 \times 2.5 = 375\) Now, substituting back to find \( r \): \(r = \frac{4.50}{375} = 0.012\) Calculate the Interest for N250.00 \) for 6 Months We need to calculate the interest for: P = N250.00, T = 6 months = 0.5yrs, r = 0.012 Using the interest formula again: \(I = P \times r \times T\) Substituting the values: \(I = 250 \times 0.012 \times 0.5\) \(I = 250 \times 0.012 = 3\) Then: \(I = 3 \times 0.5 = 1.50\) The interest on \( N250.00 \) for \( 6 \) months at the same rate is N1.5

Let x = the number of oranges The 1st received 1/3 of x = 1/3x ∴Remainder = x - 1/3x = 2x/3 The 2nd received 2/3 of 2x/3 = 2/3 * 2x/3 = 4x/3 The 3rd received 12 oranges ∴1/3x + 4x/9 + 12 = x (3x + 4x + 108)/9 = x 3x + 4x + 108 = 9x 7x + 108 = 9x 9x - 7x = 108 2x = 108 x = 54 oranges

\(\frac{81^{\frac{3}{4}}-27^{\frac{1}{3}}}{3 \times 2^3}\) = \(\frac{(3^{3-\frac{3}{4}}-3^{3-\frac{3}{4}})}{3\times 2^3}\) = \(\frac{3^3 - 3}{3 \times 8}\) = \(\frac{27-3}{24}\\=\frac{24}{24}\) =1

Log102 = 0.3010 and Log103 = 0.4771 Log104.5 = Log1041/2 = Log109/2 = Log109 - Log102 = log1032 - Log102 = 2Log103 - Log102 = 2(0.4771) - 0.3010 = 0.9542 - 0.3010 = 0.6532

\(\frac{(\sqrt{12}-\sqrt{3})}{(\sqrt{12}+\sqrt{3})}\) = \(\frac{\sqrt{4\times 3} - \sqrt{3}}{\sqrt{4\times 3}+\sqrt{3}}\) = \(\frac{2\sqrt{3}-\sqrt{3}}{2\sqrt{3}+\sqrt{3}}\) = \(\frac{\sqrt{3}}{3\sqrt{3}}\) = \(\frac{1}{3}\)

Two games played only = 5 + 7 + 3 = 15

m = 3, p = -3, q = 7 and r = \(\frac{5}{2}\) m(p+q+r) = 3(-3 + 7 + \(\frac{5}{2}\)) = 3(4 + \(\frac{5}{2}\)) = 3(4 + 2\(\frac{1}{2}\)) = 3 * 6\(\frac{1}{2}\) = 3 * \(\frac{13}{2}\) = \(\frac{39}{2}\) = 19.50

x = -2, x = -1 and x = 3 ∴x+2 = 0, x+1 = 0 and x-3 = 0 Product of the factors (x+2)(x+1)(x-3) = 0 (x2 + 3x + 2)(x-3) x(x2 + 3x + 2) -3(x2 + 3x + 2) = 0 x3 + 3x2 + 2x - 3x2 - 9x - 6 = 0 x3 - 7x - 6 = 0

t = time taken and N = number of men t ∝ 1/N t = K/N K = Nt K = 30 * 6 K = 180 ∴t = 180/N 4 = 180/N 4N = 180 N = 180/4 45 men

The weight \( W \) kg of a metal bar varies jointly as its length \( L \) meters and the square of its diameter \( d \) meters. This relationship is expressed as: \(W = k \cdot L \cdot d^2\) where \( k \) is the constant of proportionality. Given \( W = 140 \), \( d = 4\frac{2}{3} = \frac{14}{3} \), and \( L = 54 \), substitute to find \( k \): \(140 = k \cdot 54 \cdot \left( \frac{14}{3} \right)^2\) Calculate \( \left( \frac{14}{3} \right)^2 = \frac{196}{9} \): \(140 = k \cdot 54 \cdot \frac{196}{9}\) Simplify: \( 54 \div 9 = 6 \), so: \(140 = k \cdot 6 \cdot 196 = k \cdot 1176\) Thus, \(k = \frac{140}{1176} = \frac{5}{42}\) Now rearrange the original equation to solve for \( d \): \(d^2 = \frac{W}{k \cdot L}\) \(d = \sqrt{ \frac{W}{k \cdot L} } = \sqrt{ \frac{W}{\frac{5}{42} \cdot L} } = \sqrt{ \frac{42W}{5L} }\) Therefore, \( d = \sqrt{ \dfrac{42W}{5L} } \).

7x - 3 > 25 + 3x 7x - 3x > 25 + 3 4x > 28 x > 28/4 x > 7

The graph of the function f(x) is a curve that starts at (-2,0) and ends at (2,0) with a peak at (0,1). The range of values of x for which f(x) \le 0 is the interval where the function's graph is above or touches the x-axis. From the given graph, we can see that this interval is [-2, 2]. Therefore, the correct answer is x \(\leq\) - 2, 0 \(\leq\) x \(\leq\) 2

U7 = a + (7 - 1)d = a + 6d U3 = a + (3 - 1)d = a + 2d But U7 = 2(U3) ∴a + 6d = 2(a + 2d) a + 6d = 2a + 4d 2a - a + 4d - 6d = 0 a - 2d = 0 → eqn1 Sn = n/2 (2a + (n - 1)d) 42 = 4/2 (2a + (4 - 1)d) 42 = 2(2a + 3d) 21 = 2a + 3d → eqn2 eqn1 * eqn2 0 = 2a - 4d 21 = 7d ∴d = 21/7 d = 3