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To find the inverse of the matrix \(A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\) We can use the formula for the inverse of a \( 2 \times 2 \) matrix: \(A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\) where \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \). Identify \( a, b, c, d \)} In our matrix, we have: \(a = 2, \quad b = 1, \quad c = 1, \quad d = 1\) Calculate the determinant \( ad - bc \) Calculating the determinant: \(ad - bc = (2)(1) - (1)(1) = 2 - 1 = 1\) Calculate the inverse: Now substituting into the formula for the inverse: \(A^{-1} = \frac{1}{1} \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}\) Thus, the inverse of the matrix \( \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \) is \(\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}\)

The determinant of the matrix \(P = \begin{pmatrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ -1 & 0 & 1 \end{pmatrix}\) is calculated as follows: \(|P| = 1 \cdot (4 \cdot 1 - 5 \cdot 0) - 0 \cdot (3 \cdot 1 - 5 \cdot -1) + (-1) \cdot (3 \cdot 0 - 4 \cdot -1) = 4 - 0 - 4 = 0\) Thus, the determinant of the matrix \( P \) is \(|P| = 0\).

To find the values of \( x \) where the curve \(y = x^3 + 2x^2 - 5x - 6\) crosses the x-axis, we set \( y = 0 \): \(x^3 + 2x^2 - 5x - 6 = 0\) Testing for rational roots, we find that \( x = -1 \) is a root. Factoring the polynomial gives: \(x^3 + 2x^2 - 5x - 6 = (x + 1)(x^2 + x - 6)\) Solving the quadratic \( x^2 + x - 6 = 0 \) using the quadratic formula yields: \(x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot -6}}{2 \cdot 1} = \frac{-1 \pm 5}{2}\) Thus, the solutions are: \(x = -1, \quad x = 2, \quad x = -3\) Therefore, the values of \( x \) where the curve crosses the x-axis are \(x = -1, \quad x = 2, \quad x = -3\).

x = 4 and y = 2

Let the number of students that offer both subject be x hence, the number of students offering physics = 3x The number of students offering chemistry = 6x The number of students offering physics only = 3x - x = 2x The number of students offering chemistry only = 6x - x = 5x So, 2x + 5x + x = 40 8x = 40 x = \(\frac{40}{8}\) = 5. Therefore, the number of students offering physics only = 2x = 2 x 5 = 10.

To evaluate the expression \( \log_{\sqrt{2}} 4 + \log_{\frac{1}{2}} 16 - \log_{4} 32 \), we can simplify each term separately. 1. Calculate \( \log_{\sqrt{2}} 4 \) \(\log_{\sqrt{2}} 4 = \frac{\log_2 4}{\log_2 \sqrt{2}}\) We know that \( \log_2 4 = 2 \) (since \( 2^2 = 4 \)) and \( \log_2 \sqrt{2} = \log_2 2^{1/2} = \frac{1}{2} \). Thus, \(log_{\sqrt{2}} 4 = \frac{2}{\frac{1}{2}} = 4\) 2. Calculate \( \log_{\frac{1}{2}} 16 \) \(\log_{\frac{1}{2}} 16 = \frac{\log_2 16}{\log_2 \frac{1}{2}}\) We know that \( \log_2 16 = 4 \) (since \( 2^4 = 16 \)) and \( \log_2 \frac{1}{2} = \log_2 2^{-1} = -1 \). Thus, \(\log_{\frac{1}{2}} 16 = \frac{4}{-1} = -4\) 3. Calculate \( \log_{4} 32 \) \(\log_{4} 32 = \frac{\log_2 32}{\log_2 4}\) We know that \( \log_2 32 = 5 \) (since \( 2^5 = 32 \)) and \( \log_2 4 = 2 \). Thus, \(\log_{4} 32 = \frac{5}{2}\) Now, we can substitute these results back into the original expression: \(\log_{\sqrt{2}} 4 + \log_{\frac{1}{2}} 16 - \log_{4} 32 = 4 - 4 - \frac{5}{2}\) Calculating this gives: \(4 - 4 - \frac{5}{2} = 0 - \frac{5}{2} = -\frac{5}{2}\) Thus, the final result is: \(-\frac{5}{2}\) = - 2.5

You can either multiply directly in base 4 or you can decide to convert to base 10 and do the multiplication and re-convert.

Using a venn diagram, let x = number who offer both Maths and Physics. So that (32-x) offer Maths and another (24-x) offer Physics. (32-x) + (24-x) + (x) + (4 who offer neither) = 40 => 60 - x = 40 => -x = -20 Therefore x = 20.

\((\frac{1}{0.06} \div \frac{1}{0.042})^{-1}\) = \((\frac{100}{6} \div \frac{1000}{42})^{-1}\) = \((\frac{100}{6} \times \frac{42}{1000})^{-1}\) = \((\frac{7}{10})^{-1}\) = \(\frac{10}{7}\) = 1.4285 \(\approxeq\) 1.43.

C.P = N1800 S.P = (270/5) x 40 = N2160.00 Profit = S.P - C.P = N2160 - N1800 = N360.00

\(1 - (\frac{7}{3} \times \frac{5}{4}) + \frac{3}{5}\) = \(1 - \frac{35}{12} + \frac{3}{5}\) = \(\frac{60 - 175 + 36}{60}\) = \(\frac{-79}{60}\)

\(\frac{\sqrt{98} - \sqrt{50}}{\sqrt{32}}\) = \(\frac{\sqrt{49 \times 2} - \sqrt{25 \times 2}}{\sqrt{16 \times 2}}\) => \(\frac{7\sqrt{2} - 5\sqrt{2}}{4\sqrt{2}}\) = \(\frac{2\sqrt{2}}{4\sqrt{2}}\) = \(\frac{1}{2}\)

45/2% + 95/2% = 70% 100% - 70% = 30% (i.e women) => 30% = 84 1% = 84/30 95/2% = (84/30 x 95/2) = 133.

\(\frac{9^{2x - 1}}{27^{x + 1}} = 1\) \(\implies 9^{2x - 1} = 27^{x + 1}\) \((3^{2})^{2x - 1} = (3^{3})^{x + 1}\) \(2(2x - 1) = 3(x + 1) \implies 4x - 2 = 3x + 3\) \(4x - 3x = 3 + 2 \implies x = 5\)

Convert to base 10, divide, and then re-convert to base 5.