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\(\frac{3}{4}t+\frac{1}{3}(21-t) = 11; \frac{3t}{4} + \frac{7}{1} - \frac{t}{3} = \frac{11}{1}\\ \frac{3 \times 3t + 7\times 12 – 4 \times t = 11 \times 12}{12}\\ 9t + 84 – 4t = 132; 5t = 132 – 84\\ 5t = 48; t = \frac{48}{5} = 9\frac{3}{5}\)

\(1 + 2^2 = 5 + 3^2 = 14 + 4^2 = 30 + 5^2 = 55\) \(55 + 6^2 = 91 + 7^2 = 140\). 91, 140 are the next two terms.

10 – 3x - x2 = 0; 10 – 5x + 2x - x2 = 0 5(2 – x) + x(2 – x) = 0; (2 – x)(5 + x) = 0 2 – x = 0 or 5 + x = 0; x = 2 or x = -5

The general quadratic equation x2 - (sum of roots) x + (product of root) = 0 Comparing with the given equation 2x2 +3x2 - 9 = 0; \(x^2 + \frac{3}{2}x - \frac{9}{2} = 0\\ Sum \hspace{1mm} of \hspace{1mm}roots = -\frac{3}{2}\)

\(\begin{matrix} x & x-m & |x-m| \\ 6 & -2 & 2 \\ 7 & -1 & 1 \\ 8 & 0 & 0 \\ 9 & 1 & 1 \\ 10 & 2 & 2 \\ & & 6 \end{matrix} \) Mean deviation\(=\frac{\sum|x-m|}{n}=\frac{6}{5} = 1.2\)

x2 + 3x – 4 = 0 has roots value at the curve makes touches with the x – axis. The curve contact is at x = -4 and +1.

X2 - (sum of the roots)x + (product of the roots) = 0 Sum of the roots \(= -\frac{1}{2}+\frac{3}{4} = \frac{-2+3}{2}=\frac{1}{4}\) Product of the roots = \(-\frac{1}{2}\times \frac{3}{4}=\frac{-3}{8}\\ X^2-\left(\frac{1}{4}\right)x-\frac{3}{8} = 0. Taking \hspace{1mm}the\hspace{1mm} common\) \(8x^2-2x-3=0\);

\(4 \times 3^{(3 - n)}\) for the third term, n = 3 then \(4 \times 3^{(3 - 3)}\) = \(4 \times 3^{0}\) = 4 x 1 = 4.

No expanded working has been attached for this question yet. The verified answer is Option B.

No expanded working has been attached for this question yet. The verified answer is Option A.

(a) \(px + qy = 5 \) When x = 1 and y = -2, we have \(p - 2q = 5 .... (1)\) \(qx + py = -10 \) When x = 1 and y = -2, we have \(q - 2p = -10 .... (2)\) Multiplying (1) by -2, we have \(-2p + 4q = -10 ... (1a)\) \(\equiv 4q - 2p = -10\) (2) - (1a) : \(q - 4q = -10 + 10\) \(-3q = 0\) \(q = 0\) Putting q = 0 in (1a), we have \(-2p = -10 \implies p = 5\) Hence, p = 5 and q = 0. (b) \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\) \((4r - 3)(3r + 4) = (2r - 1)(6r + 1)\) \(12r^{2} + 16r - 9r - 12 = 12r^{2} + 2r - 6r - 1\) \(12r^{2} - 12r^{2} + 7r + 4r = - 1 + 12\) \(11r = 11\) \(r = 1\).

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 \(|P| = 0\).

5 + x \(\leq\) 8 and 13 + x \(\geq\) 7 5 + x \(\leq\) 8 X \(\leq\) 8 – 5 X \(\leq\) 3 And 13 + x \(\geq\) 7 X \(\geq\) 7 – 13 X \(\geq\) -6 Combining the two -6 \(\leq\) x \(\leq\) 3

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