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Area of triangle PQR = \(\frac{1}{2}\) x QR x Pc Where QR = base of triangle PQR, Pc = height of triangle PQR Area of triangle PQR = \(\frac{1}{2}\) x 20 x 13 = 10 x 13 = 130cm\(^2\).

Given that a * b = \(\frac{a + b}{\text{ab}}\) + (a - b) 3 * 2 = Given that a * b = \(\frac{3 + 2}{3.2}\) + (3 - 1) = \(\frac{5}{6}\) + 1 = \(\frac{5}{6}\) + \(\frac{6}{6}\) = \(\frac{11}{6}\)

Given: 125\(^{−\frac{1}{3}}\) \times 49\(^{−\frac{1}{2}}\) \times 10\(^0\) 125\(^{-1/3}\) = (5\(^3\))\(^{-1/3}\) = 5\(^{-1}\) = \(\frac{1}{5}\). 49\(^{-1/2}\) = (7\(^2\))\(^{-1/2}\) = 7\(^{-1}\) = \(\frac{1}{7}\). 10\(^0\) = 1. Product = \(\frac{1}{5}\) x \(\frac{1}{7}\) x 1 = \(\frac{1}{35}\).

Midpoint = [\(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2}\)] Midpoint = [\(\frac{(-3 + 5)}{2}\), \(\frac{(4 + 6)}{2}\)] = (\(\frac{2}{2}\), (\(\frac{10}{2}\)) = (1, 5).

Sin A = \(\sqrt{(1 - (\frac{12}{13})^2)}\) = \(\sqrt{(1 - \frac{144}{169}}\)) = \(\sqrt{(\frac{25}{169})}\) = \(\frac{5}{13}\). Tan A = (\(\frac{5}{13}\)) \(\div\) (\(\frac{12}{13}\)) = \(\frac{5}{12}\).

Mean = \(\frac{(0 + x + 2 + 3x + 6 + 4x + 8)}{4}\) = 4. Simplify numerator: 8x + 16. So \(\frac{(8x + 16)}{4}\) = 4. 8x + 16 = 16. 8x = 0. x = 0.

Convert the mixed number slope to an improper fraction: 3 \(\frac{1}{5} = \frac{16}{5}\) Use the point-slope form with the given point (3,15) y - 15 = \(\frac{16}{5}\)(x - 3) Multiply both sides by 5 to clear the fraction: 5(y−15) = 16(x−3) 5y − 75 = 16x − 48 Rearrange terms to standard form: 5y − 16x − 75 + 48 = 0 5y − 16x − 27 = 0

Given (\(\frac{3}{4} \div 2\frac{1}{4}\)) of 1 \(\frac{7}{11}\) (3 \(\frac{2}{3}\) - \(\frac{15}{6}\)) 1. Simplify the division in the first parentheses: \(\frac{3}{4} \div 2 \frac{1}{4} = \frac{3}{4} \div \frac{9}{4} = \frac{3}{4} \times \frac{4}{9} = \frac{3}{9} = \frac{1}{3}\). 2. Convert the mixed number: \(1 \frac{7}{11} = \frac{18}{11}\). 3. Simplify the subtraction in the second parentheses: \(3 \frac{2}{3} - \frac{15}{6} = \frac{11}{3} - \frac{5}{2} = \frac{22}{6} - \frac{15}{6} = \frac{7}{6}\). 4. The "of" indicates multiplication, so multiply the results: \(\frac{1}{3} \times \frac{18}{11} \times \frac{7}{6} = \frac{126}{198} = \frac{7}{11}\).

Total fruits = 6 + 11 + 13 = 30 Favorable (grape or banana) = 6 + 11 = 17 Probability = \(\frac{17}{30}\)

Let the first term be \( a \), and the common ratio be \( r \). Given: - Second term: \( ar = 1 \) - Fifth term: \( ar^4 = \frac{1}{8} \) From the first equation, we have: \(a = \frac{1}{r}\) Substituting into the second equation: \(\frac{1}{r} r^4 = \frac{1}{8}\) \(r^3 = \frac{1}{8}\) Taking the cube root gives: \(r = \frac{1}{2}\) Thus, the common ratio is \(\frac{1}{2}\)

Given: \(A = \frac{\theta}{360} \pi r^2 \) To make \(\theta\) the subject, isolate \(\theta\). Multiply both sides by 360: \(360A = \theta \pi r^2 \) Divide both sides by \(\pi r^2\): \( \theta = \frac{360A}{\pi r^2} \) The rearranged formula is: \( \theta = \frac{360A}{\pi r^2} \)

The function is \( y = -2x^2 - 4x + 2 \). This is a quadratic function with a negative leading coefficient (\( a = -2 < 0 \)), so the parabola opens downwards and has a maximum value (not a minimum). The minimum value would be \( -\infty \) as \( x \to \pm \infty \). Method 1: Vertex formula For \( y = ax^2 + bx + c \), the vertex occurs at \( x = -\frac{b}{2a} \). Here, \( a = -2 \), \( b = -4 \), \( c = 2 \). \( x = -\frac{-4}{2(-2)} = \frac{4}{-4} = -1 \) Substitute \( x = -1 \): \( y = -2(-1)^2 - 4(-1) + 2 = -2(1) + 4 + 2 = -2 + 4 + 2 = 4 \) Method 2: Completing the square \( y = -2x^2 - 4x + 2 = -2(x^2 + 2x) + 2 \) Complete the square inside: \( x^2 + 2x = (x + 1)^2 - 1 \) \( y = -2[(x + 1)^2 - 1] + 2 = -2(x + 1)^2 + 2 + 2 = -2(x + 1)^2 + 4 \) The maximum value is 4 (when \( x = -1 \)), and \( y \leq 4 \). Method 3: Calculus (derivative) \( \frac{dy}{dx} = -4x - 4 \) Set to zero: \( -4x - 4 = 0 \) → \( x = -1 \) Second derivative \( \frac{d^2y}{dx^2} = -4 < 0 \), confirming a maximum. \( y(-1) = 4 \) Conclusion: The function has "no minimum value" (it decreases without bound). The "maximum value" is 4.

Angle (Y) is (60\(^\circ\)). This construction forms an equilateral triangle, so all its interior angles, including (\(\angle\) Y), are (60\(^\circ\)).

Given the frequency distribution: \(\begin{array}{|c|c|} \hline \text{Class Interval} & \text{Frequency} \\ \hline 0-9 & 1 \\ 10-19 & 5 \\ 20-29 & 6 \\ 30-39 & 12 \\ 40-49 & 8 \\ 50-59 & 3 \\ \hline\end{array}\) The modal class is \(30-39\) with frequency \(f_1 = 12\), \(f_0 = 6\) (previous class), \(f_2 = 8\) (next class), and the lower boundary \(L = 29.5\) with class width \(h = 10\). Using the mode formula: \(\text{Mode} = L + \left( \frac{f_1 - f_0}{(f_1 - f_0) + (f_1 - f_2)} \right) \times h\) Substituting the values: \(\text{Mode} = 29.5 + \left( \frac{12 - 6}{(12 - 6) + (12 - 8)} \right) \times 10\) \(\text{Mode} = 29.5 + \left( \frac{6}{6 + 4} \right) \times 10 = 29.5 + \left( \frac{6}{10} \right) \times 10 = 29.5 + 6 = 35.5\)

To convert \(54_{10}\) to base 4: \(54 \div 4 = 13\) with remainder \(2\). \(13 \div 4 = 3\) with remainder \(1\). \(3 \div 4 = 0\) with remainder \(3\). Collecting the remainders from bottom to top, we have: \(54_{10} = 312_4\) Thus, \(X = 312.000\).