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The shaded portion shows the elements in Q and R, but not in P, Q n R n P'

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

5000 : 5000 * 500 5000 : 2500000 5 : 2500 1 : 500

Convert each value to decimal (base 10) \(1011_{two}\) → 11 + \(1101_{two}\) → 13 + \(1001_{two}\) → 9 - \(111_{two}\) → 7 : 11 + 13 + 9 - 7 = 26 26 → \(11010_{two}\)

\((216^{\frac{1}{3}})^{-2}\) = \(216^{\frac{-2}{3}}\) = (3\sqrt216)\(^{-2}\) = (6)\(^{-2}\) = \(\frac{1}{6^2}\) =\(\frac{1}{36}\)

LCM of 15abc and 20a\(^2\)bc = 60a\(^2\)bc Factors of 15abc = 1, 3, 5,15, a, b, c and of 20a\(^2\)bc = 1, 2, 4, 5, 10, 20, a, b, c HCF of 15abc and 20a\(^2\)bc = 5abc → LCM \div HCF = 60a\(^2\)bc \div 5abc = 12a

|AB| = 4cm |BC| = 8cm In right-angled BAC; 8\(^2\) = 4\(^2\) + |AC|\(^2\) |AC|\(^2\) = 8\(^2\) - 4\(^2\) |AC\\(^2\) = 64 - 16 → 48 |AC| = \(\sqrt{48}\)cm → \(4\sqrt{3}cm\) The area of rectangle = L x B = |AB| x |AC| = \((4 \times 4\sqrt{3}cm^2\) =\(16\sqrt{3}cm^2\)

\(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div16\frac{1}{2}; \frac{15}{2} - \left(\frac{5}{2}+\frac{3}{1}\right)\div \frac{33}{2}\\ \frac{15}{2} - \left(\frac{5+6}{2}\right)\div \frac{33}{2}; \frac{15}{2} - \frac{11}{2} \div \frac{33}{2}\\ \(\frac{11}{2}\) \div \(\frac{33}{2}\) = \(\frac{11}{2}\) * \(\frac{2}{33}\) = \(\frac{11}{33}\) or \(\frac{1}{3}\) (when simplified) \(\frac{15}{2}\) - \(\frac{1}{3}\) = \(\frac{43}{6}\) 7.166 The nearest whole number is 7

Given that \( P \propto \frac{1}{\sqrt{r}} \), we can express this relationship as: \(P = k \cdot \frac{1}{\sqrt{r}}\) for some constant \( k \). From the information provided, when \( P = 3 \) and \( r = 16 \): \(3 = k \cdot \frac{1}{\sqrt{16}}\) Since \( \sqrt{16} = 4 \), we have: \(3 = k \cdot \frac{1}{4}\) Thus, \(k = 3 \cdot 4 = 12\) Now we can write the equation as: \(P = \frac{12}{\sqrt{r}}\) Next, we need to find \( r \) when \( P = \frac{3}{2} \): \(\frac{3}{2} = \frac{12}{\sqrt{r}}\) Cross-multiplying gives: \(3\sqrt{r} = 24\) Dividing both sides by 3: \(\sqrt{r} = 8\) Squaring both sides: \(r = 64\). Thus, r = 64

first term[a] = 3 * d → 3 * 8 = 24 sixth term of the A.P = a + 5d → 24 + [5 * 8] ⇒ 24 + 40 = 64

The sum of angles at point = 360o (x+10)o + (4x+50)o + 20o + 3xo + 2xo = 360o 10x + 80o = 360o 10x = 360o - 80o = 280o \(\frac{280}{10}=28^{\circ}\)

QOP = 180 - 100(straight line angle) → 80^\circ OPQ and OQP are isosceles of 40^\circ each. ∠QRS = ∠RPQ (theorem: angle in the alternate segment are equal) ∠RPQ = 40º ∠QRS = 40º

From the diagram above, Tan \(\alpha\) = \(\frac{\text{opp}}{\text{Adj}}\) Tan \(\alpha\) = \(\frac{23}{40}\) = 0.575 \(\alpha\) = Tan\(^{-1}\) 0.575 ≈ 29.899 In degrees per minute = 29 + 0.89 x 60 = 29^\circ 54¹

Tan θ = 5√3/5 Tan θ = √3 θ = 60o

Coordinates of the line are (x1 y1)(0,2) respectively and (x2 y2)(2,0) respectively Gradient (m) of the line = (y2 - y1) / (x2 - x1) = (0-6) / (2-0) = -6/2 = -3 Equation of the line y-y1 = m(x-x1) y - 6 = -3(x - 0) y - 6 = -3x y + 3x = 6 inequality; from the y axis, the shaded region is below the line ∴ y + 3x ∠ 6 and line is a broken line