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Area of \(\Delta\)QST = 52cm\(^2\); height = 8cm b=base length; but A = 1/2 b x h =52=1/2 x b x 8cm = 104 =8b b=104/8 = 13cm; ST= 13cm; PS = 6cm TP = 13 - 6 = 7cm; area of \(\Delta\)QST = 1/2 b x h = 1/2 x 7 x 8/1 = 28cm\(^2\)

No Log 5.34 0.7275 + 67.4 1.8287 2.5562 - 2.7 0.4314 Antilog = 133.2 2.1248 \(\therefore \frac{5.34 \times 67.4}{2.7} = 133.2\)

35 - 2b - b 2; 35 - 7b + 5b - b 2 7(5 - b) + 5(5 - b) (7 + b)(5 - b)

2x 2 + x - 15; 2x 2 + 6x - 5x - 15 (2x 2 + 6x) - ( 5x + 15) 2x(x + 3) - 5(x + 3) (2x - 5)(x + 3)

% error = error/exact x 100/1 = 0.09/1.23 x 100/1 = 7.3%

At least 2 subjects = 30 + 45 + 75 + 10 = 160 students

Let the integer = y \(\therefore\) 5y - 2y = 63 3y = 63 \(\implies\) y = 21

3\(^y\) = 243; 3\(^y\) = 3\(^5\) y = 5

since it is an open cylinder; r= 3.5cm = 7/2cm h= 8cm; S.A = 2πrh + πr 2 = πr(2h + r) (16 + 3.5); 11 x 19.5 = 214.5cm2

Married = 550 + 400 = 950 Married% = \(\frac {950}{2700}\) x \(\frac {100}{1}\) =35.2%

A = {a, b, c}; B = {a, b, c, d, e}; C = {a, b, c, d, e, f} (A \(\cup\) B) = {a, b, c, d, e} (A \(\cup\) C) = {a, b, c, d, e, f} (A \(\cup\) B) \(\cap\) (A \(\cup\) C) = {a, b, c, d, e}

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

\(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) = \(\frac{3^{-1}}{3^2}\) = \(\frac{1}{27}\) In full: To simplify the expression : \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) We start by rewriting the bases in terms of powers of 3. We know that: \(9 = 3^2 \quad \text{and} \quad 27 = 3^3\) Thus, we can express \(9^{-\frac{1}{2}}\) as follows: \(9^{-\frac{1}{2}} = (3^2)^{-\frac{1}{2}} = 3^{-1}\) Next, we rewrite \(27^{\frac{2}{3}}\): \(27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^{2}\) Substitute back into the expression: Now we substitute these values back into the original expression: \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}} = \frac{3^{-1}}{3^{2}}\) This simplifies to: \(\frac{1/3}{9} = \frac{1}{3} \cdot \frac{1}{9} = \frac{1}{27}\) Thus, the simplified form of \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\) is \(\frac{1}{27}\).

from the graph, the solution sets represented by OPQ is the shaded region where y \le x, y \ge 0, and y + x \le 7

The volume of a cone = \(\frac{1}{3}\) x \(\pi\) x r\(^2\) x h = \(\frac{1}{3}\) x \(\frac{22}{7}\) x \(\frac{7}{2}\) x \(\frac{7}{2}\) x 12 = 154cm\(^3\)