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x\(^3\) - 19x - 30 test x = 5 5\(^3\) −19(5) − 30 = 125 − 95 − 30 = 0 (x - 5) is a factor. x\(^3\) - 19x - 30 divided by (x - 5) = x\(^2\) + 5x + 6 factorizing x\(^2\) + 5x + 6, we have (x+2)(x+3) (x+2)(x+3)(x - 5) = 0 x = -2, -3, and 5.

S\(_\infty\) = \(\frac{a}{1- r}\) since 1 > r S\(_\infty\) = 100, r = \(\frac{-1}{2}\), a = ? 100 = \(\frac{a}{1- \frac{-1}{2}}\) = \(\frac{a}{\frac{3}{2}}\) ( - - = +) a = 100 x \(\frac{3}{2}\) = 150 Therefore, the first term (a) = 150

T\(_7\) = ar\(^{n-1}\) = ar\(^{6}\) a = - 10, r = - 5 T\(_7\) = -10(-5)\(^{6}\) T\(_7\) = - 10 x 15625 = - 156250

p * q = 2p + pq + q (p*2) - (p*1) = 40 firstly, p = p and q = 2, secondly p = p and q = 1 then, 2p + 2p + 2 - (2p + p + 1) = 40 4p + 2 - 3p - 1 = 40 p + 1 = 40 p = 40 - 1 = 39

Given the operation defined by \( x * y = x^y \), \[x * 2 = x \implies x^2 = x\] Rearranging gives: \[x^2 - x = 0 \implies x(x - 1) = 0\] Thus, the possible values of \( x \) are: \[\boxed{0 \text{ and } 1}\]

From the diagram above, Calculate the distance between any two vertices as follows. PQ = \(\sqrt{(6 - 3)^2 + (2-2)^2}\) = \(\sqrt{9}\) = 3 QR = \(\sqrt{(6 - 4)^2 + (2-5)^2}\) = \(\sqrt{13}\) PR = \(\sqrt{(4 - 3)^2 + (5 - 1)^2}\) = \(\sqrt{10}\) \(\therefore\), perimeter = 3 + \(\sqrt{13}\) + \(\sqrt{10}\)

midpoint(x) = \(\frac{x_1 + x_2}{2}\) midpoint(y) = \(\frac{y_1 + y_2}{2}\) Given (9, 6) midpoint(x) = \(\frac{x + 10}{2}\) = 9 x + 10 = 2 x 9 = 18 x = 18 - 10 = 8.

From the diagram above: Tan 30º = \(\frac{15}{x}\) \(\frac{\sqrt{3}}{3}\) = \(\frac{15}{x}\) x\(\sqrt{3}\) = 3 x 15 = 45 x = \(\frac{45\sqrt{3}}{3}\) = 15\(\sqrt{3}\)m ( this is after rationalizing the denominator)

\(\frac{\text{dy}}{dx}\) = (5x + 1)\(^4\) = 20(5x + 1)\(^3\) let u = (5x + 1) = \(\frac{du}{dx}\) = 5 y = u\(^4\) then \(\frac{dy}{du}\) = 4u\(^3\) Therefore, \(\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx}\) = 5 x 4u\(^3\) = 20(5x + 1)\(^3\)

Cos25º - Sin 25º The differentiation of a constant = 0, But based on the Options provided Cos 25º when differentiated = -Sin 25º and Sin25º = + Cos25º Putting these together, we have - Sin25º - Cos25º = - ( Sin25º + Cos25º)

MODE = 3, the most frequent number.

Total ages of four female teachers = 4 x 40 = 160 Total ages of eight male teachers = 8 x 25 = 200 Total ages of all the teachers in the school = 160 + 200 = 360 Average ages of all the teachers = \(\frac{\text{ total ages of all teachers}}{\text{total number of teachers}}\) = \(\frac{360}{12}\) = 30.

Mean = \(\frac{ 13 + 16 + x + 18 + 21 + 2x + 35}{7}\) = 22 103 + 3x = 154 3x = 154 - 103 = 51 x = \(\frac{51}{3}\) = 17

To find the variance from the standard deviation: Given: Standard deviation = 12.34 Variance = (12.34)\(^2\) = 152.2756 Rounding to the nearest whole number: Variance ≈ 152. ​

mean = \(\frac{ 4 + 5 + 6 + 7 + 8}{5}\) = 6 x x - mean (x - mean)\(^2\) 4 4 - 6 = -2 4 5 5 - 6 = -1 1 6 6 - 6 = 0 0 7 7 - 6 = 1 1 8 8 - 6 = 2 4 Variance = \(\frac{10}{5}\) S.D = \(\sqrt{variance}\) = \(\sqrt{2}\)