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V = \pil(R\(^2\) - r\(^2\)) Divide both sides by \pil \(\frac{V}{\pil}\) = (R\(^2\) - r\(^2\)) R\(^2\) = \(\frac{V}{\pil}\) + r\(^2\) Take the square root of both sides R = \(\sqrt{\frac{V}{\pil} + r^2}\)

Given: m: Edna is respectful, n: Edna is brilliant From the implication m ⇒ n: a) Contra-positive: If Edna is not brilliant (¬n), then she is not respectful (¬m): ¬n ⇒ ¬m b) Direct: If Edna is respectful (m), then she is brilliant (n) c) Invalid: If Edna is brilliant (n), we cannot conclude she is respectful (m) Thus, the valid conclusion is: ¬n ⇒ ¬m.

Let the number added to the fraction \(\frac{1}{8}\) be y So, \(\frac{1 + y }{8 + y}\) = \(\frac{1}{2}\) cross multiply 8 + y = 2 + 2y 2y - y = 8 - 2 y = 6.

Total oranges = 60, total ratio = 5 + 3 + 7 = 15 Justina"s shared of the 60 oranges = \(\frac{3}{15}\) x 60 = 12

The roots are \(\frac{2}{3}\) and - 1 Using x\(^2\) - (\alpha + \beta)x + \alpha\beta = 0 \alpha + \beta = \(\frac{2}{3}\) + (-1) = \(\frac{2}{3}\) - 1 = \(\frac{-1}{3}\) \alpha\beta = \(\frac{2}{3}\) x (-1) = \(\frac{-2}{3}\) x\(^2\) - (\(\frac{-1}{3}\) )x + (\(\frac{-2}{3}\)) = 0 multiply through by 3 3x\(^2\) + x - 2 = 0

Perimeter of a rectangle = 2(L + B) = length of rod = 44 m The ratio of length to breadth = 7: 4 That is, \(\frac{L}{B}\) = \(\frac{7}{4}\) So, L = \(\frac{7B}{4}\) Perimeter of a rectangle = 2(\(\frac{7B}{4}\)+ B) = 44 Divide through by 2 (\(\frac{7B}{4}\)+ B) = 22 (\(\frac{11B}{4}\)+ B) = 22 cross multiply 11B = 22 x 4 = 88 B = (\(\frac{88}{11}\) = 8m.

Cos 72º = \(\frac{d}{15}\) d = Cos 72 x 15 = 4.63 m ≈ 4.6m

Line OH divides Line AB into two halves, therefore, Line AH = Line HB = 20cm Considering \(\triangle\) AOH (\(\overline{OH})^2\) = 25\(^2\) - 20\(^2\) = 625 - 400 = 225 Taking the square root of both sides (\(\overline{OH}\) = \(\sqrt{225}\) = 15 cm

Using P\(_n\) = P\(_o\)(1 - \(\frac{r}{100})^n\) = 600,000(1 - \(\frac{10}{100})^2\) = 600,000(0.90)\(^2\) = $486,000.00

Sin 60º = \(\frac{d}{25}\) d = Sin 60º x 25 = 21.6506m ≈ 21.7m

Volume of a cuboid = LBH = 1560cm\(^2\) V = 15 x 8 x H = 1560 H = \(\frac{1560}{15 \times 8}\) = 13cm Total surface area of a cuboid = 2(LB + LH + BH) = 2((15 x 8) + ( 15 x 13) + ( 8 x 13)) = 2(120 + 195 + 104) = 2(419) = 838 cm\(^2\)

Given that \( 6244_x - 1621_x = 4323_x \), find the value of \( x \). {Step 1: Convert to decimal:} \[6244_x = 6x^3 + 2x^2 + 4x + 4\] \[1621_x = 1x^3 + 6x^2 + 2x + 1\] \[4323_x = 4x^3 + 3x^2 + 2x + 3\] f{Step 2: Set up the equation:} \[(6x^3 + 2x^2 + 4x + 4) - (1x^3 + 6x^2 + 2x + 1) = 4x^3 + 3x^2 + 2x + 3\] {Step 3: Simplify the left side:} \[(6x^3 - 1x^3) + (2x^2 - 6x^2) + (4x - 2x) + (4 - 1) = 5x^3 - 4x^2 + 2x + 3\] {Step 4: Set the equation:} \[5x^3 - 4x^2 + 2x + 3 = 4x^3 + 3x^2 + 2x + 3\] {Step 5: Combine like terms:} \[5x^3 - 4x^3 - 4x^2 - 3x^2 + 2x - 2x + 3 - 3 = 0\] \[x^3 - 7x^2 = 0\] {Step 6: Factor the equation:} \[x^2(x - 7) = 0\] {Step 7: Solutions:} \[x = 7 \quad (\text{valid}), \quad x = 0 \quad (\text{not valid})\] \Thus, the base \( x \) is 7

27x\(^2\) - 48y\(^2\) = 3(9x\(^2\) - 16y\(^2\)) difference of two squares 3((3x)\(^2\) - (4y)\(^2\)) = 3((3x - 4y)(3x + 4y))

\(\frac{x - 3}{4}\) + \(\frac{x + 1}{8}\) \(\geq\) 2 multiply through by the LCM of 4, 8, and 1 = 8 8(\(\frac{x - 3}{4}\)) + 8(\(\frac{x + 1}{8}\)) \(\geq\) 8 x 2 2(x - 3) + (x + 1) \(\geq\) 16 2x - 6 + x + 1 \(\geq\) 16 3x - 5 \(\geq\) 16 3x \(\geq\) 16 + 5 = 21 x \(\geq\) \(\frac{21}{3}\) x \(\geq\) 7

Source: Myschool Past Questions Archive (Record #71824)