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hyp = 5cm, adj = \(\frac{8cm}{2}\) 4cm, opp = xcm Pythagoras theorem: hyp\(^2\) = opp\(^2\) + adj\(^2\) 5\(^2\) = x\(^2\) + 4\(^2\) x\(^2\) = 25 - 16 x = \sqrt9 x = 3cm

\((101_{two})^3\) = \((5_{ten})^3\) 5\(^3\) = 125 And 125 = \(1111101_{two}\)

Using the expansion method on (10110.011)\(_2\) (1 * 2\(^4\)) + (0 * 2\(^3\)) + (1 * 2\(^2\)) + (1 * 2\(^1\)) (0 * 2\(^0\)) + (0 * 2\(^{-1}\)) + (1 * 2\(^{-2}\)) + (1 * 2\(^{-3}\)) (1*16) + 0 (1*4) + (1*2) + 0 + 0 + (1*\(\frac{1}{4}\)) + (1*\(\frac{1}{8}\)) 16 + 4 + 2 + (0.25 + 0.125) 22 + 0.375 (10110.011)\(_2\) = 22.375\(_{10}\)

\((8^{-1})^{k+2}\) = \((8^{0})\) base 8 cancel out on both sides -1(k+2) = 0 -k -2 = 0 : k = -2

where Log\(_2\) 8\sqrt2 → Log\(_2\) \sqrt128 → Log\(_2\) 128\(^\frac{1}{2}\) = \(\frac{1}{2}\) * (Log\(_2\) 128) → \(\frac{1}{2}\) * (Log\(_2\) 2\(^7\)) = 7 * \(\frac{1}{2}\) * (Log\(_2\) 2) where (Log\(_2\) 2) = 1 → 7 * \(\frac{1}{2}\) * 1 = \(\frac{7}{2}\) or 3.5

Rationalization with conjugate \({3+\sqrt2}\) \(\frac{1}{3-\sqrt2}\) → \(\frac{1 * [3+\sqrt2]}{[3-\sqrt2][3+\sqrt2]}\) = \(\frac{3+\sqrt2}{9 -3\sqrt2 + 3\sqrt2 - \sqrt4}\) = \(\frac{3+\sqrt2}{9 - 2}\) → \(\frac{3+\sqrt2}{7}\) = \(\frac{3}{7}\) + \(\frac{\sqrt2}{7}\)

Numerator: 96 → 37.5% of 96 = 36 Decreased by 36 → 96 - 36 New numerator = 60 Denominator: 50 → 20% of 50 = 10 Decreased by 10 → 50 - 10 = 40 New Denominator = 40 New fraction = \(\frac{60}{40}\) or 1.5

5 crates = 30 eggs * 5 = 150eggs 1/2 crate = 15 eggs Total sold = 165 remainder = 200 - 165 = 35 And 35 is equivalent 17.5% of the realized eggs

It is given that the cost of the item reduces to #3800.00 after the discount on the item. Let original price = x x – (20% of x) = 100 ⇒ x – 0.2x = 100 ⇒ 0.8x = #3800 ⇒ x = \(\frac{3800}{0.8}\) ⇒ x = 4,750

-The shaded/stripped portion in the diagram is set F only. -This implies that sets E and G though present in the Universal set, are not included in the expression. -Mathematically; we have E' n F n G'

n(PuQ) = n(p) + n(q) - n(PnQ) 40 = 20 + 30 - n(PnQ) n(PnQ) = 50 - 40 = 10

|A| = (2*3) - (1*3) → 6 - 3 = 3

if 20 students = 10000 : #500 per student #1,000,000 \div 500 = 2,000 cover students

Three times a certain number (x) = 3x minus 2 = -2 is less than = < the number minus 6 = x -6 Mathematically; 3x - 2 < x - 6 collect like terms 3x - x < -6 + 2 2x < -4 x < -2 is the possible values of x

square both sides to remove the big square root → r\(^2\) = \(\frac{3v}{\pih}\) cross multiply 3v = r\(^2\) * \pih v = \(\frac{\pir^2h}{3}\)