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Let x represent the first number; Then, the other is (5 − x) , since their sum is 5 and their product is 14 x(5 − x) = 14 5x − x2 = 14 x2 − 5x − 14 = 0 (x2 − 7x) + (2x − 14) = 0 x(x − 7) + 2(x − 7) = 0 (x − 7)(x + 2) = 0 Either (x − 7) = 0 or (x + 2) = 0 x = 7 or x = − 2 The two numbers are 7 and − 2

\( 5^{(x − 2)} = (125^{−1} )^{(x + 3)}\\ 5^{(x − 2)} = (5− 3)^{x + 3}\\ 5^{(x − 2)} = 5^{− 3x − 9}\\ x + 3x = − 9 + 2\\ 4x = − 7\\ x = \frac{−7}{4} \)

Each exterior angle = 180 − 120=60 Exterior angle =360/n 60 = 360/n n =360/6 = 6

25% of N80 = 25/100 × 80 = N20.00

2.04 x 3.7 = 204 x 37 ----- 7548 = 7.548 = 7.55(2dp)

Let assume the cost price is 100% Marked up price + cost price = 20 + 100 = 120% Discount at 20% = 20/100 × 120 % of cost price Selling price = cost price − gain = (120 − 24)% of cost price = 96% of cost price Loss = (100− 96)% of cost price = 4% of cost price ∴ He will wake 4% loss

5÷ (2 − √3) Using conjugate surds [5 ÷ (2 − √3)]×(2 + √3) ÷ (2 + √3)] [5(2+√3)÷((2− √3))2] [5(2 +√3) ÷ (2)2− (√3)2] [5(2 +√3) ÷(4− 3)] = 5(2 + √3)

The trinomial = \( a^2 − 4a + 4 \\ a^2 − b^2− 4a + 4 = (a^2 − 4a + 4) − b^2 \\ (a^2 − 2a − 2a + 4) − b^2 \\ a(a − 2)− 2(a − 2)] − b^2 \\ (a − 2)2 − b^2 \\ (a − 2 + b)(a − 2 − b )\)

use trial and error method, By substituting any of x = 2, 5, 15, 18 into 7 + x = 4 (mod 9) any value of x that gives 7 + x = 4 (mod 9) is the correct value of x In this case, x = 15.

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N1.00 = $0.075 N X = $15,000 X = \(\frac{1.00 \times 15000}{0.075}\) = N200,000.00

\(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1\) \(\frac{3^{3x} \times 3^{1 - x}}{3^{2(2 - x)}} = 3^0\) \(3^{3x} \times 3^{1 - x} \div 3^{4x} = 3^0\) \(3^{(3x + 1 - x - 4x)} = 3^0\) \(3^{(1 - 2x)} = 3^0\) since the bases are equal, 1 - 2x = 0 - 2x = -1 x = \(\frac{1}{2}\)

GIVEN: 2, 5, 10, 17, 26,. The numbers are increasing by odd value. 1st term = 2 2nd term = 5 3rd term = 10 4th term = 17 5th term = 26 6th term = 37 7th term = 50 Always add 2 to the difference between the two preceding numbers and add this to the later number to get the next term.

If 2n = y, then, 2\(^{(2 + \frac{n}{3})}\) = 22 x 2\(^\frac{n}{3}\) = 4 x (2n)\(^{\frac{1}{3}}\) But y = 2n, hence 2\(^{(2 + \frac{n}{3})}\) = 4 x y\(^{\frac{1}{3}}\) = 4y\(^\frac{1}{3}\)

6ax - 12by - 9ay + 8bx = 6ax - 9ay + 8bx - 12by = 3a(2x - 3y) + 4b(2x - 3y) = (3a + 4b)(2x - 3y)