2018 NECO MATHEMATICS OBJ AND ESSAY

[04/06 10:58 am] ENGR NASER DAHER 💪🏻💪🏻💪🏻: *MATHS OBJ*. 1CDAAEABAEC 11AEDDCDCDCC 21CEBDEDCBBC 31CBEEECBDCC 41DBCBCBDBDA. 51BCBDCDBCEE *_COMPLETED_* I still remain ENGR NASER DAHER💪🏻💪🏻💪🏻 [04/06 11:14 am] ENGR NASER DAHER 💪🏻💪🏻💪🏻: (1a) Log10(20x - 10) - log10(x+3) = log10^5 Log10(20x-10/x+3)= log10^5 20x - 10/x + 3 = 5 Cross multiplying 20x - 10 = 5(x + 3) 5(4x - 2) = 5(x + 3) 4x - 2 = X + 3 4x - x = 3+2 3x = 5 X = 5/3 OR 1 whole no 2/3 (1b) Let actual amount be #X 15% of #x = #600 15x/100 = 600 X = (100/15)*600 X = 100*40 X = 4,000 Actual amount = #4,000 [04/06 11:27 am] ENGR NASER DAHER 💪🏻💪🏻💪🏻: (2a) (X^2 Y^-3 Z)^3/4/X^-1 Y^4 Z^5 = (X^2)^3/4/X^-1 * (Y^-3)^3/4/Y^4 * Z^3/4/Z^5 = X^3/2/X^-1 * Y^-9/4/Y^4 * Z^3/4/Z^5 =X^3/2+1 * Y^-9/4-4 * Z^3/4-5 =X^5/2 * Y^-25/4 * Z^-17/4 =X^10/4 * Y^-25/4 * Z^-17/4 =(X^10/Y^25 Z^17)^1/4 (2b) √2/k + √2 = 1/k - √2 Multiply both sides by (k+√2)(k-√2) √2(k-√2) = k+√2 √2k-√2 = k+√2 √2k-k = 2+√2 K(√2 -1) = 2+√2 K = 2+√2/√2-1 K = -(2+√2)/1-√2 Rationalizing K = -(2+√2) * 1+√2/1-√2 K = -(2+√2)(1+√2)/1 - 2 K = (2+√2)(1+√2) K = 2+2√2 + √2+2 K = 4+3√2 [04/06 11:29 am] ENGR NASER DAHER 💪🏻💪🏻💪🏻: 9) Log10(20x-10) - log10(x+3)= log10^5 Log10(20x-10)/x+3) =log10^5 20x-10/x+3 = 5 Cross multiply 20x-10 = 5(x+3) 5(4x-2) = 5(x+3) 4x-x = 3+2 3x= 5 X=3/5 9b)  Let actual amount be #x 15% of #x = #600 15x/100 =600 x =(100/15) * 600 x = 100 * 40 x = 4,000 Actual amount = 4,000 [04/06 11:39 am] ENGR NASER DAHER 💪🏻💪🏻💪🏻: (4) Draw the diagram (i) Arc length = Tita/360*2πr = 72/360*2*22/7*14 =1/5*44*2 =88/5 =17.6cm (ii) Perimeter of Sector = arc length +2r =17.6+2(14) =17.6+28 =45.6cm (iii) Area of sector = Tita/360*πr² =72/360*22/7*14/1*14/1 =1/5*22*2*14 =616/5 =123.2cm2 [04/06 11:42 am] ENGR NASER DAHER 💪🏻💪🏻💪🏻: 11a) x+y/2 =11 x+y= 11*2 x+y= 22 ---(1) x-y= 4 ----(11) x+y = 22----(1) - x-y= 4----(11) ____________ 2y = 18 y= 18/2 y=9 Substitute y=9 in equ 1 x+9=22 x=22-9 x=13 x=13, y=9 x+y= 13+9= 22 Sum of the two number (11b) (6x + 3) dx (6x + 3)dx (6x +3)^6 - (6x + 3)^1 (6 x + 3)^5 (7776x^5 + 243) 38,880x/6 + 243 6480 x^6 + 243x 9(720x^6 + 27x) (11c) y = x² + 5x - 3 (x = 2) y = 2² + 5(2) - 3 y = 4 + 10 - 3 y = 14 - 3 y = 11 Gradient of the curve = 11 [04/06 12:16 pm] ENGR NASER DAHER 💪🏻💪🏻💪🏻: (5a) Mode = mass with highest frequency = 35kg Median is the 18th mass = 40kg. (5b) In a tabular form Under Masses(x kg) 30,35,40,45,50,55 Under frequency(f) 5,9,7,6,4,4 Ef = 35 Under X-A -10, -5, 0, 5, 10, 15 Under F(X-A) -50, -45, 0, 30, 40, 60 Ef(X - A) = 35 Mean = A + (Ef(X - A)/Ef) = 40 + 35/35 = 40 + 1 = 41kg I still remain ENGR NASER DAHER 💪🏻💪🏻💪🏻