Advanced mathematics (Math Problem Sample)

ADVANCED MATHEMATICS III BEC3206 * Let: Heifer= x Cow= x+ 4812 x(x+423)= 8 x2 + 143x = 8 x2 + 143x + C = 8 + C C= ( b2)2 C=( 143 × 12)2 C= (146)2 x2 + 143x+ (146)2 = 8+ (146)2 (x + 146) 2 = 1219 √ (x + 146) 2 = √1219 x + 146 = ± 113 x = ± 113 - 146 x = 43 or -6 Heifer is x equivalent to 113 years Cow= 6 years. * x2= xy- 4 y = 2x- 3 let: y= 2x-3 therefore x2 = x(2x-3) – 4 x2 = 2x2 – 3x – 4 then x2- 3x- 4= 0 x2- 3x + C = 4+ C C= (b2 )2 C=( 32 )2 = 94 (x- 32 ) 2 = 4 + 94 √ (x- 32 ) 2 = √254 (x- 32 ) = ± 52 : x = ±52 + 32 When positive When negative x= 4 x= -1 y= 5y = -5 f. * Value of x = 22 given by total frequency less the sum of frequencies to individual classes. * Modal class is given by the highest frequency (35 – 44) * Median mark Median class is given by: N2 which is higher than cumulati...