Get Started with ixamBee
Start learning 50% faster. Sign in now
We have the formula, Difference = (P×R²(300+R))/100³ 512 = (P×(20)²(300+20))/((100)³) (512×100×100×100)/(20×20×320) = P P = Rs. 4000
I: x2 + 31x + 228 = 0
II: y2 + 3y – 108 = 0
I. x2 - 11x + 24 = 0
II. y² - 5y + 6 = 0
I. 8a2 – 22a+ 15 = 0
II. 12b2 - 47b + 40=0
I. p2+ 2p – 8 = 0 II. q2 – 5q + 6 = 0
'r' is one of the roots of equation (r - 1)p2 - (5r + 2)p + 12 = 0 and sum of roots is (32/5), Determine the two roots of the equation.
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the...
I. 64x2 - 64x + 15 = 0
II. 21y2 - 13y + 2 =0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 19x² - 92x + 73 = 0
Equation 2: 17y² - 76y + 59 = 0
I. 144x² - 163x - 65 = 0
II. 91y² - 128y -48 = 0
I. 10p² + 21p + 8 = 0
II. 5q² + 19q + 18 = 0
