1.

f 4x + 5y23 and xy6, find the value of 16x2 + 25%

Answer»

(4x+5y)^2=16x^2+24y^2+40xy(23)^2=16x^2+25y^2+40(6)529-240=16x^2+25y^2289=16x^2+25y^2

Answer is 289

why u take 24



Discussion

No Comment Found