given that 100.48 = x , 100.70 = y and xz = y ² , then the value of z is close to
sol . xz = y ² ⇔ ( 10 0.48 ) z ( 100.70 ) 2 ( 10 0.48 z ) = 10 ( 2 × 0.70 ) = 101.40 ⇔ 0.48 z = 1.40 ⇔ z = 140 / 48 = 35 / 12 = 2.9 ( approx ) . answer d
divide(multiply(subtract(100.7, const_100), const_2), subtract(100.48, const_100))