The following equation can be used to relate the density of liquid water to celsius temperature in the range from o°c to about 20°c, rho (g/cm3) = 0.99984 + (1.6945 x 10-2 t) – (7.987 x 10-6 t2)
1 + (1.6880 x 10-2 t)

(a) to four significant figures, determine the density of water at 10°c
(b) at what temperature does water have a density of 0.99860 g/cm3?

a) T = 10°C ⇒ ρ = 0.9997 g/cm³

b) ρ = 0.99860 g/cm³ ⇒ T = 16.7°C

Explanation:

∴ ρ = (0.99984 + (1.6945 E-2 T) - (7.987 E-6 T²))  / (1 + (1.6880 E-2 T))

a) T = 10 °C

⇒ ρ = (0.99984 + (1.6945 E-2(10)) - (7.987 E-6(10)²)) / (1 + (1.6880 E-2(10)))

⇒ ρ = (0.99984 + 0.16945 - 7.987 E-4) / ( 1 + 0.1688 )

⇒ ρ =  0.9997 g/cm³

b) ρ = 0.99860 g/cm³³

⇒ 0.99860m = (0.99984 + 1.6945 E-2T - 7.987 E-6T² ) / (1 + 1.6880 E-2 T)

⇒ 0.99860 * (1+ 1.6880 E-2T) = 0.99984 + 1.6945 E-2T - 7.987 E-6T²

⇒ 0.99860 + 0.016856T = 0.99984 + 1.6945 E-2T - 7.987 E-6T²

⇒ 7.987 E-6T² - 8.8632 E-5T - 1.24 E-3 = 0

⇒ T = 16.7°C

since electrons are negative the answer is neagtaive

i think it's letter a.