Between 0 degrees celsius and 30 degrees celsius, the volume v (in cubic centimeters) of 1 kg of water at a temperature t is given approximately by the formula v=999.87−0.06426t+0.0085043t2−0.679 t3 find the temperature at which water has its maximum density. temperature = degrees celsius

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

   = (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT =

Let dD/dT = 0 to find the critical value we will get

 = 0

Using formula of quadratic, we get the roots:

T =  79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8)    a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

24

step-by-step explanation:

i assuming that 2 sides are 6 the other 2 sides are 4 cm so 6 times 4 is 24 : )