Let s represent the initial side length of the cube. Then the area of each face of the cube is A = 6s^2 (recalling that the area of a square of side length s is s^2).

a) Now suppose we double the side length. The total area of the 6 faces of the cube will now be A = 6(2s)^2, or 24s^2 (a 24 times larger surface area),

b) tripled: A = 6(3s)^2 = 54x^2

c) quadrupled? A = 6(4s)^2 = 96s^2

6 0

3 years ago

A cold drink is poured out at 52°F. After 2 minutes of sitting in a 72°F room, its temperature has risen to 55°F. Find an equati

I am Lyosha [343]

Answer:

The model for the temperature of the drink can be written as

$T=72-20e^{-0.08t}$

Step-by-step explanation:

For a cold drink in a hotter room, we can say that the rate of change of temperature of the drink is proportional to the difference of temperature between the drink and the room.

We can model that in this way

$\frac{dT}{dt}=k*(T_r-T)$

If we rearrange and integrate

$\int\frac{dT}{(T-Tr)} =-k*\int dt\\\\ln(T-T_r)=-kt+C1\\\\T-T_r=Ce^{-kt}\\\\T=T_r+Ce^{-kt}$

We know that at time 0, the temperature of the drink was 52°F. Then we have:

$T=T_r+Ce^{-kt}\\\\52=72+Ce^0=72+C\\\\C=-20$

We also know that at t=2, T=55°F

$T=T_r+Ce^{-kt}\\\\55=72-20e^{-k*2}\\\\e^{-k*2}=(72-55)/20=0.85\\\\-2k=ln(0.85)=-0.1625\\\\k=0.08$

The model for the temperature of the drink can be written as

$T=72-20e^{-0.08t}$

7 0

3 years ago

Help I need to complete this before 3:00 pm

Oduvanchick [21]

It would be 64 x 8 x 8 =

3 0

3 years ago

One urn contains one blue ball (labeled b1) and three red balls (labeled r1, r2, and r3). a second urn contains two red balls (r

Flura [38]

Answer:

12 possibilities

Step-by-step explanation:

In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.

The same thing occurs in the second urn, as all balls have different labels.

The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).

For the first urn, we have a combination of 4 choose 2:

C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities

For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.

In total we have 6 + 6 = 12 possibilities.

3 0

3 years ago

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