The basketball teams of north middle school and south middle school played against each other twice this season. in the first game, norths score was two-thirds of souths score. in the second game, norths score increased by seven points from the first hame while souths score decreased by seven seven points from the first game. in the second game souths score was three less than norths score, what was the score for each team in both games

North team score is n, south team score is s.
first game equation:
n = (2/3)s
second game equation:
s - 7 = n + 7 - 3

So we have 2 equations and 2 unknowns to solve:
n = (2/3)s
s - 7 = n + 7 - 3
lets substitute the first into the second:
s -7 = (2/3)s + 7 - 3
multiply all equation by 3 to eliminate fraction:
3s - 21 = 2s + 21 - 9
s = 21 + 21 - 9
s = 32
that is the souths score of the first game, now substitute in the first equation:
n = (2/3)s
n = (2/3)(32)
n = 64/3
n = 21
that is the score of the norths in the first game.
For the second game the norths increased the score by seven that is 21 + 7 = 28
While the souths decreased the score by 7, 32 - 7 = 25

The largest possible volume of the box is 2000000 cubic meters.

Step-by-step explanation:

The volume (), in cubic centimeters, and surface area (), in square centimeters, of the box with a square base are described below:

(1)

(2)

Where:

- Side length of the base, in centimeters.

- Height of the box, in centimeters.

By (2), we clear within the formula:

And we apply in (1) and simplify the resulting expression:

(3)

Then, we find the first and second derivatives of this expression:

(4)

(5)

If and , then we find the critical value of the side length of the base is:

Then, we evaluate this result in the expression of the second derivative:

By Second Derivative Test, we conclude that critical value leads to an absolute maximum. The maximum possible volume of the box is:

The largest possible volume of the box is 2000000 cubic meters.