Cases Prudence has a special (cubic) die. The values on its face are the integers from 1 to 6, but they are not arranged ae in a normal die. When Prudence first tosses the die, the sum of the values on the four side faces is 15. In her second toss, the sum of these values is 12. Find what value appears in the face opposite 6 on Prudence’s special die. (Hint: what are possible values for the top and bottom face when the sum of the side faces is 12).

3

Step-by-step explanation:

first, we know that:

1 + 2 + 3 + 4 +5 +6 = 21

Now, which two numbers we should take out in order to have 15?

we can remove the 2 and the 4, or the 1 and the 5.

so here we have two possibilities, 2 and 4 are opposite, or 1 and 5 are opposite (they are located in opposite faces of the die)

in the other arrange, we have that removing two numbers we should get 12.

in order to reach 12, we should remove two numbers that add 9 together.

those can be 4 and 5, or 6 and 3.

Now, notice that in the first restriction we have that:

Or 2 and 4 are opposite,

or 1 and 5 are opposite.

So 4 and 5 can never be opposite, so we should have that 6 and 3 are opposite.

Then we can affirm that the value that appears in the face opposite to the 6, is the 3.

picture below

b. 24 mm

step-by-step explanation:

use the pythagorean theorem.

a^2 + b^2 = c^2

a^2 + 7^2 = 25^2

a^2 + 49 = 625

a^2 = 576

a = 24

answer: b. 24 mm