Answer:
See below
Step-by-step explanation:
When you roll an 8-sided die twice, the sample space is the set of all possible pairs (x,y) where x is the first outcome and y is the second outcome.
The sample space is:
![[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(1, 7),(1, 8)\\(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(2, 7),(2, 8)\\(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(3, 7),(3, 8)\\(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(4, 7),(4, 8)\\(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),(5, 6),(5, 7),(5, 8)\\(6, 1), (6, 2), (6, 3), (6, 4)(6, 5),(6, 6),(6, 7),(6, 8)\\(7, 1), (7, 2), (7, 3), (7, 4)(7, 5),(7, 6),(7, 7),(7, 8)\\(8, 1), (8, 2), (8, 3), (8, 4)(8, 5),(8, 6),(8, 7),(8, 8)]](https://tex.z-dn.net/?f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
The sample space of the product xy of each outcome forms the required possibility diagram.
This is given as:

(x-1)(x^3+x^2+x+7) is your answer.
Answer:
45.5 ft sqaured
Step-by-step explanation:
Answer:

Step-by-step explanation:
To calculate the lenght of the diagonal d across the square, we can assume that the square it is compound of two right triangles. So, we can resolve this exercise using The Pythagorean Theorem.
<em>The Pythagorean theorem</em> states that in every right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the respective lengths of the legs. It is the best-known proposition among those that have their own name in mathematics.
If in a right triangle there are legs of length a and b, and the measure of the hypotenuse is c, then the following relation is fulfilled:
a is the height, b is the base, and c is
the hypotenuse.
To obtain the value of the hypotenuse

To find the value of the lenght of the diagonal d across the square, we have:
Where a = b = 20
Substituting the values

Round the answer to 2 decimal places
