I would say the answer is 8.25
Answer:
I just do a' as a sample. You calculate b' and c'
Step-by-step explanation:
Now, calculate b x c
![\left[\begin{array}{ccc}i&j&k\\2&3&1\\1&-1&-2\end{array}\right] =](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2%263%261%5C%5C1%26-1%26-2%5Cend%7Barray%7D%5Cright%5D%20%3D%3C-5%2C%205%2C-5%3E)
i need the answer to this lol
We can start solving this problem by first identifying what the elements of the sets really are.
R is composed of real numbers. This means that all numbers, whether rational or not, are included in this set.
Z is composed of integers. Integers include all negative and positive numbers as well as zero (it is essentially a set of whole numbers as well as their negated values).
W on the other hand has 0,1,2, and onward as its elements. These numbers are known as whole numbers.
W ⊂ Z: TRUE. As mentioned earlier, Z includes all whole numbers thus W is a subset of it.
R ⊂ W: FALSE. Not all real numbers are whole numbers. Whole numbers must be rational and expressed without fractions. Some real numbers do not meet this criteria.
0 ∈ Z: TRUE. Zero is indeed an integer thus it is an element of Z.
∅ ⊂ R: TRUE. A null set is a subset of R, and in fact every set in general. There are no elements in a null set thus making it automatically a subset of any non-empty set by definition (since NONE of its elements are not an element of R).
{0,1,2,...} ⊆ W: TRUE. The set on the left is exactly what is defined on the problem statement for W. (The bar below the subset symbol just means that the subset is not strict, therefore the set on the left can be equal to the set on the right. Without it, the statement would be false since a strict subset requires that the two sets should not be equal).
-2 ∈ W: FALSE. W is just composed of whole numbers and not of its negated counterparts.
