n(A-B) denotes elements which are in A but not in B
n(Au B) denotes elements in A and B
n(AnB) denotes elements that are common in A and B
Now I will add one more set
n(B-A) which denotes elements in B but not in A
So, n(AuB) = n(A-B) + n( B-A) +n(AnB)
70 = 18 +n(B-A) + 25
70 = 43 + n(B-A)
n(B-A) = 70-43
n(B-A) = 27
So, n(B) = n( B-A) + n( AnB)
= 27+25
= 52
Answer:
The relation is 'a function that is one-to-many'.
Step-by-step explanation:
From the table, we can see that element 10 i.e. y=10 in the range, corresponds to two elements i.e. x=-5, and x=5 in the domain.
In other words, the given table represents the many-to-one function as an element of the range y = 10 corresponds to more than one element in the domain.
Therefore, the relation is 'a function that is one-to-many'.
9514 1404 393
Answer:
40·713 and 8·713
Step-by-step explanation:
When this multiplication is carried out "by hand", the usual sum of partial products is ...
8·713 + 40·713
<h3>Option C</h3><h3>The average rate of the reaction over the entire course of the reaction is:

</h3>
<em><u>Solution:</u></em>
Average rate is the ratio of concentration change to the time taken for the change

The concentration of the reactants changes 1.8 M to 0.6 M
here, the time interval given is 0 to 580 sec
Therefore,

Thus option C is correct