1.
For
(x+y)+3z=600
(x+y)+z=400
solve for p in
p+3z=600
p+z =400
From which you'll get z=100, and therefore p=(x+y)=?
2.
Hint:
|C|=15 (cardinality of C = number of elements in the set)
|P|=21
We also know that the cardinality of the union of the two sets is 25, i.e.
|C∪P|=25
To find |C∩P| (number of trays containing both), we use the principle of inclusion and exclusion:
|C∪P|=|C|+|P|-|C∩P| ...(1)
Substitute the above values into (1) and solve for |C∩P|.
3.
Hint: for
6<|x-3|<7
to be satisfied, x cannot be an integer, so there are many possible answers within a certain range.
Also, x<0 means x takes on a negative value.
First find out what |x-3| should be, and then solve for x.
x+y+3z=600
x+y+z=400
1. In the system of equations above, what is the value of x+y?
2. There are 25 trays on a table in the cafeteria. Each tray contains a cup only, a plate only, or both a cup and a plate. If 15 of the trays contain cups and 21 of the trays contain plates, how many contain both a cup and a plate?
3. If 6<lx-3l<7 and x<0, what is one possible value of lxl ?
2 answers
300
1 answer