(1 point) Consider the universal set U={1,2,3,4,5,6,7,8,9,10}, define the set A be the even numbers, the set B be the odd number
Sloan [31]
Answer:
a) AUC = {2,4,6,8,10}
b) BnC = {}
c) AnB = {}
d) B-C = B = {1,3,5,7,9}
Step-by-step explanation:
The set A is the even numbers, those that are divisible by two.
So A = {2,4,6,8,10}
B is the odd numbe.rs. An odd number is a number that is not divisible by two.
So B = {1,3,5,7,9}.
C = {4,5,6}, as the problem states
a) The union of sets is a set containing all elements that are in at least one of the sets. So the union of A and C is a set that contains all elements that are in at least one of A or C.
So AUC = {2,4,6,8,10}.
b) The intersection of two sets consists of all elements that in both sets. So, the intersection of B and C is the set that contains all elements that are in both B and C.
There are no elements that are in both B and C, so the intersection is an empty set
BnC = {}
c) Same explanation as b), there are no elements that are in both A and B, so another empty set.
AnB = {}
d) The difference of sets B and C consists of all elements that are in B and not in C. We already have in b) that BnC = {}, so:
B-C = B = {1,3,5,7,9}
Answer:
y = 1/2x + 5
sorry no explanation but pls give brainliest
The answer to ur question is 0.396
2.2% * 18 = 0.396
Answer:
The answer is "![\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%26-2%5C%5C%20-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B7%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%7D)
".
Step-by-step explanation:
![\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D)
Solve the L.H.S part:
![\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D)
After calculating the L.H.S part compare the value with R.H.S:
![\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D%20%5C%5C%5C%5C)
In equation (i) multiply by 3 and subtract by equation (iii):
put the value of c in equation (i):
In equation (ii) multiply by 3 then subtract by equation (iv):
put the value of d in equation (iv):
The final answer is "".
