Answer: W is now at (2,-3)
Step-by-step explanation: flip by the x-axis to turn (-2,3) to (-2,-3) then flip it on the y-axis to turn (-2,-3) to (2,-3).
Answer:
expected value is $4.5
Step-by-step explanation:
expected value is $4.5
Answer with Step-by-step explanation:
We are given that two matrices A and B are square matrices of the same size.
We have to prove that
Tr(C(A+B)=C(Tr(A)+Tr(B))
Where C is constant
We know that tr A=Sum of diagonal elements of A
Therefore,
Tr(A)=Sum of diagonal elements of A
Tr(B)=Sum of diagonal elements of B
C(Tr(A))=
Sum of diagonal elements of A
C(Tr(B))=
Sum of diagonal elements of B

Tr(C(A+B)=Sum of diagonal elements of (C(A+B))
Suppose ,A=![\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
B=![\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
Tr(A)=1+1=2
Tr(B)=1+1=2
C(Tr(A)+Tr(B))=C(2+2)=4C
A+B=![\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]+\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
A+B=![\left[\begin{array}{ccc}2&1\\2&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%5C%5C2%262%5Cend%7Barray%7D%5Cright%5D)
C(A+B)=![\left[\begin{array}{ccc}2C&C\\2C&2C\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2C%26C%5C%5C2C%262C%5Cend%7Barray%7D%5Cright%5D)
Tr(C(A+B))=2C+2C=4C
Hence, Tr(C(A+B)=C(Tr(A)+Tr(B))
Hence, proved.
The answer is in the image. If you convert them to improper fractions, the division is easier. Like 2/3 + 1 = 5/3
1) Seven less than twice a number, n, is 32.
ANS) B. 2n - 7 = 32