For C at 7pm they sold most and at 6 am they sold less
Answer:
Constant Term
The term in a simplified algebraic expression or equation which contains no variable(s). If there is no such term, the constant term is 0.
Example: –5 is the constant term in p(x) = 2x3 – 4x2 + 9x – 5
Answer:
6
Step-by-step explanation:
I learned this last year
Given -
y > - x + 3
To Find -
The points which is NOT a solution to the inequality
Step-by-Step Explanation -
We will put the value of each in inequality and then see if it satisfies the given condition or not.
a. ( -2,6)
So, x = -2 and y = 6 in y > - x + 3
= 6 > - (-2) + 3
= 6 > 5 (Correct Solution)
b. (5,-1)
So, x = 5 and y = -1 in y > - x + 3
= -1 > -5 + 3
= -1 > -2 (Correct Solution)
c. (2,0)
So, x = 2 and y = 0 in y > - x + 3
= 0 > -2 + 3
= 0 > 1 (Incorrect Solution)
d. (0,3)
So, x = 0 and y = 3 in y > - x + 3
= 3 > -0 + 3
= 3 > 3 (Incorrect Solution)
Final Answer -
The points which is NOT a solution to the inequality =
c. (2,0)
d. (0,3)
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.
