Answer:
yes
Step-by-step explanation:
y(x) is even or odd according as y(−x)=±y(x) . Here, #y(-x)=-(-x)^3=-(-x^3)=x^3=-y(x). So, y is an odd function of x.
1)ABCDF
2)Chord = AG
3)Tangent is AE
This is a system
x is the number of advanced tickets sold
y is the number of same day tickets sold
once you label your variables then write the equations
x+ y = 45 total tickets sold
20x + 25y = 1050 money taken in
multiply the top equation by -20 and the x will drop out
-20x - 20y = -900
20x + 25y = 1050. Add the equations result is
5y = 150, divide by 5 ,
y or same day tickets cost $30
substitute 30 into first equation
x + 30 = 45 therefore x = 15
Solve for y in 3x+7=y
y=7−3x
Substitute y=7−3x
y=7−3x into 8x−3y=30
8x−3y=30.
17x−21=30
Solve for x
x in 17x−21=30
17x−21=30.
X=3
Substitute x=3 into y=7−3x
y=-2
Therefore,
x=3
y=−2
Answer:
Verified!
Step-by-step explanation:
Upper or lower triangular matrix does not make any difference in finding eigenvalues because equalizing determinant to zero will lead to the same result.
Let's apply it for 2x2 matix:
![A = \left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0\\0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0\\-b&\lambda-c\end{array}\right]=(\lambda-a)(\lambda-c)=0](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%260%5C%5Cb%26c%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Clambda%20I%20-%20A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Clambda%260%5C%5C0%26%5Clambda%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%260%5C%5Cb%26c%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5Cdet%28%5Clambda%20I%20-%20A%29%20%3D%20det%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Clambda-a%260%5C%5C-b%26%5Clambda-c%5Cend%7Barray%7D%5Cright%5D%3D%28%5Clambda-a%29%28%5Clambda-c%29%3D0)
So, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have 
.
So, eigenvalues are 
Let's apply it for 3x3 matrix:
![A = \left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0&0\\-b&\lambda-c&0\\-d&-e&\lambda-f\end{array}\right]=(\lambda-a)(\lambda-c)(\lambda-f)=0](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%260%260%5C%5Cb%26c%260%5C%5Cd%26e%26f%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Clambda%20I%20-%20A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Clambda%260%260%5C%5C0%26%5Clambda%260%5C%5C0%260%26%5Clambda%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%260%260%5C%5Cb%26c%260%5C%5Cd%26e%26f%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5Cdet%28%5Clambda%20I%20-%20A%29%20%3D%20det%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Clambda-a%260%260%5C%5C-b%26%5Clambda-c%260%5C%5C-d%26-e%26%5Clambda-f%5Cend%7Barray%7D%5Cright%5D%3D%28%5Clambda-a%29%28%5Clambda-c%29%28%5Clambda-f%29%3D0)
So as above, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have 
.
So eigenvalues are 
