Answer:
Step-by-step explanation:
1.25 - .75 = .5 cups of flour
Answer:
x = 66
Step-by-step explanation:
Have a nice day
The area is 192 cm^2
the sides are 20 20 24 with the base being 24
to figure out the height you have to use Pythagorean theorem of the triangle cut in half so 12^2 + b^2 = 20^2
b= 16 = height
b*h* .5= area 16* 24* .5= 192
Compute the derivative dy/dx using the power, product, and chain rules. Given
x³ + y³ = 11xy
differentiate both sides with respect to x to get
3x² + 3y² dy/dx = 11y + 11x dy/dx
Solve for dy/dx :
(3y² - 11x) dy/dx = 11y - 3x²
dy/dx = (11y - 3x²)/(3y² - 11x)
The tangent line to the curve is horizontal when the slope dy/dx = 0; this happens when
11y - 3x² = 0
or
y = 3/11 x²
(provided that 3y² - 11x ≠ 0)
Substitute y into into the original equation:
x³ + (3/11 x²)³ = 11x (3/11 x²)
x³ + (3/11)³ x⁶ = 3x³
(3/11)³ x⁶ - 2x³ = 0
x³ ((3/11)³ x³ - 2) = 0
One (actually three) of the solutions is x = 0, which corresponds to the origin (0,0). This leaves us with
(3/11)³ x³ - 2 = 0
(3/11 x)³ - 2 = 0
(3/11 x)³ = 2
3/11 x = ³√2
x = (11•³√2)/3
Solving for y gives
y = 3/11 x²
y = 3/11 ((11•³√2)/3)²
y = (11•³√4)/3
So the only other point where the tangent line is horizontal is ((11•³√2)/3, (11•³√4)/3).
Answer:
Step-by-step explanation:
<u>Eigenvalues of a Matrix</u>
Given a matrix A, the eigenvalues of A, called 
are scalars who comply with the relation:
Where I is the identity matrix
![I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
The matrix is given as
![A=\left[\begin{array}{cc}3&5\\8&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%265%5C%5C8%260%5Cend%7Barray%7D%5Cright%5D)
Set up the equation to solve
![det\left(\left[\begin{array}{cc}3&5\\8&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda \end{array}\right]\right)=0](https://tex.z-dn.net/?f=det%5Cleft%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%265%5C%5C8%260%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%20%5Cend%7Barray%7D%5Cright%5D%5Cright%29%3D0)
Expanding the determinant
![det\left(\left[\begin{array}{cc}3-\lambda&5\\8&-\lambda\end{array}\right]\right)=0](https://tex.z-dn.net/?f=det%5Cleft%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3-%5Clambda%265%5C%5C8%26-%5Clambda%5Cend%7Barray%7D%5Cright%5D%5Cright%29%3D0)
Operating Rearranging
Factoring
Solving, we have the eigenvalues
