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yulyashka [42]
3 years ago
14

What is 2+2 Thank you for answering this

Mathematics
2 answers:
Olenka [21]3 years ago
6 0
The answer is will be 4.
cestrela7 [59]3 years ago
6 0
2+2=4 hope this helps
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Find the minimum and maximum of f(x,y,z)=x^2+y^2+z^2 subject to two constraints, x+2y+z=4 and x-y=8.
Alika [10]
The Lagrangian for this function and the given constraints is

L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-4)+\lambda_2(x-y-8)

which has partial derivatives (set equal to 0) satisfying

\begin{cases}L_x=2x+\lambda_1+\lambda_2=0\\L_y=2y+2\lambda_1-\lambda_2=0\\L_z=2z+\lambda_1=0\\L_{\lambda_1}=x+2y+z-4=0\\L_{\lambda_2}=x-y-8=0\end{cases}

This is a fairly standard linear system. Solving yields Lagrange multipliers of \lambda_1=-\dfrac{32}{11} and \lambda_2=-\dfrac{104}{11}, and at the same time we find only one critical point at (x,y,z)=\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right).

Check the Hessian for f(x,y,z), given by

\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}

\mathbf H is positive definite, since \mathbf v^\top\mathbf{Hv}>0 for any vector \mathbf v=\begin{bmatrix}x&y&z\end{bmatrix}^\top, which means f(x,y,z)=x^2+y^2+z^2 attains a minimum value of \dfrac{480}{11} at \left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right). There is no maximum over the given constraints.
7 0
4 years ago
The function h(t) = -4.9t2 + h0 gives the height (h), in meters, of an object t seconds after it falls from an initial height (h
sveticcg [70]
It will hit the ground after 3.499 seconds.

To solve this you first have to find the value of h(0) in this equation. That is the height from which it was dropped.

You can input any of the points into the equation and solve for the missing part. You will get 60 for the height.

The use the quadratic formula to see that it reaches the ground after 3.499 seconds.
7 0
3 years ago
Read 2 more answers
Solve the following system of equations by substitution. Show all steps.
Alinara [238K]

Answer:

Given system of equations:

\begin{cases}f(x)=-x^2+2x+3\\g(x)=-2x+3\end{cases}

To solve by substitution, equate the equations and solve for x:

\begin{aligned}f(x) & = g(x)\\\implies -x^2+2x+3 & = -2x+3\\-x^2+4x & = 0\\x^2-4x & = 0\\x(x-4) & = 0\\\implies x & = 0\\\implies x-4 & = 0 \implies x=4\end{aligned}

Therefore, the x-values of the solution are x = 0 and x = 4.

To find the y-values of the solution, substitute the found values of x into the functions:

f(0)=-(0)^2+2(0)+3=3

g(0)=-2(0)+3=3

f(4)=-(4)^2+2(4)+3=-5

g(4)=-2(4)+3=-5

Therefore, the solutions to the given system of equations are:

(0, 3) and (4, -5)

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2 years ago
Help me please free cookies if you help ✋
tensa zangetsu [6.8K]
Choclate Chip Cookies?
6 0
3 years ago
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What is the slope and y intercept
Vera_Pavlovna [14]
Slope= -2/8
Y-int= -2
5 0
3 years ago
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