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butalik [34]
3 years ago
8

Pedro had $235 in his saving account when Jim had $450 in his account. Look at the situation in the image. Which inequality show

s when the amount of money in Pedro’s account is greater than the amount of money in Jim’s account?
Mathematics
1 answer:
11111nata11111 [884]3 years ago
4 0

Answer:

$450>$235

Step-by-step explanation:

This is because, when 235 is subtracted from 450 there is a great difference thats why $450 is greater than $235

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Answer:

which one?

Step-by-step explanation:

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3 years ago
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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.
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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
If, in 7 years, Susan will be 2 times as old as she was 3 years ago, what is Susan’s present age?
pochemuha

Answer:

13

Step-by-step explanation:

Set Susan's present age as x, her age from 3 years ago as x-3, and her age in three years as 2(x-3)

Since her age in seven years = 2 times as old as she way 3 years ago, set up the equation:

x+7 = 2(x-3)

x+7 = 2x-6

x = 13

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3 years ago
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