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Aleksandr-060686 [28]
2 years ago
9

Error Analysis

Mathematics
1 answer:
aleksklad [387]2 years ago
6 0

The coefficient(s) of the expression is/are 7, 1

Use a comma to separate answers as needed.)

The constant(s) of the expression is/are 8

(Use a comma to separate answers as needed.)

What error might Jake have made?

OD. Jake did not include the coefficient 1.

Click to select your answer's

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This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex
noname [10]

L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)

L_x=10+10\lambda x=0\implies1+\lambda x=0

L_y=10+10\lambda y=0\implies1+\lambda y=0

L_z=2+4\lambda z=0\implies1+2\lambda z=0

L_\lambda=5x^2+5y^2+2z^2-42=0

\begin{cases}L_x=0\\L_y=0\\L_z=0\end{cases}\implies1+\lambda x=1+\lambda y=1+2\lambda z=0\implies x=y=2z

5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1

z^2=1\implies z=\pm1\implies x=y=\pm2

There are two critical points, at which we have

f\left(2,2,1\right)=42\text{ (a maximum value)}

f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}

3 0
3 years ago
The distance between the points (-1, -6) and (-19, -6) is units
Vladimir [108]

Answer:

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Step-by-step explanation:

\sqrt{  {( - 9 -  - 1)}^{2} +  {( - 6 -  - 6)}^{2} }

\sqrt{ {( - 18)}^{2} }

\sqrt{324}

3 0
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(3x-2)(2x^2+3x-1), after I distribute how would I get an exponent of three, for the 6x?
Aliun [14]
Because when multiplying you add the number of variables, and there are 3 X's so it would be 6x with an exponent of 3
7 0
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Help me please thanks
Svetach [21]
I think it's along the y-axis..... 
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Answer: A. 40

Step-by-step explanation:

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