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3 years ago

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Lee is in customer service. How can she best support her internal customer, Dan? O al Prioritize other tasks and customers O b)

Ask him a lot of personal questions O c) Give him her full attention when he's speaking O d) Check with him before completing basic tasks

1 answer:

VLD [36.1K]3 years ago

6 0

Answer: thx for the fr.ee points

<h2>but</h2>

BRU STILL?!

VVVVVVVVVVVVVV

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Answer 322.788 hope this helped

erma4kov [3.2K]

Answer:

322.788?

Step-by-step explanation:

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3 years ago

HELP ME PLEASE! I need to find the Interquartile range of : 0,0,0,1,1,3,3,3,4,4,4,6,6,7,7,7,8,8,9,10

Lana71 [14]

Answer:

The interquartile range is 5

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3 years ago

Helpppp<br>What is 25% of 307<br>10<br>|<br>2<br>12<br>1<br>7<br>2

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Answer:

76.75 is 25% of 307

Step-by-step explanation:

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3 years ago

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The record high temperature in Alabama is 112°F and the record low temperature is −27°F. What is the difference between the high

Minchanka [31]

The difference between the highest and lowest temperature is 85°F.

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3 years ago

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Use Lagrange multipliers to find the maximum and minimum values of

marusya05 [52]

The Lagrangian is

$L(x,y,\lambda)=x+8y+\lambda(x^2+y^2-4)$

It has critical points where the first order derivatives vanish:

$L_x=1+2\lambda x=0\implies\lambda=-\dfrac1{2x}$

$L_y=8+2\lambda y=0\implies\lambda=-\dfrac4y$

$L_\lambda=x^2+y^2-4=0$

From the first two equations we get

$-\dfrac1{2x}=-\dfrac4y\implies y=8x$

Then

$x^2+y^2=65x^2=4\implies x=\pm\dfrac2{\sqrt{65}}\implies y=\pm\dfrac{16}{\sqrt{65}}$

At these critical points, we have

$f\left(\dfrac2{\sqrt{65}},\dfrac{16}{\sqrt{65}}\right)=2\sqrt{65}\approx16.125$ (maximum)

$f\left(-\dfrac2{\sqrt{65}},-\dfrac{16}{\sqrt{65}}\right)=-2\sqrt{65}\approx-16.125$ (minimum)

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3 years ago