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ivann1987 [24]
2 years ago
10

Hey y’all. I’m so embarrassed and I want to hit my head so BAD:( I’m literally crying. So I had a test in math and I’ve always b

een cheating and I was the best math student but when the class was gonna end I told the teacher I only have two questions and she told me to still stay and when I finished the submit button didn’t work and the teacher told me to present my screen and I did but when I did the teacher face wasn’t happy because when I showed her my screen in the tabs I had tabs that help me cheat the test. But the teacher didn’t tell me anything and she told me that if I have the notes for all my questions answers and I didn’t know what to say so I said yes! But I told her if I should take pictures of the answers and she said no. And I started panicking so she told me to reload the WHOLE THING so I did and all my 18 questions answers deleted and I told the teacher I had to go to my other class and she said bye. I’m literally crying. I’m hitting my head so hard. Karma is real:( F
Mathematics
2 answers:
solniwko [45]2 years ago
7 0

Answer: OMG that is sad why did you hide the the cheating tabs

Step-by-step explanation:

Allushta [10]2 years ago
7 0

Answer:

Girl why did u cheat in the first place???

Step-by-step explanation:

Cheating is not going to get u into college or get you a scholorship. Its just a road to lies and people not trusting u. You should take this moment and reflect. Also, STUDY

sorry that happened but tbh, if u cheated then u kinda deserved that

NOT BEING MEAN

have a great day and LEARN

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madam [21]

You're looking for the extreme values of T(x,y,z)=3x+6y-6z+1 subject to x^2+y^2+z^2=9. The Lagrangian is

L(x,y,z,\lambda)=3x+6y-6z+1+\lambda(x^2+y^2+z^2-9)

with critical wherever the partial derivatives vanish:

L_x=3+2\lambda x=0\implies x=-\dfrac3{2\lambda}

L_y=6+2\lambda y=0\implies y=-\dfrac3\lambda

L_z=-6+2\lambda z=0\implies z=\dfrac3\lambda

L_\lambda=x^2+y^2+z^2-9=0

Substituting the first three solutions into the last equation gives

\dfrac9{4\lambda^2}+\dfrac9{\lambda^2}+\dfrac9{\lambda^2}=9

\implies\lambda=\pm\dfrac32

\implies x=1,y=2,z=-2\text{ or }x=-1,y=-2,z=2

At these points, we have

T(1,2,-2)=28

T(-1,-2,2)=-26

so the highest temperature the bee can experience is 28º F at the point (1, 2, -2), and the lowest is -26º F at the point (-1, -2, 2).

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3 years ago
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hi how r u....say thanks me later

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Answer:

V = 267.9 in^3

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