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akhl thumbnail
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Posted: 15 years ago

Originally posted by: Miggi

Which of the following clusters of orbitals would form the shape shown here (Part C 1 figure) and would also be possible within the valence shell of an atom?
Check all that apply.
three p orbitals
two sp orbitals and four p orbitals
three sp orbitals
two sp orbitals and two p orbitals
six sp^3d orbitals
six sp^3d^2 orbitals correct
six sp^3 orbitals
one sp orbital and two p orbitals
figure:
1076900a.jpg

Miggi thumbnail
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Posted: 15 years ago
How to draw the lewis structure of HCSNH2?
akhl thumbnail
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Posted: 15 years ago
See the image I posted in a previous answer. In that I have put Lewis structures of all the compounds you asked. In this forum, the image does not look clear. So right click on the image and the save the picture in your computer. Then open the image from your computer. You will see it clearly.
Miggi thumbnail
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Posted: 15 years ago
Use differentials (or, equivalently, a linear approximation) to estimate the given number.
sqrt100.4
I got 10.02 is it right?
Edited by Miggi - 15 years ago
Miggi thumbnail
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Posted: 15 years ago
Find the critical numbers of the function.
f(x)= x^-2lnx
I am not sure but i got e^2 and 0. Not sure which one is correct. can ny1 help?
akhl thumbnail
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Posted: 15 years ago

Originally posted by: Miggi

Use differentials (or, equivalently, a linear approximation) to estimate the given number.

sqrt100.4
I got 10.02 is it right?

Correct. But did you use linear approximation through differential? If you want, you can post the method here and I will check.
akhl thumbnail
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Posted: 15 years ago

Originally posted by: Miggi

Find the critical numbers of the function.

f(x)= x^-2lnx
I am not sure but i got e^2 and 0. Not sure which one is correct. can ny1 help?

Is only -2 the exponent over x or is whole of -2lnx the exponent over x? I mean is it
(x^-2) * (lnx)
or
x^(-2lnx)
Miggi thumbnail
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Posted: 15 years ago
yea its like (x^-2)lnx and i know my ans is wrong :( we have to find critical no's in it.
akhl thumbnail
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Posted: 15 years ago
f(x) = x^-2 * lnx
f ' (x) = x^-2 * d/dx (lnx) + d/dx (x^-2) * lnx
= x^-2 * 1/x + (-2) x^-3 * lnx
= x^-3 - 2 x^-3 lnx
= x^-3 (1 - 2 lnx)
For critical points, f ' (x) = 0
Therefore x = 0, 1-2lnx = 0
x = 0, lnx = 1/2
x = 0, sqrt(e)
Note: sqrt = square root.
Ans: x = 0, sqrt(e)
Miggi thumbnail
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Posted: 15 years ago

Originally posted by: akhl

f(x) = x^-2 * lnx

f ' (x) = x^-2 * d/dx (lnx) + d/dx (x^-2) * lnx
= x^-2 * 1/x + (-2) x^-3 * lnx
= x^-3 - 2 x^-3 lnx
= x^-3 (1 - 2 lnx)
For critical points, f ' (x) = 0
Therefore x = 0, 1-2lnx = 0
x = 0, lnx = 1/2
x = 0, sqrt(e)
Note: sqrt = square root.
Ans: x = 0, sqrt(e)

hey thnx but i have already tried this 1 and webassign says that its wrong. I put the ans as 0, and quareroot (e) but its incorrect.

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