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FromTheMoon [43]
3 years ago
14

It appears that the derivatives of sine and cosine are related. What about the second derivatives?.

Mathematics
1 answer:
Andru [333]3 years ago
5 0

Answer:

Step-by-step explanation:

\frac{d}{dx} (sin~x)=cos~x\\\frac{d}{dx}(cos~x) =-sin~x\\\frac{d^2}{dx^2} (sin~x)=\frac{d}{dx} (cos~x)=- sin~x\\\frac{d^2}{dx^2} (cos~x)=\frac{d}{dx} (-sin~x)=-cos~x

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Prove the theorem (AB )^T= B^T. A^T
Lisa [10]

Answer:

(AB)^T = B^T.A^T  (Proved)

Step-by-step explanation:

Given  (AB )^T= B^T. A^T;

To prove this expression, we need to apply multiplication law, power law and division law of indices respectively, as shown below.

(AB)^T = B^T.A^T\\\\Start, from \ Right \ hand \ side\\\\B^T.A^T = \frac{B^T.A^T}{A^T}.\frac{B^T.A^T}{B^T} (multiply \ through) \\\\                = \frac{A^{2T}.B^{2T}}{A^T.B^T} \\\\=\frac{(AB)^{2T}}{(AB)^T} \ \ (factor \ out \ the power)\\\\= (AB)^{2T-T}  \ (apply \ division \ law \ of \ indices; \ \frac{x^a}{x^b} = x^{a-b})\\\\= (AB)^T \ (Proved)

3 0
3 years ago
How do you show your work of 47 divided by 6?
saw5 [17]

Answer:

this is how you do it .

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7 0
3 years ago
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zlopas [31]
I believe the answer is feet
4 0
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The photo is now 1 inch shorter than it was originally.


3 0
3 years ago
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