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Title: ppt on mid point theorem free download
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Title: to verify de morgan theorem using ic 7400
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Title: Basic Proportionality Theorem
Page Link: Basic Proportionality Theorem -
Posted By: projectsofme
Created at: Monday 18th of October 2010 02:04:27 PM
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This article is presented by:RAJAT BAKSHI
X-A
43
Basic Proportionality Theorem



If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Aim- AD AE
DB EC
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Title: maths project on basic proportionality theorem
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Title: verification of sampling theorem using matlab program with explanation
Page Link: verification of sampling theorem using matlab program with explanation -
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Created at: Friday 02nd of November 2012 08:43:46 PM
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Title: triangle proportionality theorem ppt
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ABC is a triangle.From C a line segment CD is drawn on AB and from B a line segment BE is drawn on CA. The lines CD and BE intersect at a point O.
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Title: bernoulli s theorem and its application ppt
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Title: VERIFICATION OF SAMPLING THEOREM
Page Link: VERIFICATION OF SAMPLING THEOREM -
Posted By: seminar class
Created at: Friday 06th of May 2011 05:52:55 PM
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Procedure: -
1. Select a frequency of ‘f’ Hz
2. To generate a sine wave of F Hz, define a closely spaced time vector(to represent analog signal, at least 1000 samples pr cycle)t=0:1/1000:1;
3. Generate the sinusoid, and plot the signal x_analog=sin(2*pi*F*t)
4. Select a sampling frequency Fs=10Fsamples/sec,generate a suitable time scale for this sampling signal n=0:1/Fs:1;%vector to define the sampling instants
5. Sample the analog signal at the instants specified by ’n ’
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Title: Automated theorem proving ATP
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Title: A proof of the Heine-Borel Theorem
Page Link: A proof of the Heine-Borel Theorem -
Posted By: seminar class
Created at: Tuesday 15th of February 2011 02:58:34 PM
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A proof of the Heine-Borel Theorem
Theorem (Heine-Borel Theorem). A subset S of Ris compact if and only if S is closed and bounded.
Proof. First we suppose that S is compact. To see that S is bounded is fairly simple: Let In = (−n, n).
Then
1
[
n=1
In = R.
Therefore S is covered by the collection of {In}. Hence, since S is compact, finitely many will suffice.
S  (In1 [ · · · [ Ink ) = Im,
where m = max{n1, . . . , nk}. Therefore |x|  m for all x 2 S, and S is bounded.
Now we will show that S is closed. Suppose not. The ....etc

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