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 ’
x_discreate =sin (2*pi*F*n) plot this using separa..............etc

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can you please help me with my project on the topic to verify the basic proportionality theorem using parallel line board and triangle cut-outs..............etc

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This article is presented by:RAJAT BAKSHI
X-A
43Basic 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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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.
Here it is given that area of triangles 1)BOD=15 sq. unit 2)COE=10 sq. unit
and 3)BOC=25 sq. unit. Find area of ADOE..............etc

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arge]..............etc

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i want to make a project on basic proportionality theorem in 3 days . please help me for that. i will be highly thankful to you for this
a needy one...............etc

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Proving mathematical theorems by computer programs is the hot and latest buzz in automated reasoning (AR). The legality factor of a theorem ranges from the trivial to impossible state, based upon the logic used for its conceptualisation...............etc

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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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