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All Current Mid Term Papers Spring 2013 (25 May 2013 ~ 06 June 2013) at One Place

From 25 May 22, 2013 to 06 June 2013 Spring 2013

Current Mid Term Papers Spring 2013 Papers, May 2013 Mid Term Papers, Solved Mid Term Papers, Solved Papers, Solved Past Papers, Solved MCQs

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All solved MTH101 past midterm papers in one file

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Share Your Current mid Term Papers (Questions/Pattern) From 25 May 2013 to 06 June 2013 to help each other. Thanks 

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mth 101 midterm paper 29.05.2013 time 3:30

Only 4 to 5question from past papers and remaning from book. Paper was conceptual so read properly book.

1) Limt sinx/x=?

2)Cos (8x+5)

3)tan 5-3x option sy select karna tha

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agar concept theak ho to paper easily solve ho jata hay.

remember me in ur prayers

Mostly question on concept basis not from past paper.

Raheel thanks for sharing 

Attention Related Mid Term papers Spring 2013: All Fellows You don’t need to go any other site for current Mid Term papers Spring  2013, Because All discussed data/sharing of our members in this discussion are going from here to other sites. You can judge this at other sites yourself. So don’t waste your precious time with different links.

What are the critical points of f(x)= x^3+2x^2-4x+5?

to find the critical points of a function, we need to find the first derivative and equal it to zero as the following:

f(x) = x^3+2x^2-4x+5

f ' (x) = 3x^2 + 4x - 4 ===>equal it to zero

0 = 3x^2 + 4x - 4

-b +/- [ sqrt( b^2 - 4*a*c ) ] / [ 2a ]

-4 +/- [ sqrt( 4^2 - 4*3*-4 ) ] / [ 2*3 ]

-4 +/- [ sqrt( 16 + 48) ] / [ 6 ]

-4 +/- [ sqrt( 64) ] / [ 6 ]

[ -4+/- 8 ] / 6 =====> [ -4+8 / 6 ] = 2/3 =====> & [ -4-8 / 6 ] = -2 ( critical points )

Find the equation of the line that passes through (6,5) and (-3, -1).?

First you want to find the slope. The formula is:

y2 - y1 (divided by) 
x2 - x1 (the numbers are sub 1 and sub 2, not exponents)

in the first ordered pair (x1, y1) is (6,5). the second (x2, y2) is (-3, -1)

now fill in the formula.

y2 is -1. y1 is 5. So the numerator is -1 minus 5. Which is -6.

x2 is -3. x1 is 6. So the denominator is -3 minus 6. Which is -9.

So our slope is -6/-9 which can be simplified to 2/3.

NOW. You know the slope is 2/3. Now you must find a line with this slope that runs through those points (either set of points will work).

y - y1 = 2/3 (x - x1).
Fill in for y1 and x1 with either set of points (but be consistent. if you use from the first set for x, you must use from the first set for y).

We'll use (-3, -1)
y - -1 = 2/3 (x - -3)

When subtracting a negative, you add. So:

y + 1 = 2/3 (x + 3)

Now distribute 2/3.

y + 1 = 2/3x + 2. (2/3 times 3 is 2)

now subract 1 from both sides.

y = 2/3x + 1

All solved MTH101 past midterm papers in one file


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