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Here are some uses of the dual problem or duality

Understanding the dual problem leads to specialized algorithms for some important classes of linear programming problems.Examples include the transportation simplex method, the Hungarian algorithm for the assignment problem, and the network simplex method. Even column generation relies partly on duality.

The dual can be helpful for sensitivity analysis.Changing the primal's right-hand side constraint vector or adding a new constraint to it can make the original primal optimal solution infeasible. However, this only changes the objective function or adds a new variable to the dual, respectively, so the original dual optimal solution is still feasible (and is usually not far from the new dual optimal solution).

Sometimes finding an initial feasible solution to the dual is much easier than finding one for the primal.For example, if the primal is a minimization problem, the constraints are often of the formAx≥bAx≥b,x≥0x≥0, forb≥0b≥0. The dual constraints would then likely be of the formATy≤cATy≤c,y≥0y≥0, forc≥0c≥0. The origin is feasible for the latter problem but not for the former.

The dual variables give the shadow prices for the primal constraints.Suppose you have a profit maximization problem with a resource constraintii. Then the valueyiyiof the corresponding dual variable in the optimal solution tells you that you get an increase ofyiyiin the maximum profit for each unit increase in the amount of resourceii(absent degeneracy and for small increases in resourceii).

Sometimes the dual is just easier to solve.Aseem Dua mentions this: A problem with many constraints and few variables can be converted into one with few constraints and many variables.

(-5/17)*(51/-60)=3/15=1/5

-5/17 × 51/(-60)=3/12=1/4your answer is 1/4.

1/4 is right answer of this question

-5/17 -60 /51=3/12= 1/4

-5/17×51/(-60)3/121/4 your answer is1/4

thanks

tell,two examples,where roman numbers are used

The sign indicates it is 5 times recyclable.

this shows that the material can be 5 times recycled

30° is the correct answer to this question . √3sin30°=√3×1/2=√3/2and ,cos30°=√3/2

√3sinx=cosxcosx/sinx=√3cotx=√3cotx=cot30therefore x=30

30°is correct answer

√3sinA= cosAsinA/cosA= 1/√3tanA=tan30A=30°

answer is cos X =√3/2 and cos √3/2 Value. is 30°

cos63 = cos(90-27) = cos27 sin27 = sin(90-63) = cos63 (sin27/cos63)*(sin27/cos63) + (cos63/sin27)*(cos63/sin27) = (sin27/sin27)*(sin27/sin27) + (cos63/coa63)*(cos63/coa63) = 1+1 = 2

f(x) = |cosx - sinx|

Then,f(pi/4) = | cos pi/4 - sin pi/4|

We know, cos pi/4 = sin pi/4 = 1/sqroot (2)

= |1/sqroot (2) - 1/sqroot (2)|= 0

2x+40 + 3x+ 40 = 180=> 5x+ 80 = 180=> 5x = 100=> x= 20

m + 3x + 40 = 180=> m + 60 + 40 = 180=> m = 180 - 100=> m = 80.

n+2x+40 = 180=> n + 40 + 40 = 180=> n= 180-80=> n= 100.

Given:

Outer length(L) = 40cm Outer breadth(B) = 34cmOuter height(H) = 30 cm

Thickness= 1 cm

Volume of the box= L× B×H

Outer volume of the box = 40 cm × 34 cm × 30 cm

Outer volume of the box = 40800 cm³

Thickness = 1 cm

Inner length (l)= 40 cm -(2×1)=40-2 = 38 cmInner breadth (b)= 34 cm - (2×1)= 32 cmInner height(h) = 30 cm -(2× 1)= 28 cm

Inner volume of the box = 38 cm × 32cm × 28 cm

Inner volume of the box = 34048 cm³

Volume of wood =outer volume of box - Inner volume of box

= 40800 -34048 = 6752 cm³

Hence, the volume of wood = 6752 cm³

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The mean weight of the parcels are 50

mean weight of parcell =total weight of parcel -----------------------------------------total no of parcel

=50+65+75+90+110+120___________________________25+34+38+40+47+16

=2.55 kgper parcel

2.55 kg per parcel is the right answer.....

2.55 kg per parcel is the right one

50 kg is the mean weight of parcels

difference between 7/8 and 3/87/8-3/8(7-3/8)4/8= 1/2

a/(b-a) = 7/88a = 7b -7a15a = 7ba/b = 7/15

please help me in solving other questions

Please post the complete question as is it to verify or to prove

We know thatSin 54° = Cos 36° = ( 1/4 ) (√5 + 1)

=> Cos 54° = √( 1 - Sin² 54°) = ( 1/4 ) √[ 10 – 2 √5 ]

=> 1 – 2 Sin² 27° = ( 1/4 ) √[ 10 – 2 √5 ]

=> 2 Sin² 27° = 1/2 [ 1 - ( 1/4 ) √[ 10 – 2 √5 ]

Multiply both the sides by 8 –

=> 16 Sin² 27° = 8 - 2 √[ 10 – 2 √5 ] = [√ ( 5 + √5 ) - √ ( 3 - √5 ) ] ²

=> 4 Sin 27° = √( 5 + √5 ) - √( 3 - √5 ) ......... Proved

(Neglecting – ve value as Sin is + ve.

Let the polynomial be ax3 + bx2 + cx + d and the zeroes be α, β and γ Then, α + β + γ = -(-2)/1 = 2 = -b/a αβ + βγ + γα = -7 = -7/1 = c/a αβγ = -14 = -14/1 = -d/a∴ a = 1, b = -2, c = -7 and d = 14 So, one cubic polynomial which satisfy the given conditions will be x3 - 2x2 - 7x + 14

please accept my answer

After openinga²+2ab+b²+a²+b²-2ab2a²+2b²

=-7-8+25=-15+25=10

Let's multiply the numerator and denominator by cos(π/4)=sin(π/4):

tan^−1(sin(π/4)cos(x)−cos(π/4)sin(x))/ (cos(π/4)cos(x)+sin(π/4)sin(x)), 0<x<π

= tan^−1(sin(π/4−x)/ (cos(π/4−x)), 0<x<π

= { π/4−x if 0<x<3π/4 not defined if x=3π/4 5π/4−x if 3π/4<x<π}

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Product of zeros,P=2/3Sum of zeros=1+2/3=5/3

Polynomial=x²-Sx+P=x²-(5/3)x+2/3

zeroes are 2/3 and 1 so (x-2/3)(x-1) = 0 (3x-2)(x-1) = 0 3x*x - 3x - 2x + 2 = 0 3x*x - 5x + 2 = 0

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ACCORDING TO BODMAS ITS ANSWER IS. 2/8,-2

Any quadratic polynomial having zeroes p and q can be written as

(x - p) ( x - q) = x² - (p + q) x + pq = 0

So, quadratic polynomial is

x² - ( 2/3 + 1) x + 2/3 × 1 = 0

x² - 5x/3 + 2/3 = 0

Multiply by 3

3x² - 5x + 2 = 0

we know a quadratic polynomial is x^ 2 - (sum of roots )x + product of roots = x^2 -4x -1 ans

X^2-(sum of product)x+ (product of zeros)x^2-(1/4)"x-1

2) x^2-(√2x)+1/3

3) x^2+√5