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U l r d2 l2 d l2 u5 l2 u2 r u r2 d2 u3 r2 d2 l r u2 l2 d2 u2 r4 d2 l3 r u2 l2 d l2 d l d2 r2 d5 r2 u l d l u4 d3 r4 u2 l u l2 r d2 u r2 d2 l3 d l u2 d r2 u3 r u3 r2 u2 l2 d5 l d3 r2 u2 l r d2 l2 u2 r u l
Level 18 (36 moves) D3, R, D2, L, U4, L, D, R, U, L, U, R, D2, L, U, R5, D3, L, D2, L, D, L
Let D represent the length of the diagonal and A the area. Suppose the sides are L and B. Then A = L*B and by Pythagoras, D2 = L2 + B2 These two equations need to be solved as simultaneous equations to get the values of L and B. One way is to use the following identities: (L + B)2 = L2 + 2LB + B2 = D2 + 2A so that L + B = sqrt(D2 + 2A) and (L - B)2 = L2 - 2LB + B2 = D2 - 2A so that L - B = sqrt(D2 - 2A) which then give 2L = sqrt(D2 + 2A) + sqrt(D2 - 2A) and 2B = sqrt(D2 + 2A) - sqrt(D2 - 2A)
The formula for volume of a liquid in the pipe is V=(pi/4)D2 (L)
The passcode 524383 is stage 6. The solution is 35 moves (R-right, L-left, U-up D-down) R3,D2,R,D2,R,D, R,U,L3,U2,L,U3, R3,D2,L,U, R2, D,R,D2,R
Density = mass/volumeThe unit weight of steel is 7850 kg/m3volume of bar = (πd2/4)*Lhence mass = ((πd2/4)*L)*7850= ((3.14 *d2/4)*1)*7850 for unit length= 0.785*d2*7850= 6162.25 d2 if d is in metersor d2 /162 if d is in mm.By putting the value of diameter of rod, you can calculate the unit weight of any size tmt bar.
The moves are paired for the first 4 : R2, U2 to the orange squares. Next is R, U2, L4, D3, R2, U, L, U, R2 (done) R, D2, R, U4, L, D, L4, D2, R2, U, L (done) The code for level 19 is 839654.
The passcode to stage 6 is 524383 The passcode to stage 7 is 189493 The solution to stage 6 (35 moves): R3,D2,R,D2,R,D,R,U,L3,U2,L,U3,R3,D2,L,U,R2,D,R,D2,R
Density = mass/volumeThe unit weight of steel is 7850 kg/m3volume of bar = (πd2/4)*Lhence mass = ((πd2/4)*L)*7850= ((3.14 *d2/4)*1)*7850 for unit length= 0.785*d2*7850= 6162.25 d2 if d is in metersor d2 /162 if d is in mm.By putting the value of diameter of rod, you can calculate the unit weight of any size tmt bar.
assume:Length = L = 5.7 mWidth = W = 3.0 mDepth 1 = D1 = 1.7 mDepth 2 = D2 = 1.1 mVolume = L x W x D2 + 1/2 [ L x W x (D1 - D2)]= L x W x [ D2 + 1/2 D1 - 1/2 D2]= L x W x [D1 +D2] /2= 5.7 x 3.0 x [1.7 + 1.1]/2= 23.94 cubic meter1 cubic meter = 1000 literVolume = 23.94 x 1000 = 23940 liter
Pause the game. Hold L and R and type in the password Pause the game. Hold L and R and type in the password
1. If you measure the diameter and length in feet. the formula is below. L = length d= Diameter units are in feet # gallons = (L*pi*d2)*1.87 2. if you measure Diamter and Length in inches # gallons = (L*pi*d2)*57.75