Step 1
For a differentiable function f (x), find f '(x). Equate it to zero. Solve the equation f '(x) = 0 to get the Critical values of f (x..
Step 2:
Solve f '(x) = 0 to get the critical values for f (x). Let these values be a, b, c. These are the points of maxima or minima. Arrange these values in ascending ord..
Step 3:
Check if f (a) = f (b) If all the above condition are satisfied, then Rolle's theorem is applicable else the Rolle's theorem is not applicable. If Rolle's theorem is applicable, solve f '(c) = 0. Show that one of these roots lie in the open interval (a, b..
Step 5:
If (x 1 , y 1 ) is the point found in step 4, then x = x 1 , y = y 1 , is the optimal solution of the LPP and Z = ax 1 + by 1 is the optimal value. The above method of solving an LPP is more clear with the following exampl..
Step I :
Express the differential equation in the form f(x) dx = g(y) d..
Steps
- Carboxylation - Glycolytic reversal - Regeneration of RUBP Photorespiration Light dependent up take of O 2 by cells and release of CO 2 without producing ATP and NADPH. During high conc. of O 2 and high temperature. RUBISCO act as RUBP oxygenase. RUBP splits into PGA and phosphoglycolate...
- Carboxylation - Glycolytic reversal - Regeneration of RUBP Photorespiration Light dependent up take of O 2 by cells and release of CO 2 without producing ATP and NADPH. During high conc. of O 2 and high temperature. RUBISCO act as RUBP oxygenase. RUBP splits into PGA and phosphoglycolate...Steps
Write the symbols of the elements which form the compound. Below the symbol of each element write down its valency. Cross over the valencies of the atoms. ..
Write the symbols of the elements which form the compound. Below the symbol of each element write down its valency. Cross over the valencies of the atoms. ..First step
Sulphur dioxide reacts with water to liberate nascent hydrogen. 2H 2 O(l) + SO 2 (g) -----------------> H 2 SO 4 (aq) + 2[H]water sulphur dioxide sulphuric acid nascent hydrog..
Steps:
The first 2 electrons will go to the 1 st shell = K Shell (2n 2 ) The next shell L takes a maximum of 8 electrons (2n 2 ) In this way 2 + 8 = 10 electrons have been accommodated. The next 2 electrons go to the M Shell. K L M 2 8 2 ..
Result
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