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Solve the recurrence relation
T (K) = 2T (K-1), T (0) = 1
Ans: The following equation can be written in the subsequent form:
tn - 2tn-1 = 0
Here now successively replacing n by (n - 1) and then by (n - 2) and so on we obtain a set of equations.
The method is continued till terminating condition. Add these equations in such type of a way that all intermediate terms get cancelled. The equation can be rearranged as
Multiplying all the equations correspondingly by 20, 21, ..., 2n - 1 and then adding them together, we get
tn - 2nt0 = 0
or, tn = 2n
A chemist has one solution which is 50% acid and a second which is 25% acid. How much of each should be mixed to make 10 litres of 40% acid solution.
Solve for x , y (x + y - 8)/2 =( x + 2 y - 14)/3 = (3 x + y - 12 )/ 11 (Ans: x=2, y=6) Ans : x+ y - 8/2 = x + 2y - 14 /3 = 3x+ y- 12/11
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-x^3+6x-7
How to get the answer
The sum of the squares of two consecutive positive odd integers is 74. What is the value of the smaller integer? Let x = the lesser odd integer and let x + 2 = the greater odd
subtract 20and 10,and then mutiply by 5
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