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We can solve for x using a technique called successive approximations. step 1: if we assume that x is very small compared to 0.100 (such that 0.100 x ≈ 0.100) then our first approximation of x (let\'s call it x1) can be calculated as

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barnuts

Supposed that we are given the equation:

7.20 x 10^-5 = x(0.100+x)^2 

 

So assuming that x is very small so that:

0.100 + x ~ 0.100

and calling it x1, so:

 

7.20 x 10^-5 = x1 (0.100)^2 

Then we simply solve for x1:

 

x1 = 7.20 x 10^-5 / (0.100)^2 

x1 = 7.20 x 10^-3

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