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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Precious Rice

Mathematics

9 April, 20:07

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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Thomas · Answer 1

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