How do we find our approximation for ${2.9}^{5}$ $2.9^5$ ?

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How do we find our approximation for ${2.9}^{5}$ $2.9^5$ ?

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Answer 1 · 2016-09-24T04:15:37

See explanation

Explanation:

2.9=29/10 is a rational number of the form integer/integer..

And so,29#/1010)^5 = 20511149/100000 is rational.

The exact value in decimals is 205,11149.

Successive rounded approximations are

3-significant digits ( sd:): 205

4-sd: 205.1

5-sd: 205.11

6-sd: 205.111

6-sd: 205.1115.

You can try instead the binomial expansion for

${2.9}^{5} = {(3 - 0.1)}^{5} = 3^{5} {(1 - \frac{1}{30})}^{5}$ $2.9^5=(3-0.1)^5=3^5(1-1/30)^5$

Answer 2 · 2016-09-24T18:29:21

We can also use the approximation of $y = x^{5}$ $y=x^5$ around $x = 3$ $x=3$ .

The derivative of $y$ $y$ is $\frac{d y}{d x} = 5 x^{4}$ $dy/dx=5x^4$ , so the slope of the tangent line around $x = 3$ $x=3$ is $5 (3^{4}) = 405$ $5(3^4)=405$ .

The point it intersects is $(3, 3^{5}) = (3, 243)$ $(3,3^5)=(3,243)$ .

Thus the tangent line is $y - 243 = 405 (x - 3) \Rightarrow y = 405 x - 972$ $y-243=405(x-3)=>y=405x-972$ .

Thus, an approximation for ${2.9}^{5}$ $2.9^5$ would be:

$y = 405 (2.9) - 972 = 202.5$ $y=405(2.9)-972=202.5$ .

This compares to the actual value of ${2.9}^{5} = 205.11149$ $2.9^5=205.11149$ .

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How do we find our approximation for ${2.9}^{5}$ $2.9^5$ ?

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