anonymous
  • anonymous
use the differentials to approximate [9.02]to power 1.5,plus [1divide by [9.02]raised to power 0. 5]-26
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
please use equation...
anonymous
  • anonymous
Assuming you mean,\[(9.02)^{3/2}\left( \frac{1}{(9.02)^{1/2}} \right)^{-26}\]you can consider the following. The differential of y (i.e., dy) is approximately equal to the exact change in y (i.e., Δy), provided that the change in x (Δx) is relatively small. Since for small changes in x,\[\frac{\Delta y}{\Delta x}\approx \frac{dy}{dx} \rightarrow \Delta y \approx \frac{dy}{dx}\Delta x\]you can approximate a small change in a function y as,\[y+\Delta y \approx y + \frac{dy}{dx}\Delta x\]The function you need to consider is \[y=x^{1/2}\]since you're looking at combinations involving \[9.02^{1/2}\]
anonymous
  • anonymous
For \[x=9 \rightarrow (9)^{1/2}=3\]which is easy. But you'll notice that x=9.02 is the old x-value plus a *small* change, so we can approximate the actual value of the function by the linear approximation we derived above:\[(x+\Delta x)^{1/2} \approx x^{1/2}+ \frac{1}{2}x^{-1/2}\Delta x\]Our x-value is 9 and the small change is 0.02. So,\[(9+0.02)^{1/2} \approx (9)^{1/2}+\frac{1}{2} (9)^{-1/2}(0.02)=3+\frac{1}{300}=\frac{901}{300}\]

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anonymous
  • anonymous
You can now use this result in your equation:\[(9.02)^{3/2}\left( \frac{1}{(9.02)^{1/2}} \right)^{-26} \approx \left( \frac{901}{300} \right)^3\left( \frac{1}{\left( \frac{900}{300} \right)} \right)^{-26}\]\[=\left( \frac{901}{300} \right)^3\left( \frac{901}{300} \right)^{26}=\left( \frac{901}{300} \right)^{29}\]
anonymous
  • anonymous
If you have any difficulty with what I've written, let me know.
anonymous
  • anonymous
Incidentally, to nine significant figures, your original expression gives,\[7.08752811 \times 10^{13}\]while the approximate solution found yields,\[7.087654705 \times 10^{13}\]You can see that they're pretty close.

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