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Q-tlte Neuton rphson Method is #pplied to fiud Ilte root of f(~)-> with a iwitinl guess Xo Fo.5 #d within specified eror 0f 3 Tlue Her" ofiteration Heeded o...

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Q-tlte Neuton rphson Method is #pplied to fiud Ilte root of f(~)-> with a iwitinl guess Xo Fo.5 #d within specified eror 0f 3 Tlue Her" ofiteration Heeded oftlie perceut "pproxintion error to fall helow Ihe specilied error isEnter your answiet

Q-tlte Neuton rphson Method is #pplied to fiud Ilte root of f(~)-> with a iwitinl guess Xo Fo.5 #d within specified eror 0f 3 Tlue Her" ofiteration Heeded oftlie perceut "pproxintion error to fall helow Ihe specilied error is Enter your answiet



Answers

Use the data in FERTIL3 to verify that the standard error for the LRP in equation $(10.19)$ is about $.030 .$

So if we know that angle, a eyes are missing. Angle inside. A is 19 so 19 was on the left side of the equal sign. The other two sides are 21 squared and 11th where those air the A and and those the pian the seaside's. So those are also the side that are being multiplied against the coastline of angle eight on the other side. So we see the site B and C are no longer the same as 21 in 11. This 19 has somehow gotten into that multiplication. So there should actually say 11 times 21 not 1941. If we type this into our calculator and we get co sign of A is equal to about 0.44 to find the inverse coastline of both sides is how we can get that way by itself. And a is equal to about 64 degrees angle, eh?

In this problem, I can write the reaction age. Just look at it carefully Jordan also. Food ECU was Yes. Any to co three ACT was will give the product age I need to S. 04 ac was plus Jordan CP three solid and I can also write DDX an edge. Just look at it carefully. Yeah, I didn't. Also for it was blessing do any H C 03 Oculus will give will give any two S 04 Backwash plus Jordan CO three solid plus slash two plus CO two. So according to the option in this problem, option beach, correct option B. Each got it.

We want to use Taylor's theorem to obtain an upper bound for the error of the approximation is in calculate the exact error. So we know that the fifth derivative. So just to point this out first, we are given FX's co sign of X. X is equal to zero point three. He is equal to zero And then is equal to four. That means the 5th derivative at Z is equal to negative sign of C. So this is always going to be less than or equal to one because of how sign is bounded. So Taylor's theorem is yes to the N plus one at sea over in plus one factorial Times X -C to the N Plus one. Okay, and so this means that we have our four at that is less than or equal to F five at sea over five factorial 0 -0.3 the fifth. So we know that this is less than or equal to zero point. We can drop the absolute value 0.3 to the 5th over five. Factorial. So this is approximately equal to 2.0 25 times 10 the negative 50. So this is the upper bound and then the Actual error is the absolute value of coastline of 0.3 minus The approximation, so -1 plus 0.3 squared over two, factorial -0.3 the fourth over, or factorial, which is equal to ah 10 to the -6.

So the work here shows how many degrees one interior angle measures, but the question was to find an exterior angle. So to do so, we have to use 360 degrees, as are some of all exterior angles and divide by the number of angles on the outside, which is five for a Pentagon that would leave us with 72 degrees for each exterior angle.


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