oh? and the correct answer is????

From: "DD730"

Wrong again.

JAXAshby wrote:
7.071067812

donny don't even know what the sq rt of 50 is.

actually, the square root of 50 is an irrational number (meaning
never end, never repeating) that is just a rough estimate. a more
accurate answer to 1,000 decimal places is:

5 * 1.4142135623 73095048801688724209
69807856967187537694807317667973799073247846210703 88503875343276415727
350138462309122970249248360558 507372126441214970999358314132
226659275055927557999505011527 820605714701095599716059702745
345968620147285174186408891986 095523292304843087143214508397
626036279952514079896872533965 463318088296406206152583523950
547457502877599617298355752203 375318570113543746034084988471
603868999706990048150305440277 903164542478230684929369186215
805784631115966687130130156185 689872372352885092648612494977
154218334204285686060146824720 771435854874155657069677653720
226485447015858801620758474922 657226002085584466521458398893
944370926591800311388246468157 082630100594858704003186480342
194897278290641045072636881313 739855256117322040245091227700
226941127573627280495738108967 504018369868368450725799364729
060762996941380475654823728997 180326802474420629269124859052
181004459842150591120249441341 728531478105803603371077309182
869314710171111683916581726889 419758716582152128229518488472 08969

Note: if a thousand digits is not accurate enough, let me know and I
can do the math quickly up to maybe 5 million decimal places. More
than 5 mil will take me a bit longer.

We know you're a bumpass. What's your point?

"Horvath" wrote in message
om...



If you're going to forge posts from me,
at least learn to spell dumbass.


" D U M B A S S "



This signature is now the ultimate power in the universe

You missed a digit.

"JAXAshby" wrote in message
...
7.071067812

donny don't even know what the sq rt of 50 is.

actually, the square root of 50 is an irrational number (meaning never
end,
never repeating) that is just a rough estimate. a more accurate answer to
1,000 decimal places is:

5 * 1.4142135623 73095048801688724209
69807856967187537694807317667973799073247846210703 88503875343276415727
350138462309122970249248360558 507372126441214970999358314132
226659275055927557999505011527 820605714701095599716059702745
345968620147285174186408891986 095523292304843087143214508397
626036279952514079896872533965 463318088296406206152583523950
547457502877599617298355752203 375318570113543746034084988471
603868999706990048150305440277 903164542478230684929369186215
805784631115966687130130156185 689872372352885092648612494977
154218334204285686060146824720 771435854874155657069677653720
226485447015858801620758474922 657226002085584466521458398893
944370926591800311388246468157 082630100594858704003186480342
194897278290641045072636881313 739855256117322040245091227700
226941127573627280495738108967 504018369868368450725799364729
060762996941380475654823728997 180326802474420629269124859052
181004459842150591120249441341 728531478105803603371077309182
869314710171111683916581726889 419758716582152128229518488472 08969

Note: if a thousand digits is not accurate enough, let me know and I can
do the
math quickly up to maybe 5 million decimal places. More than 5 mil will
take
me a bit longer.

"JAXAshby" wrote in message
...
donny, your math wouldn't get you promoted from 10th grade to 11th.

(what the hell do *you* think the sq rt of 50 is, for the krist sakes?)

As it happens, I don't have an opinion about the square root of 50.

Now, perhaps you could tell us why you think that my mathematical skills are
lacking??

While you are at it, you can also tell us about the malicious code that I
sent you in an e-mail?


I haven't forgotten about your lying accusation! I think that you should
admit that you might have made a mistake

Think about it!


Regards


Donal
--

yeah, maybe. but what the hey. one digit in a thousand.


You missed a digit.

"JAXAshby" wrote in message
...
7.071067812

donny don't even know what the sq rt of 50 is.

actually, the square root of 50 is an irrational number (meaning never
end,
never repeating) that is just a rough estimate. a more accurate answer to
1,000 decimal places is:

5 * 1.4142135623 73095048801688724209
69807856967187537694807317667973799073247846210703 88503875343276415727
350138462309122970249248360558 507372126441214970999358314132
226659275055927557999505011527 820605714701095599716059702745
345968620147285174186408891986 095523292304843087143214508397
626036279952514079896872533965 463318088296406206152583523950
547457502877599617298355752203 375318570113543746034084988471
603868999706990048150305440277 903164542478230684929369186215
805784631115966687130130156185 689872372352885092648612494977
154218334204285686060146824720 771435854874155657069677653720
226485447015858801620758474922 657226002085584466521458398893
944370926591800311388246468157 082630100594858704003186480342
194897278290641045072636881313 739855256117322040245091227700
226941127573627280495738108967 504018369868368450725799364729
060762996941380475654823728997 180326802474420629269124859052
181004459842150591120249441341 728531478105803603371077309182
869314710171111683916581726889 419758716582152128229518488472 08969

Note: if a thousand digits is not accurate enough, let me know and I can
do the
math quickly up to maybe 5 million decimal places. More than 5 mil will
take
me a bit longer.

As it happens, I don't have an opinion about the square root of 50.

opinion? how the hell can one had an **opinion** re sq rt 50????

Now, perhaps you could tell us why you think that my mathematical skills are
lacking??

ah ..... maybe see above???

Since no one seems willing to do this probably, here are the answers:
Starting with heading of 190 and speed through water 5, to have a COG of 90 and
SOG of 5,
current must be 50 degrees, speed of 7.66

Starting with heading of 190 and speed through water 5, to have a COG of 90 and
SOG of 2.5,
current must be 34.37 degrees, speed of 5.96



For any oblique triangle with angles A, B, and C, and opposite sides a,b, and, c
then:
the law of sines says:

a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle

and the law of cosines says:
c^2 = a^2 + b^2 - 2*a*b* cos C



"SkitchNYC" wrote in message
...
Say you are sailing a course of 190 and making 5 kn. An adverse current
suddenly gets you and you are now making a COG of 90. Assume the new COG is
at
the same speed (5kn) and again at 2.5 kn. What direction and speed must the
current be to produce either of these results? Can such a current exist in a
Gulf Stream eddie?

jeffie, the answer was posted three days ago. And it was posted out to 1,000
decimal places.

Earth to jeff, Earth to jeff ...


Since no one seems willing to do this probably, here are the answers:
Starting with heading of 190 and speed through water 5, to have a COG of 90
and
SOG of 5,
current must be 50 degrees, speed of 7.66

Starting with heading of 190 and speed through water 5, to have a COG of 90
and
SOG of 2.5,
current must be 34.37 degrees, speed of 5.96



For any oblique triangle with angles A, B, and C, and opposite sides a,b,
and, c
then:
the law of sines says:

a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle

and the law of cosines says:
c^2 = a^2 + b^2 - 2*a*b* cos C



"SkitchNYC" wrote in message
...
Say you are sailing a course of 190 and making 5 kn. An adverse current
suddenly gets you and you are now making a COG of 90. Assume the new COG
is
at
the same speed (5kn) and again at 2.5 kn. What direction and speed must
the
current be to produce either of these results? Can such a current exist in
a
Gulf Stream eddie?

Why give an answer to 1000 places when it wrong in the second place? The point
of nav problems is to be able to get the correct answer, not something within
10%.



"JAXAshby" wrote in message
...
jeffie, the answer was posted three days ago. And it was posted out to 1,000
decimal places.

Earth to jeff, Earth to jeff ...


Since no one seems willing to do this probably, here are the answers:
Starting with heading of 190 and speed through water 5, to have a COG of 90
and
SOG of 5,
current must be 50 degrees, speed of 7.66

Starting with heading of 190 and speed through water 5, to have a COG of 90
and
SOG of 2.5,
current must be 34.37 degrees, speed of 5.96



For any oblique triangle with angles A, B, and C, and opposite sides a,b,
and, c
then:
the law of sines says:

a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle

and the law of cosines says:
c^2 = a^2 + b^2 - 2*a*b* cos C



"SkitchNYC" wrote in message
...
Say you are sailing a course of 190 and making 5 kn. An adverse current
suddenly gets you and you are now making a COG of 90. Assume the new COG
is
at
the same speed (5kn) and again at 2.5 kn. What direction and speed must
the
current be to produce either of these results? Can such a current exist in
a
Gulf Stream eddie?

ah, it seems you didn't understand the ramifications of the question.

not surprising.


Why give an answer to 1000 places when it wrong in the second place? The
point
of nav problems is to be able to get the correct answer, not something within
10%.



"JAXAshby" wrote in message
...
jeffie, the answer was posted three days ago. And it was posted out to
1,000
decimal places.

Earth to jeff, Earth to jeff ...


Since no one seems willing to do this probably, here are the answers:
Starting with heading of 190 and speed through water 5, to have a COG of
90
and
SOG of 5,
current must be 50 degrees, speed of 7.66

Starting with heading of 190 and speed through water 5, to have a COG of
90
and
SOG of 2.5,
current must be 34.37 degrees, speed of 5.96



For any oblique triangle with angles A, B, and C, and opposite sides a,b,
and, c
then:
the law of sines says:

a/ sinA = b / sinB = c / sinC = diameter of circumscribed circle

and the law of cosines says:
c^2 = a^2 + b^2 - 2*a*b* cos C



"SkitchNYC" wrote in message
...
Say you are sailing a course of 190 and making 5 kn. An adverse current
suddenly gets you and you are now making a COG of 90. Assume the new
COG
is
at
the same speed (5kn) and again at 2.5 kn. What direction and speed must
the
current be to produce either of these results? Can such a current exist
in
a
Gulf Stream eddie?