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| 25 Feb 2012 09:20 PM |
2 arcs of circles with the same center are 25 degrees wide
<))
if the arc length of one is 5, and the other is 6, how far apart are they!? |
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| 25 Feb 2012 10:29 PM |
| NOBODEH KNOWS ITS 2.2918311805233!? |
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| 25 Feb 2012 11:12 PM |
My turn
2y+sqrt3y = 1 find y |
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| 25 Feb 2012 11:23 PM |
4y^2 + 3y = 1 4y^2 + 3y - 1 = 0 y = (-b +- (b^2 - 4ac)^0.5)/2a y = (-3 +- (9 - 4*4*-1)^0.5)/8 y = (-3 +- 25^0.5)/8
y = -7/2, 11/4
Sincerely, Noobertuber |
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| 25 Feb 2012 11:24 PM |
sqrt(3y) or sqrt(3)y
for sqrt(3y) 2y-1=-(3y)^0.5 (2y-1)^2=3y 4y^2-2y+1=3y 4y^2-5y+1=0 4(y-0.625)^2-0.5625=0 y=0.625+(0.5625/4)^0.5 and 0.625-(0.5625/4)^0.5
y=1 and 0.25
for sqrt(3)y (2+3^0.5)y=1 y=1/(2+3^0.5)
y=0.26794919243112 |
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| 25 Feb 2012 11:26 PM |
Put your solutions into the equation:
2y + sqrt3y = 1
2 + sqrt3 = 1 FALSE
0.5 + sqrt0.75 = 1 FALSE
Quadratic formula for the win.
Sincerely, Noobertuber |
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| 25 Feb 2012 11:29 PM |
If you square the whole thing, it would be (2y + (3y)^0.5)^2 4y^2+4y*2(3y)^0.5+3y :P FOILed, brother. |
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| 25 Feb 2012 11:31 PM |
| can't take the square root of a negative |
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| 25 Feb 2012 11:32 PM |
2*(11/4)+sqrt(3*(11/4))=8.37228132326901
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| 25 Feb 2012 11:34 PM |
| both of us, not both of your answers, just to clarify |
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| 25 Feb 2012 11:44 PM |
found my mistake 2y-1=-(3y)^0.5 (2y-1)^2=3y v 4y^2-4y+1=3y 4y^2-7y+1=0 4(y-0.875)^2-2.0625=0 y=0.875+(2.0625/4)^0.5 and 0.875-(2.0625/4)^0.5 y=1.59307033081726 and 0.15692966918274
but only 0.15692966918274 works.
Now do my problem! |
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| 26 Feb 2012 12:20 AM |
What is your problem?
Sincerely, Noobertuber |
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| 26 Feb 2012 03:41 AM |
Let x,y be the radii of the inner and outer circle respectively.
(also π = pi, if your font is weird like mine is)
2πx * 25/360 = 5 50πx = 1800 x = 36/π
2πy * 25/360 = 6 50πy = 2160 y = 43.2/π
The distance between the two circles = y - x = 43.2/π - 36/π = 7.2/π
= 36/5π |
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| 26 Feb 2012 05:29 AM |
| good sir, I do believe that 4 in base 3.5, to 50 decimal places, is 10.12202100121030302211303102030300130221223010001010. |
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| 26 Feb 2012 05:33 AM |
now another problem similar to yours: what is 3 in base phi? phi=(1+5^0.5)/2 |
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Oysi
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| Joined: 06 Jul 2009 |
| Total Posts: 9058 |
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Oysi
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| Joined: 06 Jul 2009 |
| Total Posts: 9058 |
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| 26 Feb 2012 06:00 AM |
local answer="10.12202100121030302211303102030300130221223010001010" local start=answer:find(".",1,true)-1 local result=0 local count=0 for d=1,#answer do local digit=tonumber(answer:sub(d,d)) if digit then start=start-1 result=result+digit*3.5^start end end print(result)
>4 |
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| 26 Feb 2012 06:02 AM |
| (wikipedia): Non-integer_representation |
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