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It has a laserwriter resource named "RM1140" registered on port 250. The third line is another reply to the same request saying host techpit has laserwriter "techpit" registered on port 186. ATP packet formatting is demonstrated by the following example: jssmag.209.165 > helios.132: atp-req 12266 0xae030001helios.132 > jssmag.209.165: atp-resp 12266:0 (512) 0xae040000helios.132 > jssmag.209.165: atp-resp 12266:1 (512) 0xae040000helios.132 > jssmag.209.165: atp-resp 12266:2 (512) 0xae040000helios.132 > jssmag.209.165: atp-resp 12266:3 (512) 0xae040000helios.132 > jssmag.209.165: atp-resp 12266:4 (512) 0xae040000helios.132 > jssmag.209.165: atp-resp 12266:5 (512) 0xae040000helios.132 > jssmag.209.165: atp-resp 12266:6 (512) 0xae040000helios.132 > jssmag.209.165: atp-resp*12266:7 (512) 0xae040000jssmag.209.165 > helios.132: atp-req 12266 0xae030001helios.132 > jssmag.209.165: atp-resp 12266:3 (512) 0xae040000helios.132 > jssmag.209.165: atp-resp 12266:5 (512) 0xae040000jssmag.209.165 > helios.132: atp-rel 12266 0xae030001jssmag.209.133 > helios.132: atp-req* 12267 0xae030002 Jssmag.209 initiates transaction id 12266 with host helios by requesting up to 8 packets (the ''). The hex number at the end of the line is the value of the 'userdata' field in the request. Helios responds with 8 512-byte packets. The ':[digit]' following the transaction id gives the packet sequence number in the transaction and the number in parens is the amount of data in the packet, excluding the atp header. The '*' on packet 7 indicates that the EOM bit was set. jssmag.209 then requests that packets 3 & 5 be retransmitted. Helios resends them then jssmag.209 releases the transaction. Finally, jssmag.209 initiates the next request. The '*' on the request indicates that XO ('exactly once') was not set. IP fragmentation Fragmented Internet datagrams are printed as: (frag id:size@offset+)(frag id:size@offset) The first form indicates there are more fragments. The second indicates this is the last fragment. Id is the fragment id. Size is the fragment size (in bytes) excluding the IP header. Offset is this fragment's offset (in bytes) in the original datagram. The fragment information is output for each fragment. The first fragment contains the higher level protocol header and the frag info is printed after the protocol info. Fragments after the first contain no higher level protocol header and the frag info is printed after the source and destination addresses. For example, here is part of an ftp from arizona.edu to lbl-rtsg.arpa over a CSNET connection that doesn't appear to handle 576 byte datagrams: arizona.ftp-data > rtsg.1170: . 1024:1332(308) ack 1 win 4096 (frag 595a:328@0+)arizona > rtsg: (frag 595a:204@328)rtsg.1170 > arizona.ftp-data: . ack 1536 win 2560 There are a few things to note here: First, addresses in the 2nd line don't include port numbers. Because the TCP protocol information is all in the first fragment and we have no idea what the port or sequence numbers are when we print the later fragments. Second, the tcp sequence information in the first line is printed as if there were 308 bytes of user data when, in fact, there are 512 bytes (308 in the first frag and 204 in the second). If you are looking for holes in the sequence space or trying to match up acks with packets, this can fool you. A packet with the

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A+(9−1)d−8 = a+8d ………………………………………………… (ii)On subtracting equation (i) from (ii), we will get here,−12 = 6dd = −2From equation (i), we can write,4 = a+2(−2)4 = a−4a = 8Let nth term of this A.P. be zero.an = a+(n−1)d0 = 8+(n−1)(−2)0 = 8−2n+22n = 10n = 5Hence, 5th term of this A.P. is 0.10. If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.Solution: We know that, for an A.P series;an = a+(n−1)da17 = a+(17−1)da17 = a +16dIn the same way,a10 = a+9dAs it is given in the question,a17 − a10 = 7Therefore,(a +16d)−(a+9d) = 77d = 7d = 1Therefore, the common difference is 1.11. Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54th term?Solution:Given A.P. is 3, 15, 27, 39, …first term, a = 3common difference, d = a2 − a1 = 15 − 3 = 12We know that,an = a+(n−1)dTherefore,a54 = a+(54−1)d⇒3+(53)(12)⇒3+636 = 639a54 = 639+132=771We have to find the term of this A.P. which is 132 more than a54, i.e.771.Let nth term be 771.an = a+(n−1)d771 = 3+(n −1)12768 = (n−1)12(n −1) = 64n = 65Therefore, 65th term was 132 more than 54th term.Or another method is;Let nth term be 132 more than 54th term.n = 54 + 132/2= 54 + 11 = 65th term12. Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?Solution:Let, the first term of two APs be

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In a RGB color space, hex #848484 is composed of 51.8% red, 51.8% green and 51.8% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 0% magenta, 0% yellow and 48.2% black. It has a hue angle of 0 degrees, a saturation of 0% and a lightness of 51.8%. #848484 color hex could be obtained by blending #ffffff with #090909. Closest websafe color is: #999999. R 52 G 52 B 52 RGB color chart C 0 M 0 Y 0 K 48 CMYK color chart ● #848484 color description : Dark gray. The hexadecimal color #848484 has RGB values of R:132, G:132, B:132 and CMYK values of C:0, M:0, Y:0, K:0.48. Its decimal value is 8684676. Hex triplet 848484 #848484 RGB Decimal 132, 132, 132 rgb(132,132,132) RGB Percent 51.8, 51.8, 51.8 rgb(51.8%,51.8%,51.8%) CMYK 0, 0, 0, 48 HSL 0°, 0, 51.8 hsl(0,0%,51.8%) HSV (or HSB) 0°, 0, 51.8 Web Safe 999999 #999999 CIE-LAB 55.148, -0, -0.005 XYZ 21.931, 23.074, 25.127 xyY 0.313, 0.329, 23.074 CIE-LCH 55.148, 0.005, 266.929 CIE-LUV 55.148, -0.003, -0.007 Hunter-Lab 48.035, -2.566, 2.611 Binary 10000100, 10000100, 10000100 #5e5e5e #5e5e5e rgb(94,94,94) #6b6b6b #6b6b6b rgb(107,107,107) #777777 #777777 rgb(119,119,119) #848484 #848484 rgb(132,132,132) #919191 #919191 rgb(145,145,145) #9e9e9e #9e9e9e rgb(158,158,158) #aaaaaa #aaaaaa rgb(170,170,170) Monochromatic Color Below, you can see some colors close to #848484. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice. #7c7c7c #7c7c7c rgb(124,124,124) #7f7f7f #7f7f7f rgb(127,127,127) #818181 #818181 rgb(129,129,129) #848484 #848484 rgb(132,132,132) #878787 #878787 rgb(135,135,135) #898989 #898989 rgb(137,137,137) #8c8c8c #8c8c8c rgb(140,140,140) Similar Colors Text with hexadecimal color #848484 This text has a font color of #848484. Text here #848484 background color This paragraph has a background color of #848484. Content here #848484 border color This element has a border color of #848484. Content here

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10, 2011 132 0 18,680 #3 so i will let prime95 run about 8hrs so i can sleep how will i know if it fails? Dec 10, 2011 132 0 18,680 #4 and does it matter if temp goes two high when i'm using prime95 some times it gets to 70-9xC? so does it matter if i run it for 8hrs Mar 29, 2011 1,034 0 19,360 #5 If it crashed (read: bsod or freezes) or gives you an error message, it failed. Otherwise it's stable.As festerovic said, blend for 24 hours if you want to make sure it's 100% stable. If you don't care that there's a small chance it might crash, run it for less time (it's debatable how healthy long prime95 sessions are).Also do 20 minutes of small ffts to get max temperature. If its over 80C after 20 min of small ffts, it's too hot. Though at 4ghz and that cooler, you should have no problem keeping it under 70C.edit: what's your voltage? those temps are too high for a 4GHz oc. Prime won't kill it overnight though. Jul 16, 2008 2,504 1 21,165 #6 You shouldnt have to use voltage higher than 1.30 volts most of the time. Dec 10, 2011 132 0 18,680 #7 Clock BCLk/pcie frequen 100.00mhz, cpu ration x40.0, gt frequency 1100mhz voltage for all of the valtage of w.e this areCpu voltage offset +0VDram valtage 1.500vpch valtage 1.059vtt valtage 1.051vcpu pll valtage 1.832vvccsa vaktage 0.925vigpu valtage offset 0 mVthats what it shows on my asrock extreme tuning utility didn't want to restart pc and yea the ram i'm using is a cosair vengeance 4x2* stick Mar 29, 2011 1,034 0 19,360 #8 There are some excellent ocing guides here, use them to get the lowest voltage possible.basically, reduce the voltage in small steps while testing in prime for 5 minutes until the voltage is too low and it crashes. then up the voltage a bit and test overnight. Mar 29, 2011 1,034 0 19,360 #9 does that say 1.832v??? download cpuz and send the voltage from that.First though, restart the computer and make sure the core cpu voltage is not above 1.4 at the most. if it is at 1.8, u could be seriously (likely are) damaging the cpu.edit: nvm, cpu pll valtage is not core voltage. give us your cpuz reading with and wihtout furmark running. Dec 10, 2011 132 0 18,680 #10 k downloaded its showstechnology 32mm core valtage 1.176v ran prime95 really fast so the cpu to go to 4hz and the valtage went to 1.2v max Dec 10, 2011 132 0 18,680 #11 and no the temps are not going over 5xC now its stable at 5x. download netcut 3 1 - download netcut 3 - download netcut 2 0 - download netcut 3 00 - download netcut 3 5 - netcut 3 0 download - download free netcut 3 0 - netcut 3 0 full - netcut 3 0 downloadnetcut - download netcut 3 win7 64bit

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Multiply 124726132 by 5 by first halving each even grouping in the first number and then appending zero:12 4 72 6 132 x 5 = 6 2 36 3 066 0 or 8 32 6 31 x 5 = 4 16 3 15 5Multiplying a number by 15 can be done by multiplying by 10 and adding half the result. We can think of adding a zero, and then adding half of each even grouping to itself, working left to right and keeping the same number of digits in the grouping as it started with. If a grouping ends up with an additional digit, the upper digit is added to the grouping to the left. The presentation below makes the calculation look more difficult than it actually is—the result is generated smoothly from left to right, with perhaps a correction for a carry from the next grouping, as with the carry of 1 from the (72+36) grouping below to the group on its left:12 4 72 6 132 x 15 = (12+6) (4+2) (72+36) (6+3) (132 + 66) 0 = 18 7 08 9 198 0Multiplication by 25, or 100/4, can be thought of as appending two zeros and dividing by 4. Multiplying by 50 can be done as 100/2, 75 as 300/4, 125 as 1000/8, and so forth.These are reasonable and readily understood concepts that involve looking at the whole number rather than individual digits. This is a mental shift that is subtle but critical in developing a number sense. Methods like these are also more general than they seem at first, because if they almost apply, we can use them on nearby numbers and then apply a correction at the end.These methods all involve thinking about the properties of numbers, so they appeal to me as methods for