nyuuzyou/kompy huggingface dataset:
Figure 38 – Unbuffered mapped throughput over Gigabit Ethernet

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Figure 38 – Unbuffered mapped throughput over Gigabit Ethernet
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bet | 46/47 | Sana | 30.03.2021 | Hajmi | 1,33 Mb. | | #13789 |
Figure 38 – Unbuffered mapped throughput over Gigabit Ethernet. CIFS read performance is very good: at 37.5MBps it reaches PCI bus saturation on the server. It is likely that with a faster PCI bus, the throughput would be greater. Unbuffered writes showed poorer performance, with peak write performance less than half peak read performance. The curves of the write throughput graph are similar to those of mapped writes in Figure 35.
Summary
To summarize:
Win2K and both SP6 and SP3 of NT4 show similar one-disk sequential I/O performance.
NT4SP6 has no performance penalty for small-unbuffered requests. NT4SP3 shows significant penalties for both unbuffered synchronous reads and unbuffered asynchronous requests.
Basic and dynamic volumes in Win2K produce similar results.
For sequential workloads, 32bit 33MHz PCI is now the major bottleneck. PCI bus saturation can be achieved with only three disks, a configuration not uncommon on workstation machines.
Random workloads limited by the PCI bus. A PCI bus can handle about 10,000 random IOs per second, approximately the capacity of 100 disks.
Single disk IDE performance is very good, especially relative to their much more expensive SCSI cousins. Good performance is easy to attain; the rules for optimizing performance on SCSI drives still hold true for IDE. Although the 3ware card we tested was hampered by its PCI bus interface and unable to take full advantage of all four disks, second generation 3ware cards show good PCI performance and no longer have this limitation.
Single disk IDE drives are the clear winner on bang for the buck. For sequential loads, they come very near the performance of SCSI drives that are more than three times as expensive. On random IO, IDE drives have a 25% better IOs per second per dollar ratio than SCSI drives.
3ware’s IDE adapter card offers a linear improvement in sequential throughput for reads and good scalability for writes up to three IDE drives. Filling the last slot with a fourth drive gives little to no improvement in sequential IO performance on our outer band measurements due to limitations in our 3ware card’s PCI bus interface. Second generation 3ware cards have no such limitation. Random IO performance is still good, as the throughputs involved are much lower so four drives don’t come anywhere near saturating the card or bus.
IDE drives show similar overhead to that of SCSI drives, unless PIO is used. Avoid PIO as it severely degrades performance while dramatically increasing processor overhead. DMA is needed for good performance.
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Figure 37 – Unbuffered UNC and mapped overhead

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Figure 37 – Unbuffered UNC and mapped overhead
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bet | 45/47 | Sana | 30.03.2021 | Hajmi | 1,33 Mb. | | #13789 |
Figure 37 – Unbuffered UNC and mapped overhead. UNC and mapped IO are expensive, with up to ten times as much overhead per MB compared with local drives. UNC showed lower overhead than mapped drives for small requests, however this difference in overheads diminishes with increased request size. This implies a higher fixed overhead for mapped drives. We’ve truncated the left graph’s read overhead at the 16KB request size for reads. This is due to the large performance degradation that’s seen with UNC reads with large request sizes. As the number of IOs per second dropped to single digits, the error bound on our soaker became larger than the results, so we were unable to measure that overhead.
CIFS read throughput over Gigabit Ethernet was very good. We were able to achieve over 37.5MBps using sequential 32KB read requests. At 1GBps, the PAP of Gigabit is 125MBps. In round numbers, the half-power point is 63MBps. Although we were unable to achieve the half-power point, this is likely due to the PCI bus as our machines did not reach processor or disk saturation. At 37.5MBps, the server must read from the SCSI controller at 37.5MBps and then write to the Gigabit Ethernet adapter at 37.5MBps. The total bandwidth being used by both reads and writes is 37.5 * 2 or 75MBps, which is nearly the maximum throughput we were able to achieve using local disks.
Unbuffered write throughput was less than half the read throughput. Using 16KB requests, we were only able to achieve 15MBps. With requests larger than 16KB, a performance hit similar to the local UNC case shown in Figure 34 is seen. This indicates that the poor write performance may be due in part to the drive mapping translation. In addition, unlike reads requests that stream data from the server with little acknowledgement from the client, writes require more handshaking as we observed much higher bi-directional network traffic with writes.
Processor utilization for reads is less per MB than that for writes. Whichever system was having data pushed to it had the higher overhead. For reads from the server, processor utilization on the server was 28% compared to 32.2% on the client for 2KB four-deep read requests. For 128KB requests, the server processor utilization was 14% while the client processor utilization was 23.4%. For writes to the server, the server processor utilization was 11.6% for 2KB writes and 7% for 128KB writes, while the client’s processor consumption was 6% and 9.3% respectively.
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Windows 95 drivers readme. TXT IOMEGA DRIVES AND windows 95

WINDOWS 95 DRIVERS - README.TXT
IOMEGA DRIVES AND WINDOWS 95
(Source Iomega FaxBack document #: 2049 9/8/95
Windows 95 has 2 modes of operation Real(MSDOS) Mode and Protected Mode.
Real (MSDOS) Mode - Real mode operation is Windows 95 MSDOS
compatibility mode. This is the slower of the two modes. This mode is
used when devices are configured with MSDOS device drivers and
Protected Mode drivers are not available.
Protected Mode - Protect mode operation is the fastest mode of
operation under Windows 95, it is also the preferred mode of operation.
This mode uses built in 32-bit drivers, sometimes called Mini-Port drivers,
to support devices including: Hard Drive Controllers, Video Controllers,
SCSI Controllers, etc..
Any device that does not have built in support, or the manufacturer does not
supply protected mode drivers can use the Real Mode (current MSDOS) drivers.
Iomega has protected mode drivers available for the PC2, PC1600, PC1616,
PC800, PPA-3, and the Zip Zoom SCSI adapters.
Iomega Products and Protected Mode Operation
Windows 95 will support Iomega drives in protected mode if there is a protected
mode driver for the SCSI adapter. Windows 95 has built in protected mode
drivers for 3 Iomega SCSI adapters. The PC1600, the Zip Zoom SCSI
accelerator, and the PC2. Windows 95 will load the protect mode drivers
automatically if these adapters are installed previous to Windows 95. If the
protect mode drivers do not load automatically you will have to install and
configure these drivers manually. To add a supported SCSI adapter use the
Add New Hardware program in the Control Panel and follow the instructions later
in this document. The drives should work attached to any SCSI adapter that has
a built in protected mode driver.
If you use the protect mode drivers with Iomega drives, you will not be able to
run the DOS/Windows versions of Zip Tools or Iomega utilities. A Windows 95
version of the Iomega Tools is required (See Windows 95 Upgrade Kit below).
Figure 36 – A comparison between UNC paths and mapped drives accessing a local Quantum Atlas 10K drive

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Figure 36 – A comparison between UNC paths and mapped drives accessing a local Quantum Atlas 10K drive
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bet | 44/47 | Sana | 30.03.2021 | Hajmi | 1,33 Mb. | | #13789 |
Figure 36 – A comparison between UNC paths and mapped drives accessing a local Quantum Atlas 10K drive. Accessing files using both UNC and mapped drives adds additional system layers and overhead to each request. Compared with single disk results, this results in degraded throughput. Although write performance is the same for both UNC and mapped drives, read performance differs substantially. For good performance, mapped drives should be used.
Accessing local files through UNC and mapped drive paths forces the IO to go through the network stack despite the fact that the IO is local. This combined with some sub-optimal buffering and serialization choices result in poor performance. Since programs will increasingly be using network-transparent names from the Active Directory, this will be an increasingly important issue.
The primary cost of UNC and mapped drives is processor time. Compared with single disk overhead, requests to UNC and mapped drives can cost ten times more processor time per MB. On our machine, 2KB unbuffered mapped reads consumed over 71% of both processors. On small request sizes, UNC showed significantly less overhead for small 2KB writes than on mapped drives. Mapped drives had a fixed overhead of 563μs per write request while UNC paths had a fixed overhead of 454μs per write. 2KB reads showed a smaller impact. As the request size increases, both UNC and mapped drives converge to 27ms per MB.
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Figure 33 – Win2K two-disk hardware RAID1 sequential throughput

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Figure 33 – Win2K two-disk hardware RAID1 sequential throughput
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bet | 41/47 | Sana | 30.03.2021 | Hajmi | 1,33 Mb. | | #13789 |
Figure 33 – Win2K two-disk hardware RAID1 sequential throughput. One of the advertised features of the 3ware card is a technology called “TwinStor” which offers RAID1+0 read performance using two drives. However, in order to achieve good rest performance, we had to make deep requests at large request sizes. It is not until there was a total of 256KB of requests outstanding did TwinStor give us any benefit. On writes, slightly less than single disk performance was achieved.
The 3ware card shows good random read IO performance with its hardware RAID1. Figure 34 shows both RAID0 and RAID1 unbuffered random performance with 8KB requests. As the graph on the right indicates, the additional request scheduling being done by the 3ware controller has improved random read IO performance significantly over that of a RAID0 stripe set. At two deep, the 3ware card shows almost linear gains from the second disk. The additional performance is due to both drives containing the same data. In the case of a RAID0 stripe set, there is a 50% chance that both random IOs will be for only one of the disks with two requests outstanding. This means for the second request, half of the time one of the disks will be idle while the other is busy servicing both outstanding requests. This can be seen in the left graph as the single disk performance of 66 IO/s only reaches 99 IO/s (66 IO/s + .5 * 66 IO/s) with two requests outstanding. The RAID1 TwinStor drive however can efficiently dispatch the second IO 100% of the time to the second disk, making full use of the second set of arms. This can be seen on the right graph as two-deep requests result in 119 IO/s or close to two times the 66 IO/s performance of a single disk.
With deep queue depths, total random read throughput is higher with TwinStor than with RAID0 for a different reason: the controller reorders the requests in a way that minimizes disk arm movement on both drives. This produces a more than doubling of random IO performance on two drives compared to one.
Write performance with TwinStor was similar to that of a single disk. This is due to the cost of mirroring: writing two copies of the bits uses both drives at the same time. The additional parallelism offered by the second drive is lost. RAID0 is not subject to this constraint as it lacks data redundancy, so it is able to take advantage of the second drive.
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Figure 34 - Win2K two disk hardware RAID1 random IOs per second

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Figure 34 - Win2K two disk hardware RAID1 random IOs per second
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bet | 42/47 | Sana | 30.03.2021 | Hajmi | 1,33 Mb. | | #13789 |
Figure 34 - Win2K two disk hardware RAID1 random IOs per second. By scheduling read requests to both drives, a two disk TwinStor RAID1 array shows better performance than a two-disk RAID0 array. However, write performance is only that of a single disk as both disks are utilized during writes in order to mirror the bits, so the parallelism offered by the second drive is lost. The RAID0 array is not subject to this constraint, so it is able to take advantage of the additional disk.
As shown in Figure 35 below, the overhead measured for hardware RAID1 is similar to the overhead for a single disk. This is as expected as hardware RAID1 offloads the additional RAID1 processing onto the 3ware card requiring additional processor cycles. The fixed cost for a single disk was 57s, while the marginal cost was .53s per KB read and .42s per KB written. A RAID1 set had a fixed cost of 57s, and a marginal cost of .55s per KB written and .60s per KB read.
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Figure 35 – Win2K hardware RAID1 overhead. Hardware RAID1 has similar overhead to that of a single disk. This is as expected as hardware RAID1 offloads all processing onto the 3ware card rather than burdening the processor.
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To repeat, as suggested in Table 6, the next generation of 3Ware host-bus adapters and drivers improve on many of these results.
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Internet ma’lumotlari - Mundarija kirish I bob. Uzoq muddatli majburiyatlar hisobi bo'yicha umumiy ma’lumotlar

Mundarija kirish I bob. Uzoq muddatli majburiyatlar hisobi bo'yicha umumiy ma’lumotlar
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Bog'liq
Uzoq muddatli majburiyatlar hisobi
Internet ma’lumotlari
http://www.iia.org.uk
http://www.iasc.org.uk
http://www.aicpa.rog/index.htm
http://www.accounting.com/
http://www.accounting.rutgars.edu
1
Pardayev M.Q., Abdukarimov I.T, Isroilov V.I. Iqtisodiy audit. – T.: Mehnat, 2004. – 416 b., – 418 b.
2
Pardayev M.Q., Abdukarimov I.T, Isroilov V.I. Iqtisodiy audit. – T.: Mehnat, 2004. – 416 b., – 418 b.
3
Karimov A.A., Islomov F.R., Avloqulov A.Z. Buxgalteriya hisobi. – T.: Sharq, 2004.
4
Karimov A.A., Islomov F.R., Avloqulov A.Z. Buxgalteriya hisobi. – T.: Sharq, 2004.
5
Karimov A.A., Islomov F.R., Avloqulov A.Z. Buxgalteriya hisobi. – T.:
6
Gadoyev E.F., Hasanov N.X., Yugay L.P. Boshlovchilar uchun buxgalteriya hisobi. – T.:
Iqtisodiyot va huquq dunyosi
, 1996.
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Mundarija kirish I bob. Uzoq muddatli majburiyatlar hisobi bo'yicha umumiy ma’lumotlar
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SELECT MIN(id) FROM STUDENT so‘rovi nimani qaytaradi 
* STUDENTning minimal indentifikatori 
218

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SELECT MIN(id) FROM STUDENT so‘rovi nimani qaytaradi
* STUDENTning minimal indentifikatori
218
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bet | 14/45 | Sana | 07.01.2024 | Hajmi | 109,51 Kb. | | #131376 |
Bog'liq Savollar Javoblar (2)217. SELECT MIN(id) FROM STUDENT so‘rovi nimani qaytaradi
* STUDENTning minimal indentifikatori
218. SELECT MAX(id) FROM STUDENT so‘rovi nimani qaytaradi
* STUDENTning maksimal indentifikatori
219. Qaysi operatorlardan biri faqatgina bitta qator ustida operatsiya bajarishi mumkin
*Barchasi bir nechta qatorlar bilan ishlaydi
220. Quyidagi so`rov qanday javob qaytaradi SELECT * FROM
students s INNER JOIN Groups g ON s.group_id g.id
* groups va students jadvalidagi s.group_id g.id tengligi orqali
bog‘lanadigan barcha qatorlarni chiqaradi
221. SELECT SUBSTR(`familiya`,1,1) FROM `xodimlar`; so`rov natijasini toping
Xodimlar familyalarining bosh harflarini chiqaradi
222. Quyidagi savol qanday javob qaytaradi SELECT * FROM
students s FULL JOIN Groups g ON s.group_id g.id
* NULL qiymatiga ega bo‘lmagan s.group_id g.id tengligi orqali
bog‘lanadigan groups va students jadvalidagi barcha qatorlarni chiqaradi
223. Ma’lumotlarni qaysi tipiga MIN, MAX, AVG, SUM agregat funksiyalarni qo‘llash
mumkin emas
*Qatorlar
224. LIKE to`g`ri qollanilgan so`rovni ko`rsating
SELECT * FROM
Products WHERE PName
LIKE ‘%gizmo%’
225. Select * from exam Where mark in (4,5); so`rov natijasini toping
Imtixondan 4 yoki 5 olganlar hadiqa ma`lumot
226. Lower operatorining vazifasi nima?
berilgan ustunni kichik harflarga almashtirib beradi
227. SELECT LEFT(`familiya`,1) FROM `xodimlar`; so`rov natijasini toping
Xodimlar familyalarining bosh harflarini chiqaradi
228. Upper operatorining vazifasi nima?
berilgan ustunni katta harflarga almashtirib beradi
229. Initcap operatorining vazifasi nima?
berilgan ustunni har bir qatoridagi ma`lumotni
birinchi harfini bosh harfga o`zgartiradi
230. SQLda so`rov orqali jadval ustunlari o`rnini almashtirish mumkinmi
mumkin
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Ma`lumotlar bazasidagi ma`lumotlarni foydalanuvchilarga ko`rinishi qaysi tavsivlashga 
tegishli 
*logik 
15

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Ma`lumotlar bazasidagi ma`lumotlarni foydalanuvchilarga ko`rinishi qaysi tavsivlashga
tegishli
*logik
15
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bet | 2/45 | Sana | 07.01.2024 | Hajmi | 109,51 Kb. | | #131376 |
Bog'liq Savollar Javoblar (2)14.
Ma`lumotlar bazasidagi ma`lumotlarni foydalanuvchilarga ko`rinishi qaysi tavsivlashga
tegishli
*logik
15.
Bitta MBBTda nechtagacha MB bo`lishi mumkin?
*Bir nechta
16.
MBBT arxitekturasining bosqichi bo‘lmagan javob variantini ko‘rsating
* Logik
17.
MBBT arxitekturasining nechta bosqichi mavjud
*3 ta
18.
MBBT dagi MBning barcha mantiqiy strukturasini ko‘rsatuvchi abstraksiya bosqichini
ko‘rsating
*konseptual
19.
MBBTning vazifasiga nimalar kirmaydi
*Ma’lumotlar ortiqcha xajmini qisqartirish
20.
Ma’lumotlar bazasi tizimi nimalarni o‘z ichiga oladi
*Ma’lumotlar, qurilma ta’minoti, dasturiy ta’minot
foydalanuvchilari
21.
Ob’еkt atributideganda nima tushuniladi?
jadval ustuni
22.
Mohiyat aloqa diagrammasida qanday uchraydigan shaklni toping
ellips
23.
Mohiyat aloqa diagrammasida qanday shakl qo`llanilmaydi
parallelogram
24.
Relyatsion algebra amallarini ko`rsating
*Kesishuv, birlashtirish, ayirish, dekard ko`paytma
25.
MBBT dagi axborotni jismonan saqlashga yaqin abstraksiya bosqichini ko‘rsating
* ichki
26.
Ma’lum bir sust strukturaga ega va qiyin shakllanuvchi fan sohasining sun’iy intellekt
tizimi nima deb ataladi
*Ekspert tizimi
27.
Ma’lumotlar bazasining klassifikatsiyasiga tegishli bo‘lmagan javobni ko‘rsating
Ma’lumotlarni saqlash va ularga murojaat qilishni
tashkil qilish bo‘yicha
28.
Saqlanayotgan axborot turi bo‘yicha klassifikatsiya qaysi variantda tasvirlangan
*hujjatli, faktografik, leksikografik
29.
Ma’lumotlarni taqdim etish modellari bo‘yicha klassifikatsiya qaysi variantda tasvirlangan
*ierarxik, tarmoqli, relyatsion, ob’ektga yo‘naltirilgan
30.
Ma’lumotlar saqlashni va ularga murojaat qilishni tashkillishtirish bo‘yicha klassifikatsiya
qaysi variantda tasvirlangan
*lokal, tarmoqli, taqsimlangan
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II. Tashkilotning uzoq muddatli majburiyatlari. Balansning uzoq muddatli majburiyatlari. Avanslar va depozitlar

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II. Tashkilotning uzoq muddatli majburiyatlari. Balansning uzoq muddatli majburiyatlari. Avanslar va depozitlar
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bet | 8/16 | Sana | 14.05.2024 | Hajmi | 271,31 Kb. | | #233149 |
Bog'liq Uzoq muddatli majburiyatlar hisobiII. Tashkilotning uzoq muddatli majburiyatlari. Balansning uzoq muddatli majburiyatlari. Avanslar va depozitlar.
Balansda uzoq muddatli majburiyatlar.
Asosiy uzoq muddatli majburiyatlarga quyidagilar kiradi:
Tashkilotning to'lov qobiliyatini baholashda barcha mavjud uzoq muddatli majburiyatlar odatda ikkita toifaga bo'linadi: hisobot sanasidan keyin bir yil o'tmasdan to'lanishi kerak bo'lgan uzoq muddatli qarzning bir qismi va uzoq muddatli qarzning bir qismi. hisobot sanasidan keyin bir yil o'tgach to'lanishi kerak bo'lgan qarz.
Tashkilot balansida uzoq muddatli majburiyatlarning qiymatini aks ettirish uchun hisobot sanasida ma'lum hisobvaraqlarning kredit balansi to'g'risidagi ma'lumotlardan foydalanish kerak.
Bu erda balansda uzoq muddatli majburiyatlar ko'rsatkichlarini hisoblash algoritmi (Moliya vazirligining 2000 yil 31 oktyabrdagi 94n-son buyrug'i). E'tibor bering, masalan, "K67" hisobot sanasida 67-schyotning kredit qoldig'ini anglatadi.
Ko'rsatkich nomi
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Kod
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Qaysi hisob tafsilotlari buxgalteriya hisobi ishlatiladi
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Ko'rsatkichni hisoblash algoritmi
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Qarzga olingan mablag'lar
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1410
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67 "Uzoq muddatli kreditlar va qarzlar bo'yicha hisob-kitoblar"
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K67 (hisobot sanasiga 12 oydan ortiq muddatga ega bo'lgan qarz bo'yicha)
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Joriy majburiyatlar - Mundarija kirish I bob. Uzoq muddatli majburiyatlar hisobi bo'yicha umumiy ma’lumotlar

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Mundarija kirish I bob. Uzoq muddatli majburiyatlar hisobi bo'yicha umumiy ma’lumotlar
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bet | 4/16 | Sana | 14.05.2024 | Hajmi | 271,31 Kb. | | #233149 |
Bog'liq Uzoq muddatli majburiyatlar hisobiJoriy majburiyatlar quyidagi schyotlarda aks ettiriladi:
6000 «Mol yetkazib bеruvchilar va pudratchilarga to'lanadigan schyotlar»;
6100 «Ajratilgan bo’linmalar, sho'ba va qaram xo’jalik jamiyatlariga to’lanadigan schyotlar»;
6200 «Muddati kеchiktirilgan majburiyatlarni hisobga oluvchi schyotlar»;
6300 «Olingan bonaklarni hisobga oluvchi schyotlar»;
6400 «Byudjеtga to’lovlar bo’yicha qarzlarni hisobga oluvchi schyotlar»;
6500 «Sug’urta va maqsadli davlat jamg’armalariga to’lovlar bo’yicha qarzlarni hisobga oluvchi schyotlar»;
6600 «Ta'sischilarga bo’lgan qarzlarni hisobga oluvchi schyotlar»; 6700 «Mеhnat haqi bo’yicha xodimlar bilan hisoblashishlarni hisobga oluvchi schyotlar»;
6800 «Qisqa muddatli krеditlar va qarzlarni hisobga oluvchi schyotlar»;
6900 «Turli krеditorlarga bo’lgan qarzlarni hisobga oluvchi schyotlar».
Uzoq muddatli majburiyatlarga to’lov muddati bir yildan oshadigan majburiyatlar kiradi. Ularning qiymati foiz hisoblab yoziladigan asosiy majburiyat summasi bilan aniqlanadi. Qarz summasi hisobda kеlgusida pul to’lanadigan to’lovlar qiymatini diskontlash yo’li bilan olingan mazkur to’lovlarning joriy qiymati ko’rinishida aks ettiriladi. Diskontlash stavkasi inflyatsiya darajasiga, moliya bozori konyukturasi va mazkur majburiyatni bajarmaslik ehtimoli bilan bog’liq, xatar darajasiga bog’liq hamda pullarning vaqtinchalik qiymati nazariyasiga muvofiq hisob-kitob qilinadi. Uzoq muddatli majburiyatlar quyidagi schyotlarda aks ettiriladi:
7000 «Mol yetkazib bеruvchilar va pudratchilarga to’lanadigan uzoq muddatli schyotlar»;
7100 «Ajratilgan bo’linmalar, sho'ba va qaram xojalik jamiyatlariga bo’lgan uzoq, muddatli qarzlar»;
7200 «Kеchiktirilgan uzoq muddatli majburiyatlarni hisobga oluvchi schyotlar»;
7300 «Xaridor va buyurtmachilardan olingan bonaklarni hisobga oluvchi schyotlar»;
7800 «Uzoq muddatli krеditlar va qarzlarni hisobga oluvchi schyotlar»;
7900 «Turli krеditorlarga bo’lgan uzoq muddatli qarzlarni hisobga oluvchi schyotlar».
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Figure 30 – Hardware and Software RAID0 unbuffered overhead

Figure 30 – Hardware and Software RAID0 unbuffered overhead. Windows 2000’s software RAID0 and 3ware’s hardware RAID0 show almost identical results.
When file system buffering is used, Windows coalesces smaller requests into 64KB chunks and issues one deep requests to the disk. Since software RAID0 peaks at 20MBps for one-deep, 64KB requests while hardware RAID0 peaks at 30MBps, the peak buffered throughputs matched correspondingly. Riedel found that host-based RAID controller cards provided performance generally poorer than software RAID. However, given the 3ware card’s hardware RAID0 has slightly less overhead, almost identical unbuffered performance, and much better buffered throughput, it is advantageous to use the hardware RAID0 provided on the controller.
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Topshiriq
F=a-(B\a) - Mustaqil ish 1 Mavzu: Buyruqlar tizimining Bajardi: kis21-05 Jo’lliyev Shaxzod Qabul qildi

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Mustaqil ish 1 Mavzu: Buyruqlar tizimining Bajardi: kis21-05 Jo’lliyev Shaxzod Qabul qildi
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bet | 2/2 | Sana | 17.05.2024 | Hajmi | 24,67 Kb. | | #240220 |
Bog'liq Jo\'lliyev shaxzod 13Topshiriq
F=a-(B\a)
Dastur kodi
PYTHON
def calculate_F(a, B):
F = a - (B / a)
return F
# Example usage
a = 8
B = 4
F = calculate_F(a, B)
print(f"The value of F is: {F}")
C++
def calculate_F(a, B):
F = a - (B / a)
return F
# Example usage
a = 8
B = 4
F = calculate_F(a, B)
print(f"The value of F is: {F}")
Xulosa
Vazifalarni yanada samarali parallel holda bajarish jarayonini amalga oshirish uchun ikki va undan ortiq yadroni bitta mikroprotsessorda integrallashdir. Bitta kristalldagi bunday ko’p yadroli konfiguratsiya ko’p protsessor tizimlarida tashqi shinalar, kommutatorlar va hokazolarda foydalanishdan ko’ra yadrolar orasidagi almashinishning eng yuqori tezligini ta’minlaydi. Ko’p yadroli arхitektura protsessorlar unumdorligini oshirish uchun ikkita yoki undan ko’p resurslarni to’liq funksional to’plamidan iborat bo’ladi. Ko’p yadrolik va 6,5 nm teхnologik jarayon elektr iste’molini yetarlicha iqtisod qilishga va unumdorlikni oshirishga imkon berdi.
Foydalanilgan adabiyotlar
1. A.A. Xalikov. Raqamli sxemotexnika. – Toshkent, 2006.
2. Вайсман Г.М. Автоматика и телемеханика в метеорологии. – Ленинград, 1987.
3. B.M. Xolmatjanov, Z.N. Fatxullayeva. Meteorologik axborotni qayta ishlash va tahlil qilish usullari. – Toshkent, 2011.
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Fani: Kiber xavfsizlik. Guruh: 810 -22. Talaba: Daliyev Sanjar Amaliy ish

Fani: Kiber xavfsizlik. Guruh: 810 -22. Talaba: Daliyev Sanjar Amaliy ish
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Fani: Kiber xavfsizlik. Guruh: 810 -22. Talaba: Daliyev Sanjar Amaliy ish
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Is Windows xp professional x64 Edition Right for Me? White Paper Charlie Russel

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Is Windows xp professional x64 Edition Right for Me? White Paper Charlie Russel
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bet | 1/2 | Sana | 26.12.2019 | Hajmi | 240,5 Kb. | | #5246 |
Is Windows XP Professional x64 Edition Right for Me?
White Paper
Charlie Russel
Microsoft MVP
Author of Microsoft Windows Server 2003 Administrator’s Companion 2nd Edition (Microsoft Press, 2006)
Published: May 2006
For the latest information, please see http://www.microsoft.com/windowsxp/64bit/default.mspx
Abstract
This whitepaper outlines the factors that users should consider before adopting Windows XP Professional x64 Edition. The paper starts by defining what 64-bit computing is, how it’s different from 32-bit computing, and its benefits. The paper then details who the key target audience is in the short and longer terms, and finishes with a discussion of the potential roadblocks to adoption.
The information contained in this document represents the current view of Microsoft Corporation on the issues discussed as of the date of publication. Because Microsoft must respond to changing market conditions, it should not be interpreted to be a commitment on the part of Microsoft, and Microsoft cannot guarantee the accuracy of any information presented after the date of publication.
This White Paper is for informational purposes only. MICROSOFT MAKES NO WARRANTIES, EXPRESS, IMPLIED OR STATUTORY, AS TO THE INFORMATION IN THIS DOCUMENT.
Complying with all applicable copyright laws is the responsibility of the user. Without limiting the rights under copyright, no part of this document may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying, recording, or otherwise), or for any purpose, without the express written permission of Microsoft Corporation.
Microsoft may have patents, patent applications, trademarks, copyrights, or other intellectual property rights covering subject matter in this document. Except as expressly provided in any written license agreement from Microsoft, the furnishing of this document does not give you any license to these patents, trademarks, copyrights, or other intellectual property.
© 2006 Microsoft Corporation. All rights reserved.
Microsoft, Windows XP Professional x64 Edition, Windows Server, and Windows Vista are either registered trademarks or trademarks of Microsoft Corporation in the United States and/or other countries.
All other trademarks are property of their respective owners.
Contents
Introduction 1
What Is 64-Bit Computing? 2
What does 64-bit mean? 2
What are the benefits? 4
Who Is It Right For? 7
Immediately 7
Soon 7
What Do I Need to Consider? 9
System requirements 9
Applications 9
Devices 12
Evaluation 13
Windows Vista readiness 15
64-bit readiness checklist 15
Conclusion 17
Glossary 18
Memory Terms 18
Processor Terms 19
Related Links 20
Acknowledgements
Greg Amrofell, Microsoft
John Clavin, Microsoft
Serena Ho, Microsoft
Shawn Maloney, Microsoft
Padmanand Warrier, Microsoft
Lisa Pere, Studio B
Elsa Rosenberg, Studio B
David Talbott, Studio B
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17- mavzu Yarimo‘tkazgich diodlar Reja Trigger ta’rifi Bistabil yacheykalar

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17- mavzu Yarimo‘tkazgich diodlar Reja Trigger ta’rifi Bistabil yacheykalar
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bet | 1/3 | Sana | 17.12.2023 | Hajmi | 211,5 Kb. | | #121112 |
Bog'liq 6IMoUpI8oFaNBnzEczKXi5YuvZGufOp2484ISodd - 17- mavzu
- Yarimo‘tkazgich diodlar
Reja - Trigger ta’rifi
- Bistabil yacheykalar
- Trigger turlari
- Registrlar
- Hisoblagichlar
REJA Taʼrifi va shartli belgisi - Yarimoʼtkazgich diod deb bir (yoki bir necha) elektr oʼtishlarga ega ikki elektrodli elektron asbobga aytiladi. Diodlar radioelektron qurilmalarda ishlatilishi va bajaradigan vazifasiga muvofiq tasniflanadilar.
Yarimoʼtkazgichli diod VАXi - Yarimoʼtkazgich diodning toʼgʼri va teskari yoʼnalishlaridagi qarshiliklari bir – biridan keskin farq qiladi: toʼgʼri yoʼnalishda siljitilgan diodning qarshiligi qiymati kichik, teskari siljitilgan diodniki esa – katta boʼladi. Shu sababdan diod bir tomonga elektr tokini yaxshi oʼtkazadi, ikkinchi tomonga esa – yomon oʼtkazadi.
Diodni asosiy parametrlari - 1. Statik qarshilik
- [Ом] ;
- 2. Differentsial qarshilik
-
- [Ом] ;
- 3. Xarakteristika tikligi
- [А/В] .
REJA - Teskari kuchlanish maʼlum chegaraviy qiymatga UChEG yetganda, teskari tok keskin ortib ketadi, oʼtishning elektr teshilishi yuz beradi.
- Oʼtishning teshilish turlari ikki guruhga boʼlinadi: elektr va issiqlik.
- Elektr teshilishining ikki mexanizmi mavjud: koʼchkisimon va tunnel teshilish.
p-n oʼtish teshilish turlari - Koʼchkisimon teshilish nisbatan keng p-n oʼtishlarda sodir boʼladi. Bunday oʼtishda teskari kuchlanishda elektron va kovaklar zarba ionizatsiyasi uchun yetarli boʼlgan energiya oladilar va natijada qoʼshimcha elektron-kovak juftlar hosil boʼladi. Bu juftliklarning har bir tashkil etuvchisi, oʼz navbatida, elektr maydonida tezlashib, yana yangi juftlikni yuzaga keltiradi va x.z. Zaryad tashuvchilarning bunday koʼchkisimon koʼpayishi natijasida oʼtishdagi tok keskin ortadi.
- Tor p-n oʼtishga ega boʼlgan yarim oʼtkazgichlarda tunnel effektiga asoslangan tunnel teshilish sodir boʼladi. UTES UChEG yetganda zaryad tashuvchilarning bir sohadan ikkinchisiga energiya sarf qilmasdan oʼtishiga imkon yaratiladi (tunnelь effekti). UChEGning yanada ortishi bilan shuncha koʼp zaryad tashuvchilar tunnel oʼtishi sodir etadilar va teskari tok keskin ortib boradi.
- p-n oʼtishda issiqlik teshilishi teskari tok oʼtish natijasida oʼtishning qizishi hisobiga sodir boʼladi. Teskari tok, issiqlik toki boʼlib, u ortgan sari qizish ham ortadi. Bu holat tokning koʼchkisimon ortishiga olib keladi, natijada p-n oʼtishda issiqlik teshilishi yuz beradi va u ishdan chiqadi.
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Dastur yodamida tarmoq paketlarini ushlaydi va tahlil

Biz Ethernet ni tanladik va analiz jarayoni ketdi
Zaxvat jarayoni boshlandi
Biz ximoyalanmagan saytni tanlaymiz va uni maʼlumotlarini tutish ishini olib boramiz
Ximoyalanmagan sayt yaʼni http da joylashgan saytni tutishga xarakat qilamiz
http://ferlibrary.uz/about ushbu sayt http:// da joylashganligi tufayli
sayt trafigini tutamiz
Shu saytga kirib login parolni terib kuramiz
Mavzu: Muloqotning tuzilishi. Muloqotning psixologik vositalari va darajalari. Reja

Mavzu: Muloqotning tuzilishi. Muloqotning psixologik vositalari va darajalari. Reja
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Mavzu: Muloqotning tuzilishi. Muloqotning psixologik vositalari va darajalari. Reja
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Tuxumdan tayyorlangan 8 xil eng zo’r nonushtalar

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Tuxumdan tayyorlangan 8 xil eng zo’r nonushtalar
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bet | 13/16 | Sana | 18.05.2024 | Hajmi | 4,56 Mb. | | #242299 |
Bog'liq Абраева ЛолаTuxumdan tayyorlangan 8 xil eng zo’r nonushtalar
Tuxum, mutlaqo universal mahsulot bo’lib, u mavsumga qaramay foydali va dunyoning istalgan nuqtasida, barcha qit’alarda mavjud.
Tuxumdan – qovurilgan tuxum, pishirilgan tuxum, yumshoq qaynatilgan tuxum, pashot tuxumi … Va qancha asl taomlarni tayyorlash mumkin! Agar g’oyalaringiz tugab qolsa, bizning ro’yxatimizni oching va har kungi tongingizni mazali tarzda boshlang!
Feta pishlog’i va pomidordan quymoq
Dunyodagi eng mashhur nonushtalardan biri. Bu yerda hamma narsa juda oddiy: ko’p pomidor, tuxum, ko’katlar, shuningdek bizning retseptimizda feta pishlog’i ham bor.
Pishloqni eritib olish uchun pishirgandan keyin bu omletni pechda bir necha daqiqa ushlab turish yaxshi maqsadga muvofiq.
Nonushtabop tuxumli skrembl
Bir necha daqiqada pishiradigan nonushta. Tuxum, yog ‘, tuz, murch va yaxshi kayfiyat sizga kerak bo’lgan narsadir.
Tuxumli skrembl – bu nonushtada tayyorlaniladigan ana’anaviy yevropacha taom bo’lib, tuxum va sariyog’dan pishiriladi. Xohishga qarab bunday yegulikni faqatgini tuxumning o’zidan tayyorlashingiz, yoki qo’shimchalardan ham solishingiz mumkin bo’lib, masalan, sabzavotlar, qo’ziqorin, pishloq, ko’klatlar va ziravorlar.
Lavash ichiga solib pishirilgan tuxum
Nega bu nonushtani tanlash kerak. Birinchidan, bu o’ziga xos nonushta. Ikkinchidan, juda chiroyli – kamdan-kam odamlar bunday nonushtadan bosh tortishadi. Uchinchidan, lavash xamiri ichiga pishirilgan tuxumva ba’zi bir foydali sabzavotlarni bolalarga yashirincha solib berishni ajoyib usuli.
Menemen – turkcha qovurilgan tuxum
Menemen – dam olish kunlari nonushtasining qiroli. Turk oshxonasidan juda mazali omlet.
Menemenni odatda yangi non bilan dasturxonga tortishadi va nonni mazali tuxum va sabzavotli aralashmag botirib, yeyishadi.
Benedikt tuxumi
Tayyorlanishiga ko’p vaqt va ma’lum bir mahorat talab qiladigan ajoyib retsept.
Ammo xozir ushbu noodatiy taomn tayyorlashni o’rganishning ayni vaqti.
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Noinersial sanoq sistemalari

Noinersial sanoq sistemalari.
Yo’lning to’g’ri chiziqli gorizontal qismida harakatlanayotgan vagon ichidagi jismlar vaziyatini tekshiraylik. Buning uchun Yer sirti bilan bog’langan К sanoq sistemasida va vagon bilan bog’langan К' sanoq sistemasida turgan kuzatuvchilar nuqtai nazaridan fikr yuritaylik.
V agon harakatlanmayotgan holda К va К' sanoq sistemalaridagi kuzatuvchilarning fikri aynan bir xil bo’ladi (5.1 rasm). Vagonning gorizontal polida turgan sharning og’rlik kuchi polning reaktsiya kuchi bilan muvozanatlanganligi uchun Nyutonning birinchi qonuniga asosan, shar o’zining tinch holatini saqlaydi. Vagon to’g’ri chiziq bo’lib tekis harakatlangan
holda ham sharning vaziyati o’zgarmasligini К va К' sanoq sistemalaridagi kuzatuvchilar qayd qiladilar. Ma‘lumki, Yer sirti bilan bog’langan sanoq sistemasini taqriban inertsial sanoq sistemasi deb hisoblash mumkin edi. Shuning uchun К’ sanoq sistemasi К sanoq sistemasiga (ya‘ni inertsial sanoq sistemasiga) nisbatan tinch turgan yoki to’g’ri chiziqli tekis harakat qilayotgan hollarda inertsial sanoq sistemasi deb hisoblanadi.
K' sanoq sistemasidagi kuzatuvchi esa sharni chap tomonga qarab tezlanuvchan harakat qilayotgailigini qayd qiladi (5.3 rasm). Jism tezlanishga ershishi uchun unga, Nyutonning ikkinchi qonuniga asosan, biror kuch ta‘sir etishi lozim, albatta. Shuning uchun К' sanoq sistemasidagi kuzatuvchi sharga mazkur kuch bilan ta‘sir etayotgan jismni axtaradi, lekin topa olmaydi. Natijada kuzatuvchi quyidagi xulosaga keladi: К’ sanoq sistemasidagi sharga to’liq jismlar ta‘sir etmayotgan bo’lsada, u o’zining
tinch vaziyatini yoki to’g’ri chiziqli tekis harakatini saqlamayapti, ya‘ni inertsiya qonuni bajarilmayapti. Shuning uchun K’ sistemadagi kuzatuvchi mazkur sistemani noinertsial sanoq sistemasi deb hisoblashi, tabiiydir. Aylanuvchi sanoq sistemalaridagi jismlar uchun ham inertsiya qonuni bajarilmaydi.
Bunga quyidagi tajriba asosida ishonch hosil qilish mumkin. 5.4 а rasmda tasvirlangan disk ustiga П-simon sterjen o’rnatilgan, sterjenga esa sharlar osilgan. Disk tinch turganida sharlar osilgan barcha iplar vertikal ravishda yo’nalgan. Disk aylanish o’qi atrofida burchak tezlik bilan aylantirilganda (5.4 b rasm) sharlar (ipi OZ bilan ustma-ust tushgan shardan boshqalari) tashqi tomonga og’adi. Bu holda ham sharlarga to’liq jismlar ta‘sir etmayapti. Lekin sharlar tezlanish oladi. Shuning uchun aylanuvchi sanoq sistemasini ham noinertsial sanoq sistemasi deb hisoblanadi.
5.4a,b rasm
D emak tezlanuvchan ilgarilanma harakat qilayotgan sanoq sistemasidagi ixtiyoriy jismga ta‘sir etadigan inertsiya kuchining yunalishi sanoq sistemasining harakati o’ynalishiga teskari, kuchning moduli esa jism massasi bilan sanoq sistemasi tezlaniishning ko’paytmasiga teng. Endi, inertsiya kuchi tushunchasidan foydalanib noinertsial sanoq sistemalari uchun harakat tenglamasini yozaylik. Tabiiyki, bu holda jismga ta‘sir etuvchi kuchlarning vektor yig’indisiga nyuton kuchlari bilan bir qatorda inertsiya kuchi ham hissa qo’shadi:
YOKI
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Axborot xavfsizligi ta’minlash usullari. Risqulov Bahodir - bet 3

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Axborot xavfsizligi ta’minlash usullari. Risqulov Bahodir
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bet | 3/4 | Sana | 18.01.2024 | Hajmi | 0,7 Mb. | | #140175 |
Bog'liq Axborot xavfsizligi ta’minlash usullari.pptmLogin shaxsning, o'zini axborot kommunikatsiya tizimiga tanishtirish jarayonida qo'llaniladigan belgilar ketma-ketligi bo'lib, axborot kommunikatsiya tizimidan foydalanish huquqiga ega bo'lish uchun foydalaniluvchining maxfiy bo'lmagan qayd yozuvi hisoblanadi. Parol uning egasi haqiqiyligini aniqlash jarayonida tekshiruv axboroti sifatida ishlatiladigan belgilar ketma- ketligi. U kompyuter bilan muloqot boshlashdan oldin, unga klaviatura yoki identifikatsiya kartasi yordamida kiritiladigan harfli, raqamli yoki harfli-raqamli kod shaklidagi mahfiy so'zdan iborat. Login Parol Avtorizatsiya foydalanuvchining resursdan foydalanish huquqlari va ruxsatlarini tekshirish jarayoni. Bunda foydalanuvchiga hisoblash tizimida ba'zi ishlarni bajarish uchun muayyan huquqlar beriladi. Avtorizatsiya shaxs harakati doirasini va u foydalanadigan resurslarni belgilaydi. Registratsiya foydalanuvchilarni ro'yxatga olish va ularga dasturlar va ma'lumotlarni ishlatishga huquq berish jarayoni. Ayrim veb-saytlar foydalanuvchilarga qo'shimcha xizmatlarni olish va pullik xizmatlarga obuna bo'lish uchun ro'yxatdan o'tishni, ya'ni o'zi haqida ayrim ma'lumotlarni kiritishni (anketa to'ldirishni) hamda login va parol olishni taklif qiladilar. Foydalanuvchi ro'yxatdan o'tgandan so'ng tizimda unga qayd yozuvi (account) yaratiladi va unda foydalanuvchiga tegishli axborotlar saqlanadi. Avtorizatsiya Registratsiya Kompyuter virusi Kompyuter virusi - bu o'z-o'zidan ko'payuvchi, kompyuter tarmoqlari va axborot tashuvchilari orqali erkin tarqaluvchi, hamda kompyuter va unda saqlanayotgan axborot va dasturlarga zarar etkazuvchi dastur kodi yoki komandalar ketma-ketligi hisoblanadi. Kompyuter viruslari quyidagi xossalarga ega: o'zidan nusxa ko'chirish, axborotdan ruxsatsiz foydalanishni amalga oshirish. U o'zining nusxalarini kompyuterlarda yoki kompyuter tarmoqlarida qayta ko'paytirib va tarqatib, hamda qonuniy foydalanuvchilar uchun nomaqbul harakatlarni bajaradi. Virus, aksariyat hollarda nosozlik va buzilishlarga sabab bo'ladi va biror hodisa yuz berishi bilan, masalan, aniq kunning kelishi bilan ishga tushirilishi mumkin.
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Antivirus Antivirus - Axborot xavfsizligi ta’minlash usullari. Risqulov Bahodir

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Axborot xavfsizligi ta’minlash usullari. Risqulov Bahodir
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bet | 4/4 | Sana | 18.01.2024 | Hajmi | 0,7 Mb. | | #140175 |
Bog'liq Axborot xavfsizligi ta’minlash usullari.pptmAntivirus Antivirus
Hozirgi kunda kompyuter viruslarini aniqlash va ulardan himoyalanish uchun maxsus dasturlarning bir necha xillari ishlab chiqilgan bo'lib bu dasturlar kompyuter viruslarini aniqlash va yo'qotishga imkon beradi. Bunday dasturlar virusga qarshi dasturlar yoki antiviruslar deb yuritiladi. Antivirus dasturlariga AVP, Doctorweb, Nod32 va boshqalar dasturlarini kiritish mumkin. Umuman barcha virusga qarshi dasturlar zaharlangan dasturlar va yuklama sektorlarning avtomatik tarzda tiklanishini ta'minlaydi.
1.Muntazam profilaktika ishlarini, ya'ni virusga tekshiruv ishlarini olib borish.
2. Taniqli virusni zararsizlantirish
3. Taniqli bo'lmagan virusni zarasizlantirish
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Asboblar paneli - Fani: Kiber xavfsizlik. Guruh: 810 -22. Talaba: Daliyev Sanjar Amaliy ish

Asboblar paneli: Wireshark asboblar panelidagi chap tomonda joylashgan ko'k sariq
sindirish tugmasini bosing
Haqiqiy ta'qib qilish jarayoni endi boshlanadi va paketli ma'lumotlar Wireshark oynasida
saqlanadi. Yozib olishni to'xtatish uchun quyidagi harakatlardan birini bajaring.
Klaviatura: Ctrl + E tugmalarini bosing
Asboblar paneli: Wireshark asboblar panelidagi köpekbalığı finining yonida joylashgan
qizil tugmachani bosing
Paket tarkibini ko'rish va tahlil qilish
Scott Orgera
Endi ba'zi tarmoq ma'lumotlarini yozib olganingizdan so'ng, olingan paketlarni ko'rib
chiqish vaqti keldi. Yuqoridagi rasmda ko'rsatilganidek, olingan ma'lumotlar interfeysi
uchta asosiy bo'limni o'z ichiga oladi: paketlar ro'yxati paneli, paket tafsilotlari paneli va
paket baytlar paneli.
Paket ro'yxati Oynaning yuqori qismida joylashgan paketlar ro'yxati paneli faol suratga olish faylida
joylashgan barcha paketlarni ko'rsatadi. Har bir paket o'z navbatida va unga mos
keladigan raqamga ega bo'lib, ushbu ma'lumotlar nuqtalarining har biri bilan birga.
Vaqt: Paketning qachon qo'lga kiritilganligi haqidagi vaqt belgilari bu ustunda
ko'rsatiladi, bu maxsus ta'qib qilish fayllari birinchi yaratilganligi uchun standart
formatga soniyalar soni (yoki qisman soniyalar) kiradi. Ushbu formatni biroz ko'proq
foydali bo'lishi mumkin bo'lgan, masalan, kunning haqiqiy vaqti kabi o'zgartirish
uchun asosiy interfeysning yuqori qismida joylashgan
Wireshark
ko'rinishi menyusidan
Time Display Format- ni tanlang.
Manba: Bu ustun paketni tashkil etadigan manzilni (IP yoki boshqa) o'z ichiga oladi.
Nishon: Bu ustun paketga yuborilgan manzilni o'z ichiga oladi.
Centos and Ubuntu Linux distributions to assign ip address automatically to a client machine. How Does dhcp work?

DHCP Server and Client on
CentOS and Ubuntu
I will explain how to configure a DHCP server in
CentOS
and
Ubuntu
Linux distributions to assign IP address automatically to a client machine.
How Does DHCP Work?
The following is a quick description of how DHCP actually works:
Once a client (that is configured to use DHCP) and connected to a network
boots up, it sends a
DHCPDISCOVER
packet to the DHCP server.
When the DHCP server receives the
DHCPDISCOVER
request packet, it
replies with a
DHCPOFFER
packet.
Then the client gets the
DHCPOFFER
packet, and it sends
a
DHCPREQUEST
packet to the server showing it is ready to receive the
network configuration information provided in the
DHCPOFFER
packet.
Finally, after the DHCP server receives the
DHCPREQUEST
packet from
the client, it sends the
DHCPACK
packet showing that the client is now
permitted to use the IP address assigned to it.
In this article, I will show you how to setup a DHCP server in
Ubuntu/Debian Linux, and we will run all the commands with the sudo
command to gain root user privileges.
Installing DHCP Server in CentOS and
Ubuntu
The
DCHP
server package is available in the official repositories of mainstream
Linux distributions, installing is quite easy, simply run the following command.
# yum install dhcp #CentOS
$ sudo apt install isc-dhcp-server #Ubuntu
Once the installation
is complete, configure the interface on which you want the
DHCP
daemon to
serve requests in
the configuration file
/etc/default/isc-dhcp-server
or
Ikkinchi xili (5-rasm) - Mahsulot tayyorlash texnologiyasi

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Mahsulot tayyorlash texnologiyasi
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bet | 15/16 | Sana | 18.05.2024 | Hajmi | 4,56 Mb. | | #242299 |
Bog'liq Абраева ЛолаIkkinchi xili (5-rasm):
Kеrakli masalliqlar:
1 banka baliq konsеrvasi, 1 dona qaynatilgan tuxum, 1 dona pishirilgan sabzi, 2 osh qoshiq konsеrvalangan ko‘k no‘xot, 1–2 osh qoshiq mayda to‘g‘ralgan ko‘k piyoz, 2 osh qoshiq mayonеz, ko‘katlar. Ishni bajarish tartibi
1. Tuxum qaynatib pishiriladi, sovigach, po‘chog‘idan tozalab, sarig‘i oqidan ajratib olinadi.
2. Tuxumning oqi va pishirilgan sabzi kubik shaklida to‘g‘raladi.
3. Konsеrva bankasini ochib, baliq moyi bilan tovoqqa solinadi va sanchqida eziladi.
4. Salat idishga hamma masalliq solinib, unga ko‘k no‘xot qo‘shiladi va mayonеz solib aralashtiriladi.
5. Bu massa ustiga tayyorlangan baliqning yarmi, uning ustiga piyoz va kеyingi qatlamga qolgan baliq solinadi.
6. Hammasining ustiga mayonеz solib, tayyor salat sovitkichda 1,5–2 soat sovitib, dasturxonga tortishdan avval maydalangan tuxum sarig‘i sеpiladi va ko‘katlar bilan bеzatiladi.
Xulosa
Xulosa o`rnida shuni aytish mumkinki, Inson organizmi deyarli barcha
kerakli moddalarni oziq-ovqat va suv orqali oladi. Oziq-ovqat mahsulotlarining
tarkibi va ularning xususiyatlari salomatlik, jismoniy rivojlanganlik, mehnat
qobiliyati, hissiy holat va umuman, umr ko’rish davomiyligi va sifatiga bevosita
ta’sir qiladi. Inson organizmiga shu qadar ta’sir qilishga qodir bo’lgan yana boshqa bir omilni topish qiyin.
To’g’ri ovqatlanish borasida ko’plab nazariyalar mavjud va deyarli har kuni
yangi ma’lumotlar paydo bo’ladi. Ularning ba’zilari go’shtni cheklash kerak desa,
boshqa birlari faqat ma’lum bir mahsulotlardan iborat taomnoma har qanday kasalliklardan forig’ bo’lish kaliti deyishadi. Aslida inson tanasi juda murakkab
tizimdir, uning normal ishlashi uchun ham go’sht tuxum, ham sabzavotli taomlarni talab qiladi. Shu sababli to’g’ri ovqatlanish qoidalarini bilish juda muhim sanaladi. Bu borada shubhali tavsiyalarga amal qilish salomatlikda jiddiy muammolar keltirib chiqarishi mumkin.
To’g’ri ovqatlanish bilan bir qatorda ovqatlanish vaqtini ham aniq hisoblash
kerak, chunki faqat o’z vaqtida ovqatlanish energiya sarfini to’liq qoplaydi.
Sog’lom odam uchun kuniga to’rt marta ovqatlanish eng oqilona hisoblanadi.
Bunda ovqat hazm qilish tizimi tushadigan yuk teng taqsimlanadi, oziq-ovqat
yaxshiroq hazm bo’ladi va o’zlashtiriladi.
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Uni asosida kuyidagi postulatlar yotadi

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Uni asosida kuyidagi postulatlar yotadi
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bet | 6/6 | Sana | 19.01.2024 | Hajmi | 22,44 Kb. | | #141228 |
Bog'liq 1-mustaqil ish reja-fayllar.orgUni asosida kuyidagi postulatlar yotadi :
1. Barcha inеrtsial sanok sistеmalarida bir xil sharoitda olingan xamma fizik xodisalar (mеxanik, elеktromagnit, optik va x.k.) bir xilda ruy bеradi.
2. Vakuumdagi yoruglik tеzligi s barcha in-еrtsial sanok sistеmalarida bir xil bulib uzgar-mas absolyut kattalikdir, ya'ni, u xam inеrtsial sistеmaga nisbatan invariantdir.
Eynshtеynning bu postulatlari katta tеzlik bilan xarakatlanuvchi jismlar dinamikasini urganuvchi rеlyativistik mеxanika uchun Galilеy nisbiylik printsipining davomi va umumlashgan ifodasidir. Utkazilgan kator tajribalar va maxsus nisbiylik nazariyasidan kеlib chikadigan natijalarning tajribaga mos kеlishi bu prin-tsipning tugriligini tasdikladi, ammo uni tula -tukis isbotlash imkoni yuk, chunki xali ochilma-gan tabiat xodisalari juda kup
Xulosa
Qator texnik qiyinchiliklar tufayli reaktiv va raketa texnikasining keng-ko'lamda o'sishi faqat ikkinchi jahon urushi davrida va asosan urush tamom bo'lgandan keyin boshlandi. Reaktiv dvigatellarni aviatsiyada qo'llash samolyotlarning tezligini, uchish uzoqligini va yuk ko'tarish kabi xususiyatlarini bir necha marta ko'paytirish imkonini berdi.
Raketa texnikasi, uning asosida Yer sun'iy yo'ldoshlari, boshqariladigan kosmik kemalar, arbital va planetalararo stantsiyalarni uchirish mumkinligiga asos bo'lib qoldi.
Adabiyotlar.
1.Strеlkov S.P. Mеxanika, T.,1977.
2.Sivuxin D.V. Umumiy fizika kursi. 1-tom. T.,1981.
3.Raxmatullaеv M. Umumiy fizika kursi. Mеxanika. 1995.
www.pedagog.uz
www.ziyonet.uz
http://fayllar.org
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