Ethereum EDC
Publish: 2021-03-25 22:21:02
1. 1. What is E & E net system
e & E net system adopts the world's first proprietary communication protocol EDC (electronic disk Communications), which concentrates all the hard disk storage resources in the network on the Ethernet network hard disk manager E & E, and virtualizes multiple electronic hard disks for client computers. It not only has the flexibility of ordinary PC, but also has the characteristics of easy management and security of thin client computers
economically, the investment in operating system and application software is reced while the cost of system hardware is reced. The installation and implementation of the whole system is more simple and fast, and the requirement for maintenance personnel is very low - only having single machine maintenance experience can significantly rece the total cost of ownership of the system
2. What is the essence of E & E
e & E is not an online network disk drive, but a physical C hard disk recognized by the operating system of the edisk network terminal. The meaning of E & E includes the following five points:
e & E stands for Ethernet hard disk
e & E stands for extensible hard disk
e & E stands for external hard disk
e & E stands for simple hard disk
e & E stands for efficient hard disk
3. What is the EDC (TM) agreement< EDC (TM) protocol is a set of disk management and service supply protocol, which can be used as the basis of various application service management and supply in enterprise IT environment
the basic principle of EDC (TM) is to use a set of disk based computing architecture, treat various service suites as disks on the computing platform (that is, terminal computers in client SAN), and engage in service management and supply; Its function is just like the disk used in the traditional personal computer to store the application services that can be executed on the computer
e DC (TM) protocol is basically the presentation layer protocol in OSI 7-layer network mode. It contains multiple groups of protocol instructions, which can be used to perform disk access and management (such as construction, deconstruction, capacity adjustment, replication, access control, etc.)
all kinds of tools provided by any manufacturer can be easily built by using EDC (TM) protocol, so that all kinds of services designated for specific users in IT environment can be managed with high precision
the client San is equipped with a set of EDC (TM) disk emulator on each terminal computer, which is used for data transmission with the EDC (TM) service supply server (usually used to store the EDC (TM) management mole). This set of EDC (TM) emulator presents the form of disk on the terminal computer after the server obtains the disk section
the server of EDC (TM) is responsible for managing a set of shared data storage space and constructing virtual disks for terminal computers that require services in a dynamic way. This EDC (TM) server will first verify the legitimacy of the user's request when receiving the service request, and then provide disk access service for the verified user.
e & E net system adopts the world's first proprietary communication protocol EDC (electronic disk Communications), which concentrates all the hard disk storage resources in the network on the Ethernet network hard disk manager E & E, and virtualizes multiple electronic hard disks for client computers. It not only has the flexibility of ordinary PC, but also has the characteristics of easy management and security of thin client computers
economically, the investment in operating system and application software is reced while the cost of system hardware is reced. The installation and implementation of the whole system is more simple and fast, and the requirement for maintenance personnel is very low - only having single machine maintenance experience can significantly rece the total cost of ownership of the system
2. What is the essence of E & E
e & E is not an online network disk drive, but a physical C hard disk recognized by the operating system of the edisk network terminal. The meaning of E & E includes the following five points:
e & E stands for Ethernet hard disk
e & E stands for extensible hard disk
e & E stands for external hard disk
e & E stands for simple hard disk
e & E stands for efficient hard disk
3. What is the EDC (TM) agreement< EDC (TM) protocol is a set of disk management and service supply protocol, which can be used as the basis of various application service management and supply in enterprise IT environment
the basic principle of EDC (TM) is to use a set of disk based computing architecture, treat various service suites as disks on the computing platform (that is, terminal computers in client SAN), and engage in service management and supply; Its function is just like the disk used in the traditional personal computer to store the application services that can be executed on the computer
e DC (TM) protocol is basically the presentation layer protocol in OSI 7-layer network mode. It contains multiple groups of protocol instructions, which can be used to perform disk access and management (such as construction, deconstruction, capacity adjustment, replication, access control, etc.)
all kinds of tools provided by any manufacturer can be easily built by using EDC (TM) protocol, so that all kinds of services designated for specific users in IT environment can be managed with high precision
the client San is equipped with a set of EDC (TM) disk emulator on each terminal computer, which is used for data transmission with the EDC (TM) service supply server (usually used to store the EDC (TM) management mole). This set of EDC (TM) emulator presents the form of disk on the terminal computer after the server obtains the disk section
the server of EDC (TM) is responsible for managing a set of shared data storage space and constructing virtual disks for terminal computers that require services in a dynamic way. This EDC (TM) server will first verify the legitimacy of the user's request when receiving the service request, and then provide disk access service for the verified user.
2.
How to use the Ethernet upload program to upload the touch screen mt8100ie
you need to first see that the IP address needs to be consistent with the internal address of the touch screen
how to upload wintone touch screen program< br />://g. hiphotos://c. : / / g.hiphotos.. JPG "it knowledge ESRC =" http., uploaded content.. hiphotos.com / / wh% 3d600% 2c800 / sign = /: / / g.com / / wh% 3d600% 2c800 / sign = /. COM / / PIC / item /. JPG "/ & gt; Step 5. COM / / PIC / item / hiphotos://g.jpg ＂ esrc=＂http.hiphotos. hiphotos://e.hiphotos.jpg ＂ esrc=＂http..jpg＂ target=＂_ Blank "title =" click
how to set such a switch in the wintone touch screen editing software_ 100
reset mode is reset when the hand is released. You can select the type of switch, click on and then off, or set it to "set to on", and the closing condition is set in PLC
3. Example 1 is known: as shown in Figure 1-1, in the quadrilateral ABCD, BC > Ba, ad = CD, BD bisects ∠ ABC,
verification: ∠ a + C = 180 °
analysis: because the horizontal angle is equal to 180 °， Therefore, we should consider transforming two angles which are not together into equal angles by congruence, and there is no congruent triangle in the graph. Therefore, the key to solve the problem is to construct isosceles triangle, which can be realized by "truncation and complement method". "
prove: intercept be = AB on BC, connect De, and then take the midpoint m of EC, Connecting DM
∵ AB = be
and ∵ BD bisecting ∠ ABC a d
∵ abd = ∠ EBD
in △ abd and △ EBD,
AB = be
∠ abd = ∠ EBD
BD = BD B e M C
{abd ≌ △ EBD (SAS) is shown in Figure 1-1
} ad = ed ∠ a = ∠ bed,
∵AD = DC ， ∴ED = DC ∴∠ C = ∠DEC
∴∠A + ∠C = ∠BED +∠DEC = 180 °< As shown in Figure 2-2, AE / / BC, ad and BD share & eth equally; EAB、Ð CBA, EC, point D
verification: ab = AE + BC
analysis 1: to prove that ab = AE + BC, observe that AD and BD are bisectors, so the Daed can be folded along a, so it is necessary to add an auxiliary line to intercept BF = BC on AB, and only need to dece AF = AE, then the problem can be solved, so how to dece AF = AE becomes the key to solve the problem. Because the sum of internal angles of Daed, DADB and DBD is 180 °， And & eth; EDC=180 °， Because AE / / be, therefore & eth; E+Ð C=180 ° Thus & eth; EAB+Ð CBA=180 °， Ad and BD are bisectors of angles, and we can dece & eth; 1+Ð 4=90 °， Thus, we can dece & eth; ADB=90 °， Thus & eth; 6+Ð 8=90 ° If we can dece & eth; 7=Ð 8, then we only need to dece Daed ≌ dafd, thus we can dece AE = AF, BC = BF, & eth; 1=Ð 2. BD is a common edge, so we can dece dbfd ≌ DBCD, then & eth; 5=Ð 6, e to & eth; 5+Ð 7=90 ° So & eth; 6+Ð 7=90 °， Due to & eth; 6+Ð 8=90 °， Thus & eth; 7=Ð Therefore, ad can be the common edge, & eth; 3=Ð 4 dece Daed ≌ dafd, thus the idea is smooth, dece AE = AF, and dece AB = AE + BC by equivalent substitution
proof 1: intercept BF = BC on AB and link DF
BD is & eth; Bisector of ABC & eth; 1=Ð 2
in dbdf and DBDC,
(common side)
{dbdf ≌ DBDC (SAS) is shown in Figure 2-2
} & eth; 5=Ð 6 (the corresponding angles of congruent triangles are equal)
! & eth; 3+Ð 8+Ð E=Ð 4+Ð 1+Ð 5+Ð 7=Ð 2+Ð 6+Ð C=180 ° The sum theorem of internal angles of triangle)
{eth; E+Ð EAB+Ð ABC+Ð C+Ð EDC=540 °
and ﹣ AE / / BC ﹣ eth; E+Ð C=180 ° Two lines are parallel and complementary)
and ∵ & eth; EDC=180 ° ∴Ð 1+Ð 2+Ð 3+ Ð 4=180 °
ad is & eth; The bisector of EAB & eth; 3=Ð 4
∴Ð 1+Ð 4=90 ° ∴Ð 5+Ð 7=90 ° The sum theorem of internal angles of triangle)
{eth; 6+Ð 8=90 ° ∵Ð 5=Ð 6 ∴ Ð 7=Ð 8
in Daed and dafd,
{Daed ≌ dafd (ASA)
} AE = AF (the corresponding sides of congruent triangles are equal)
≌ AF + FB = AB
} AE = FB = AE + BC = AB
that is, ab = AE + BC
analysis 2: extend the extension line of BC intersecting ad to F. To prove AB = AE + BC, we only need to prove BF = AB and dece CF = AE. In order to prove CF = AE, we only need to dece two triangles Daed ≌ dfcd with CF and AE e to & eth; 5 = 6, AE / / BC, so & eth; 3=Ð F. If we want to dece ad = FD, it becomes the key to solve the problem °，& ETH; E+Ð BCE=180 °， Therefore, it can be known that & eth; EAB+Ð CBA=180 °， Ad and BD are & eth; EAB、Ð CBA's bisector, thus can be introced & eth; 1+Ð 4=90 °， Therefore & eth; ADB=90 °， Then & eth; EDB=90 °， At this point, they can dece dabd ≌ dfbd according to ASA by observing the figure, and thus dece ad = FD, forming the idea
proof 2: as shown in Fig. 2-3, the sum of the internal angles of the three triangles in Daed, DADB and DBDC is 540 ° The sum theorem of internal angles of triangles)
and ∵ & eth; EDC=180 ° Horizontal angle definition & eth; E+Ð C+Ð EAB+Ð ABC=180 °
AE / / BC (two lines are parallel and complementary)
■ & eth; 3+Ð 4+Ð 1+Ð 2=180 °
AD and BD are & eth respectively; EAB、Ð The bisector of ABC
& eth; 3=Ð 4，Ð 1=Ð 2 (angular bisector definition)
﹣ eth; 1+Ð 4=90 ° ∴Ð ADB=90 ° The sum theorem of internal angles of triangle)
{eth; BDF=90 ° In DADB and dbdf,
≌ DADB ≌ dbdf (ASA)
≌ ad = FD, ab = FB, & eth; 4=Ð F (opposite sides of congruent triangle, corresponding angles are equal) as shown in Fig. 2-3
in Daed and dfcd, it is known that ad is the angular bisector of △ ABC, AB > AC, as shown in Fig. 3-1,
verification: AB-AC > bd-dc
analysis: in order to prove AB-AC > bd-dc, we need to transform the difference between AB and AC, the difference between BD and DC or their equal quantity into the edge of the same triangle, and then use the relationship between the three sides of the triangle to prove it
proof: Method 1: intercept AE = AC on AB and connect ed. A
∫ ad bisection ∠ BAC ∫ bad = ∠ DAC
in △ ade and △ ADC, E
AE = AC
ead = ∠ DAC B D C
ad = ad as shown in Figure 3-1
{△ ade ≌ △ ADC (SAS)} de = D C
in △ abd, Be > bd-de (the difference between the two sides of the triangle is less than the third side)
that is, ab-ae > bd-dc
Ψ AB-AC > bd-dc (equivalent substitution)
method 2: short complement method
extend AC to point E, so that AE = ab, Connecting de a
∫ ad bisection ∠ BAC ∫ bad = ∠ DAC
in △ bad and △ ead,
AB = AEC
∠ bad = ∠ DAC B D E
ad = ad
{ade ≌ delta ADC (SAS)} D B = D E, as shown in figure 3-2
in △ abd, EC > de-dc (the difference between the two sides of the triangle is less than the third side)
that is ae-ac > de-dc AB-AC > bd-dc
example 4 shows that, as shown in Figure 4-1, in △ ABC, ∠ C = 2 ∠ B, ∠ 1 = 2.
verification: ab = AC + CD.
analysis: from the conclusion analysis, "truncation" or "complement" can realize the transformation of the problem, that is, extending AC to e makes CE = CD, It is proved that: Method 1 (short complement method)
prolongs AC to e, so that DC = CE, then ∠ CDE = ∠ CED, as shown in Figure 4-2
Figure 4-2
achb = 2 ∠ e,
∫ ACB = 2 ∠ B, ∫ B = e,
in △ abd and △ AED,
{abd ≌ △ AED (AAS)
, ≌, And the AE = AC + CE = AC + DC, < br /
< br /
< br /
< br / < br / < br / < br /
< br /
< br / < br / < br / < method 2 (truncation method) < br / < br /
in the Figure 4-3, as shown in Figure 4-3 < br / < br > < br / < br /
in △ AFD and △ ACD, < br /
< br /
< br / < br / < br > < br /
< br / < br / < br > < br / < br / < br > < br / < br / < br > < br / < br > < br / < br / < br > < br / < br / < br / < in △ AFD 87808780 8780 87808780b,
∴FD=FB. ∵AB=AF+FB=AC+FD，
∴AB=AC+CD.
verification: ∠ a + C = 180 °
analysis: because the horizontal angle is equal to 180 °， Therefore, we should consider transforming two angles which are not together into equal angles by congruence, and there is no congruent triangle in the graph. Therefore, the key to solve the problem is to construct isosceles triangle, which can be realized by "truncation and complement method". "
prove: intercept be = AB on BC, connect De, and then take the midpoint m of EC, Connecting DM
∵ AB = be
and ∵ BD bisecting ∠ ABC a d
∵ abd = ∠ EBD
in △ abd and △ EBD,
AB = be
∠ abd = ∠ EBD
BD = BD B e M C
{abd ≌ △ EBD (SAS) is shown in Figure 1-1
} ad = ed ∠ a = ∠ bed,
∵AD = DC ， ∴ED = DC ∴∠ C = ∠DEC
∴∠A + ∠C = ∠BED +∠DEC = 180 °< As shown in Figure 2-2, AE / / BC, ad and BD share & eth equally; EAB、Ð CBA, EC, point D
verification: ab = AE + BC
analysis 1: to prove that ab = AE + BC, observe that AD and BD are bisectors, so the Daed can be folded along a, so it is necessary to add an auxiliary line to intercept BF = BC on AB, and only need to dece AF = AE, then the problem can be solved, so how to dece AF = AE becomes the key to solve the problem. Because the sum of internal angles of Daed, DADB and DBD is 180 °， And & eth; EDC=180 °， Because AE / / be, therefore & eth; E+Ð C=180 ° Thus & eth; EAB+Ð CBA=180 °， Ad and BD are bisectors of angles, and we can dece & eth; 1+Ð 4=90 °， Thus, we can dece & eth; ADB=90 °， Thus & eth; 6+Ð 8=90 ° If we can dece & eth; 7=Ð 8, then we only need to dece Daed ≌ dafd, thus we can dece AE = AF, BC = BF, & eth; 1=Ð 2. BD is a common edge, so we can dece dbfd ≌ DBCD, then & eth; 5=Ð 6, e to & eth; 5+Ð 7=90 ° So & eth; 6+Ð 7=90 °， Due to & eth; 6+Ð 8=90 °， Thus & eth; 7=Ð Therefore, ad can be the common edge, & eth; 3=Ð 4 dece Daed ≌ dafd, thus the idea is smooth, dece AE = AF, and dece AB = AE + BC by equivalent substitution
proof 1: intercept BF = BC on AB and link DF
BD is & eth; Bisector of ABC & eth; 1=Ð 2
in dbdf and DBDC,
(common side)
{dbdf ≌ DBDC (SAS) is shown in Figure 2-2
} & eth; 5=Ð 6 (the corresponding angles of congruent triangles are equal)
! & eth; 3+Ð 8+Ð E=Ð 4+Ð 1+Ð 5+Ð 7=Ð 2+Ð 6+Ð C=180 ° The sum theorem of internal angles of triangle)
{eth; E+Ð EAB+Ð ABC+Ð C+Ð EDC=540 °
and ﹣ AE / / BC ﹣ eth; E+Ð C=180 ° Two lines are parallel and complementary)
and ∵ & eth; EDC=180 ° ∴Ð 1+Ð 2+Ð 3+ Ð 4=180 °
ad is & eth; The bisector of EAB & eth; 3=Ð 4
∴Ð 1+Ð 4=90 ° ∴Ð 5+Ð 7=90 ° The sum theorem of internal angles of triangle)
{eth; 6+Ð 8=90 ° ∵Ð 5=Ð 6 ∴ Ð 7=Ð 8
in Daed and dafd,
{Daed ≌ dafd (ASA)
} AE = AF (the corresponding sides of congruent triangles are equal)
≌ AF + FB = AB
} AE = FB = AE + BC = AB
that is, ab = AE + BC
analysis 2: extend the extension line of BC intersecting ad to F. To prove AB = AE + BC, we only need to prove BF = AB and dece CF = AE. In order to prove CF = AE, we only need to dece two triangles Daed ≌ dfcd with CF and AE e to & eth; 5 = 6, AE / / BC, so & eth; 3=Ð F. If we want to dece ad = FD, it becomes the key to solve the problem °，& ETH; E+Ð BCE=180 °， Therefore, it can be known that & eth; EAB+Ð CBA=180 °， Ad and BD are & eth; EAB、Ð CBA's bisector, thus can be introced & eth; 1+Ð 4=90 °， Therefore & eth; ADB=90 °， Then & eth; EDB=90 °， At this point, they can dece dabd ≌ dfbd according to ASA by observing the figure, and thus dece ad = FD, forming the idea
proof 2: as shown in Fig. 2-3, the sum of the internal angles of the three triangles in Daed, DADB and DBDC is 540 ° The sum theorem of internal angles of triangles)
and ∵ & eth; EDC=180 ° Horizontal angle definition & eth; E+Ð C+Ð EAB+Ð ABC=180 °
AE / / BC (two lines are parallel and complementary)
■ & eth; 3+Ð 4+Ð 1+Ð 2=180 °
AD and BD are & eth respectively; EAB、Ð The bisector of ABC
& eth; 3=Ð 4，Ð 1=Ð 2 (angular bisector definition)
﹣ eth; 1+Ð 4=90 ° ∴Ð ADB=90 ° The sum theorem of internal angles of triangle)
{eth; BDF=90 ° In DADB and dbdf,
≌ DADB ≌ dbdf (ASA)
≌ ad = FD, ab = FB, & eth; 4=Ð F (opposite sides of congruent triangle, corresponding angles are equal) as shown in Fig. 2-3
in Daed and dfcd, it is known that ad is the angular bisector of △ ABC, AB > AC, as shown in Fig. 3-1,
verification: AB-AC > bd-dc
analysis: in order to prove AB-AC > bd-dc, we need to transform the difference between AB and AC, the difference between BD and DC or their equal quantity into the edge of the same triangle, and then use the relationship between the three sides of the triangle to prove it
proof: Method 1: intercept AE = AC on AB and connect ed. A
∫ ad bisection ∠ BAC ∫ bad = ∠ DAC
in △ ade and △ ADC, E
AE = AC
ead = ∠ DAC B D C
ad = ad as shown in Figure 3-1
{△ ade ≌ △ ADC (SAS)} de = D C
in △ abd, Be > bd-de (the difference between the two sides of the triangle is less than the third side)
that is, ab-ae > bd-dc
Ψ AB-AC > bd-dc (equivalent substitution)
method 2: short complement method
extend AC to point E, so that AE = ab, Connecting de a
∫ ad bisection ∠ BAC ∫ bad = ∠ DAC
in △ bad and △ ead,
AB = AEC
∠ bad = ∠ DAC B D E
ad = ad
{ade ≌ delta ADC (SAS)} D B = D E, as shown in figure 3-2
in △ abd, EC > de-dc (the difference between the two sides of the triangle is less than the third side)
that is ae-ac > de-dc AB-AC > bd-dc
example 4 shows that, as shown in Figure 4-1, in △ ABC, ∠ C = 2 ∠ B, ∠ 1 = 2.
verification: ab = AC + CD.
analysis: from the conclusion analysis, "truncation" or "complement" can realize the transformation of the problem, that is, extending AC to e makes CE = CD, It is proved that: Method 1 (short complement method)
prolongs AC to e, so that DC = CE, then ∠ CDE = ∠ CED, as shown in Figure 4-2
Figure 4-2
achb = 2 ∠ e,
∫ ACB = 2 ∠ B, ∫ B = e,
in △ abd and △ AED,
{abd ≌ △ AED (AAS)
, ≌, And the AE = AC + CE = AC + DC, < br /
< br /
< br /
< br / < br / < br / < br /
< br /
< br / < br / < br / < method 2 (truncation method) < br / < br /
in the Figure 4-3, as shown in Figure 4-3 < br / < br > < br / < br /
in △ AFD and △ ACD, < br /
< br /
< br / < br / < br > < br /
< br / < br / < br > < br / < br / < br > < br / < br / < br > < br / < br > < br / < br / < br > < br / < br / < br / < in △ AFD 87808780 8780 87808780b,
∴FD=FB. ∵AB=AF+FB=AC+FD，
∴AB=AC+CD.
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