Which of the following commands will prompt the user for their name?

name = input('Enter your name: ')

name = read('Enter your name: ')

name = getInput('Enter your name: ')

name = print('Enter your name: ')

What does the remark statement do in Python?

It is used to add comments in the code.

Print and input statements in python

Year 8 - Python print, input and int

Python basic print, input keyword commands and variable names

In this video we take a look at the 0:02 fetch to code 0:03 execute cycle including its effect on 0:06 the various registers we've previously 0:12 [Music] 0:14 discussed a computer is defined Definition 0:17 as an electronic device that takes an 0:20 input 0:22 processes data 0:25 and delivers output 0:29 in this simple example you can see we're 0:31 taking the input 5 0:35 we're multiplying it by 2 that's our 0:37 process 0:39 and we're outputting 10. 0:44 but this could be way more complex for 0:46 example of a game console 0:48 the input could be the buttons you press 0:50 on a controller 0:53 the processes would then be carried out 0:55 by the console itself 0:59 and the output would be some form of 1:01 update to a monitor 1:02 and sound out for a speaker possibly 1:04 vibration feedback through the 1:06 controller 1:10 to process data a computer follows a set 1:13 of instructions 1:14 known as a computer program 1:18 if we take the lid off a typical desktop 1:20 computer we can identify 1:22 two critical components the memory 1:26 that stores the program and the central 1:29 processing unit or processor 1:31 which is under this large fan and 1:33 carries out the instructions 1:37 a computer carries out its function by 1:40 fetching 1:41 instructions decoding them and then 1:43 executing them 1:44 in a continuous repetitive cycle 1:46 billions of times a second 1:48 let's look at each of these stages in a 1:50 little more detail Fetch 1:53 so let's start with the fetch stage the 1:55 very first thing that happens 1:57 is the program counter is checked as it 2:00 holds the address 2:01 of the next instruction to be executed 2:07 the address stored is then copied into 2:09 the memory address register 2:14 the address is then sent along the 2:16 address bus to main memory 2:18 where it waits to receive a signal from 2:21 the control 2:22 bus so it knows what to do 2:27 as we want to read the data that's 2:29 stored in memory address 2:30 0 0 0 0 the control unit sends 2:34 a read signal along the control bus to 2:36 main memory 2:41 now main memory knows the data needs to 2:44 be read 2:45 the content stored in memory address 000 2:49 can be sent along the data bus to the 2:51 memory data register 2:56 now as we're currently in the process of 2:58 fetching an instruction 3:00 the data received by the memory data 3:03 register gets copied 3:04 into the current instruction register 3:11 the instruction effectively has now been 3:14 fetched from memory 3:16 just before we proceed to the decode 3:18 phase we now 3:19 increment the program counter so that 3:22 the address it contains 3:24 points to the address of the next 3:26 instruction which will need to be 3:30 executed 3:32 the instruction now being held in the 3:33 current instruction register 3:35 is ready to be decoded 3:39 now as we mentioned in the previous 3:41 video the instruction is made up of two 3:43 parts 3:44 we have the op code that's what it is we 3:47 need to do 3:50 and we have the operand what are we 3:53 going to do it to 3:55 now the operand could contain the actual 3:57 data 3:58 or indeed it could contain an address of 4:01 where the data is to be found 4:06 by decoding this instruction we can see 4:08 the operation we need 4:10 is a load operation so we need to load 4:14 the contents of memory location0101 4:18 into the cpus accumulator 4:25 in the exam a simple model will be used 4:27 to describe the 4:29 structure of any given instruction 4:32 you're not going to be expected to 4:34 define how an opcode is made up 4:36 but simply to interpret opcodes in the 4:39 given context of an exam 4:40 question in the example here 4:44 you can see there's a total of 16 4:46 different opcodes available 4:48 and this is because we're using four 4:50 bits for our representation 4:56 so now we've fetched the instruction and 4:59 we've decoded it so we know what we need 5:00 to do 5:01 we're finally ready to execute it 5:05 so we now send address 0101 5:08 to the memory dress register 5:13 now we're in the memory address register 5:15 we can finally send the address 5:18 down the address bus to main memory 5:24 this time we want to read the data 5:26 that's stored in memory 5:28 and so the control unit again sends a 5:30 read signal along the control bus 5:36 so main memories now receive an address 5:38 and a read signal 5:40 so the content stored at memory location 5:43 0101 5:44 can now be sent along the data bus back 5:46 to the cpu 5:47 and into the memory data register 5:54 finally the contents of the memory data 5:56 register are copied to the accumulator 5:59 and this is one of a number of general 6:00 purpose registers found in the cpu 6:04 this first instruction is now complete Branching 6:11 so what does this program actually do 6:14 you should be able to work it through 6:16 carefully and figure it out 6:19 we're now pointing instructions zero 6:21 zero zero one in the program counter 6:23 and we're ready to fetch the second 6:25 instruction 6:27 at the end of this video we're gonna 6:29 provide you with the answer 6:34 so let's talk a second about programs 6:37 that branch 6:40 on the left here we have a very simple 6:42 piece of pseudo code 6:44 line zero says first execute this line 6:46 of code 6:47 line 1 now execute this line and then 6:50 line 2 says 6:52 if the age is greater than 18 then 6:56 we're going to execute lines 3 and 4 6:58 otherwise 6:59 we're going to execute lines six and 7:02 seven 7:03 so this program doesn't necessarily 7:05 follow strictly in sequence from line 7:07 zero through to seven there's a chance 7:10 here the program may branch and jump 7:14 around 7:16 so we're going to pretend that this 7:17 program has been loaded into memory 7:20 each line of code on the left here has 7:23 ended up 7:24 as a location in memory now this is not 7:27 strictly how this would happen in this 7:28 one-to-one way 7:29 but for the purpose of example it's 7:31 absolutely fine 7:35 so the program counter starts by 7:37 pointing to memory address zero 7:39 and we fetch the first instruction 7:41 decode it and execute it 7:44 it then updates and tells us the next 7:47 instruction 7:48 is zero zero zero one because remember 7:50 the program counter is being incremented 7:52 so we fetch it decode it and we execute 7:55 line one of our program 7:59 we then fetch line two which in binary 8:01 is one 8:02 zero 8:06 now at this point depending on what 8:10 happens during the execution 8:11 of line two the program may be required 8:15 to fetch line three from memory or 8:18 line five from memory 8:25 so let's look at how this actually works 8:27 because we've said the program counter 8:28 simply gets incremented 8:31 well in the current instruction register 8:33 we have an instruction with the op code 8:36 0 1 1 0. 8:41 now when we look this up in the decode 8:43 unit we discover that this 8:45 code means branch always 8:51 this replaces the value held in the 8:54 program counter 8:56 with the contents of the operand that's 8:58 the second part of the instruction 9:01 from the current instruction register so 9:03 this case 9:04 one zero zero one 9:09 now when the next fetch cycle begins the 9:12 program counter is obviously checked 9:14 and as its contents have been previously 9:16 updated to a new memory location 9:19 and not simply incremented the program 9:22 effectively is able to jump 9:24 around memory 9:28 so having watched this video you should 9:30 be able to answer the following key 9:32 question 9:33 how does a cpu work 9:39 okay so let's um answer the question we 9:41 posed 9:42 earlier what did that program actually 9:48 do 9:50 so this is the first fetch to code 9:53 execute cycle 9:55 and this is the one that we ran through 9:57 in detail earlier 9:58 it effectively loaded the contents of 10:01 the memory 10:02 stored at location location0101 10:05 into the accumulator in other words 10:08 the dna number 3 is moved 10:11 from memory into the cpu 10:18 we then proceed onto the second fetch 10:20 decode execute cycle 10:23 now this one adds the contents of memory 10:27 located at 0 1 1 0 10:30 to the current contents of the 10:32 accumulator 10:34 so in other words the dna number one 10:38 because that's what's stored at address 10:40 zero one one zero 10:43 is added to the number three that was in 10:45 the accumulator 10:46 the results are stored back over the 10:48 accumulator 10:49 so effectively we've done three plus one 10:53 equals four 10:58 the third fetch to code execute cycle 11:00 stores the contents which are in the 11:02 accumulator 11:03 into memory location zero one one one 11:07 and that's because the op code the first 11:09 part of this current instruction 11:10 zero zero one one is the command to 11:13 store when we look it up in the decoder 11:15 unit 11:16 so in other words the result of the 11:17 previous calculation three plus one 11:19 equals four 11:20 is now written back into main memory 11:28 the fourth fetch decode execute cycle 11:30 outputs the contents of the accumulator 11:33 remember they were copied into main 11:34 memory but they're still held in the 11:35 accumulator 11:37 so in this simple abstraction the number 11:40 four is now 11:41 output to the user so they can see the 11:43 result of the calculation 11:49 the fifth and final fetch code execute 11:51 cycle 11:52 brings a halt to the current program 11:58 so this very simple program which has 12:01 five 12:02 fetch decode execute cycles has 12:04 performed the calculation 12:06 three plus one is then stored the result 12:09 in main memory 12:10 and displayed the result four to the 12:12 user 12:13 and in a high-level language this may 12:15 look something very similar to the 12:17 following two lines of code 12:20 sum variable equals num1 plus num2 12:24 print sum to the user 12:27 so you can start to get an appreciation 12:29 here of how the high level code you 12:32 write actually ends up being fetched 12:34 decoded 12:35 and executed inside a processor 12:38 of course your processor is doing 12:40 billions and billions of these 12:42 operations a second 12:43 which when you think about it is really 12:45 very impressive 12:52 [Music] 13:03 you. make 10 questions for a standerd of a level

PASSIVE TRANSPORT Cell membranes help organisms maintain homeostasis by controlling what substances may enter or leave cells. Some substances can cross the cell membrane without any input of energy by the cell in a process known as passive transport. DIFFUSION The simplest type of passive transport is diffusion. Diffusion is the movement of molecules from an area of higher concentration to an area of lower concentration. This difference in the concentration of molecules across a distance is called a concentration gradient. Consider what happens when you add a sugar cube to a beaker of water. As shown in Figure 5-1, the sugar cube sinks to the bottom of the beaker. This sinking makes the concentration of sugar mole- cules greater at the bottom of the beaker than at the top. As the cube dissolves, the sugar molecules begin to diffuse slowly through the water, moving towards the lower concentration at the top. Diffusion is driven entirely by the molecules’ kinetic energy. Molecules are in constant motion because they have kinetic energy. Molecules move randomly, traveling in a straight line until they hit an object, such as another molecule. When they hit some- thing, they bounce off and move in a new direction, traveling in another straight line. If no object blocks their movement, they con- tinue on their path. Thus, molecules tend to move from areas where they are more concentrated to areas where they are less concentrated, or “down” their concentration gradient. In the absence of other influences, diffusion will eventually cause the molecules to be in equilibrium—the concentration of molecules will be the same throughout the space the molecules occupy. Returning to the example in Figure 5-1, if the beaker of water is left undisturbed, at some point the concentration of sugar molecules will be the same throughout the beaker. The sugar concentration will then be at equilibrium. SECTION 1 OBJECTIVES ● Explain how an equilibrium is established as a result of diffusion. ● Distinguish between diffusion and osmosis. ● Explain how substances cross the cell membrane through facilitated diffusion. ● Explain how ion channels assist the diffusion of ions across the cell membrane. VOCABULARY passive transport diffusion concentration gradient equilibrium osmosis hypotonic hypertonic isotonic contractile vacuole turgor pressure plasmolysis cytolysis facilitated diffusion carrier protein ion channel Sugar Water 1 2 3 FIGURE 5-1 Sugar molecules, initially in a high concentration at the bottom of a beaker, , will move about randomly through diffusion, , and eventually reach equilibrium, . At equilibrium the sugar concentration will be the same throughout the beaker. Diffusion occurs naturally because of the kinetic energy the molecules possess. 3 2 1 Copyright © by Holt, Rinehart and Winston. All rights reserved. 98 CHAPTER 5 It is important to understand that even at equilibrium the ran- dom movement of molecules continues. But because there is an equal concentration of molecules everywhere, molecules are just as likely to move in one direction as in any other. The random movements of many molecules in many directions balance one another, and equilibrium is maintained. Diffusion Across Membranes Cell membranes allow some molecules to pass through, but not others. If a molecule can pass through a cell membrane, it will diffuse from an area of higher concentration on one side of the membrane to an area of lower concentration on the other side. Diffusion across a membrane is also called simple diffusion, and only allows certain molecules to pass through the membrane. The simple diffusion of a molecule across a cell membrane depends on the size and type of molecule and on the chemical nature of the membrane. A membrane can be made, in part, of a phospho- lipid bilayer, and certain proteins can form pores in the membrane. Molecules that can dissolve in lipids may pass directly through the membrane by diffusion. For example, because of their nonpolar nature, both carbon dioxide and oxygen dissolve in lipids. Molecules that are very small but not soluble in lipids may diffuse across the membrane by moving through the pores in the membrane.

Stages in the Sale of a Property Stage 1 – Getting to Instruction • Initial contact with the vendor: need to check the following: type of property, contact details of vendor, address of property/Eircode and purpose of the contact - sale or valuation? If a sale, does the vendor need a quick sale? Qualify the lead i.e. is the vendor buying another property? If an investment property, is the tenant in situ? Check if there is a folio number available and confirm the ownership of the property. Schedule the viewing. • Pre-viewing: Set up a file & record all info from initial contact on CRM system. Check the Property Price Register to help get a general idea of property valuation (subject to viewing, helps to display knowledge of area/market and set expectations for the vendor). Nature of property may affect pricing e.g. starter home vs. larger property with vendor seeking to downsize. Consideration for comparables may include similar/same location, size and condition of property, availability and type of parking, layout of property, plot size, orientation of garden, extensions undertaken etc. Nature of market conditions, state of wider economy, cost of capital and availability of credit may also be factors. • Appraisal/viewing: Bring an advertising pack/sales & marketing brochures. Walk through property with client, note nice features/selling points for the brochure, let the client talk about upgrades/specific features of the property. It is very important to listen to the vendor and build rapport. Confirm property details e.g. condition and layout, plot size, orientation of garden. Check for certificates of compliance for any extensions, planning permissions for conversions, right of way if applicable etc. Check if a BER available/provide details for approved assessors. Demonstrate your/the practice’s professional expertise, justify why you should get the instruction, discuss recent local sales and give your potential valuation. Discuss the sales fee, marketing fee and any additional charges e.g. professional photography, drone footage, virtual tours (walkthrough video, Matterport etc.) Ask how the vendor heard about you/your practice and why are they considering you for the sale. Where appropriate offer advice to help vendor increase potential sales price. (If possible, leave with signed Property Services Agreement/Letter of Engagement.) Thank you, send/email market appraisal, any queries/questions do get in touch and let the vendor know that we’ll be in touch in coming days. • Post appraisal – letter sent that pm/next morning with market appraisal; diary note to follow up. Check that market appraisal letter received and check for questions. If did not get sale, find out why not/debrief. If get the sale, email confirmation of instruction. Once PSRA sent and LOE returned signed = stage 2. Other details required – ID, proof of address, proof of ownership/title, solicitor details, BER certificate (refer to assessor if not available). All these should be uploaded to CRM. Stage 2 – Getting to ‘Sale Agreed’ Set up appointment to measure & photograph, note any special features e.g., upgraded kitchen, south-facing garden. Provide ideas for improving sales potential (declutter, painting, tidy garden etc. Check if has vendor potential buyers in mind already e.g., relations, friends, other parties interested. Seek vendor approval for photos/text of brochure. Check for access (tenants in situ/working from home etc) and confirm viewing times. If given a key for viewings – tag it! Check alarm codes & whether a sign is allowed on the property. Bring to market – upload to all websites e.g., daft/my home, in house websites and create window display. Match the property against your internal database of potential purchasers /CRM system. Set up appointments for viewings on CRM or arrange for open viewings. Confirm viewings with vendor & purchaser. Turn on lights, open windows, secure valuables, leave out brochures & business cards, bring viewings sheets to keep record of attendees. Introduce yourself and get attendee details. Let people view the property and address any questions. Point out key features. Record questions to be answered and any feedback from viewers. Ask are they selling property? Let viewers know of offers already received. Lock up/alarm property/close windows. Provide vendor with feedback on viewings - number of viewers / questions raised/overall reaction to property. Offers should be confirmed in writing & upload to on CRM/ offers will be input by bidders onto online bidding platforms ‘Proof of funds’ required for offers in some practices. Successful bidder will be chosen by vendor, who might want quick sale/no chain or prefer the highest bidder. Booking deposit will be sought from successful bidder. The amount varies by practice but must cover fees. Sales Advice Notice/letter should be sent to both solicitors (and may be cc’d to vendor/buyer or notify both that SAN have gone out). Booking deposit receipt should be issued. The BER certificate and report should go to the solicitor. Send requests for docs/info to successful bidder including steps they need to take to progress sale e.g., organise the bank valuation and/or schedule the survey. Once the deposit is paid the property is Sale Agreed, inform other bidders, and update all websites/sales board etc. Stage 3 – Getting to closing Access should be organised for the bank valuation/survey. Stay in touch with both solicitors ‘contract-chasing’ i.e., check when contracts are issued, signed and queries answered. Legal searches undertaken by the solicitors may include checking boundaries, land registry, title, rights of way, compliance certs etc. When contracts are signed 10% purchase price/booking deposit should be sent to the vendor’s solicitor. Once all queries satisfied = drawdown of mortgage/funding, house/life insurance in place. Title deeds will be requested once contract is signed. Decide final closing date. Check that the property taxes have been paid. Check that vendor has vacated the property. When vacant, conduct the final walkthrough and take final readings (MPRNs ). Check with solicitor if the drawn down funds h, and once received the solicitor gives authorisation to the estate agent to release the keys. The agent will do up invoice, send the balance of funds to solicitor and provide gift to purchaser. Finally remove sign, mark as sold on CRM, seek testimonials, upload to social media and close a/c on CRM

Print and pattern with vegetables

9i Print and Variable Names

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