HSC HEAD START LECTURE
HSC HEAD START LECTUREStandard MathematicsSaturday 5th October 2019Presented by:KRYSTELLE VELLA
First of all WELCOME TO YOURHSC YEAR!
By the end of today’s lecture You will:learn study techniques to prepare yourself for asuccessful year aheadlearn the first few topics of Standard Maths!q Financial Mathsq Measurementq Networksask questions about anything you are unsure of!
STUDY TECHNIQUES
How to study for Standard MathsAlthough you don’t have exams for a while, it is a great idea tolearn study techniques for Standard Maths now!q Start the year with a fresh book, and keep it up-to-date and neatthroughout.q If a concept confuses you now, seek extra help before the year goeson (i.e. asking for extra help, YouTube videos etc!)q Get into the habit of showing all your working.q As you learn new formulas, create flashcardsto help you practice and learn them off byheart.q Begin thinking about your exam techniquetoday.
Topic 1:Financial Mathematics
In Year 11, you would have learnedhow to:ü calculate simple interest (I Prn)ü calculate percentage increases and decreasesü use the straight-line method of depreciationü calculate rates of pay (salaries, wages, overtime, commission etc!)ü calculate income tax (PAYG)ü use budgets and calculate household expensesSound familiar?
Now in Year 12:The two BROAD topics we cover include:1. Investments and Loans2. Annuities
1. INVESTMENTS & LOANSInvestments are used to GROW money.Future value of loansWhen calculating the future value of loans, compound interest iscalculated on the initial amount borrowed/invested plus any interestcharged or earned.To calculate the future value of a loan:𝐹𝑉 𝑃𝑉 (1 𝑟)*where: FV - future value of the investment (final balance)PV - present value of the principal (initial quantity of money)r - rate of interest per compounding time period (note: this is expressed as a decimal)n - number of compounding time periods
FUTURE VALUE OF LOANSExample question:Brendan invests 14,000 over 6 years compounding annually.What is the future value of the investment if the compound interest rateis 5% p.a.?Solution:Remember the figures required to fulfil the future value formula! Present Value (PV) 14 000Rate (r) 5% p.a. (0.05)Number of periods (n) 6 𝐹𝑉 14 000 (1 0.05)E𝑭𝑽 𝟏𝟖 𝟕𝟔𝟏. 𝟑𝟒(𝑡𝑜 2 𝑑𝑒𝑐𝑖𝑚𝑎𝑙 𝑝𝑙𝑎𝑐𝑒𝑠)
FUTURE VALUE OF LOANSWhen solving future value questions, it is so important that youcorrectly find the value of the RATE and NO. OF PERIODS!For example:Jocelyn invests 5000 over 2 years compounding quarterly.Calculate the future value of the investment if the compound interestrate is 6% p.a.?How many compounding periods are there?What is the interest rate per compounding period?𝐹𝑉 5632.46
PRESENT VALUE OF LOANSWe can also use the future-value formula to find the presentvalue of an investment.To find present value:𝑭𝑽𝑷𝑽 (𝟏 𝒓)𝒏
PRESENT VALUE OF LOANSExample question:Calculate the present value of an investment that has a future value of 25 000 after 8 years and earns 9.6% compound interest,compounding bi-annually.Solution:Let’s break down the question to obtain our required values before weinsert them into the formula. Future Value (FV) 25 000 Rate (r) 9.6% p.a. compounding bi-annuallyØ 4.8% (0.048) per compounding period Number of periods (n) 8 years compounding bi-annuallyØ 16 compounding periods
PRESENT VALUE OF LOANSSolution continued.Therefore, the present value of this loan is equal to:25 000𝑃𝑉 (1 0.048)LE𝑷𝑽 𝟏𝟏 𝟖𝟎𝟕. 𝟓𝟒
FINDING THE INTEREST EARNEDNOTE!Some questions may ask you to find the interest earned only.In this case, we use the formula:𝑰 𝑭𝑽 𝑷𝑽Simply meaning that the interest earned is equal to the future valueof the investment minus the present value (principal) of theinvestment.Careful not to confuse this with finding the interest rate.
FINDING THE INTEREST RATETo try and trick students, NESA may ask you to find the interest rateused to calculate the present or future value of an investment.In these questions, we simply rearrange the future value formula.Example question:The present value of Henry’s investment is 31 250. After 4years, this has grown to 35 000. If interest compounds annually,calculate the compound interest rate at which Henry’s investmentis growing, correct to two decimal places.
FINDING THE INTEREST RATESolution:To begin, let’s insert the values we know into the future valueformula.𝑭𝑽 𝑷𝑽(𝟏 𝒓)𝒏In Henry’s investment: Future Value (FV) 35 000 Present Value (PV) 31 250 Number of periods (n) 4 Rate (r) ?𝟑𝟓 𝟎𝟎𝟎 𝟑𝟏 𝟐𝟓𝟎(𝟏 𝒓)𝟒Now, we must use our algebraic knowledge to find the value of r.
FINDING THE INTEREST RATESolution:1.35 000 31 250 (1 𝑟)S31 250 31 250Remove the 31 250 from both sidesby division.2.1.12 (1 𝑟)SSimplify.3.T4.1.028737345 1 𝑟Simplify.5.𝑟 0.028737345 100%Subtract 1 from either side, leaving ther on its own. Multiply by 100% to get apercentage answer.1.12 T“Fourth-root” both sides to eliminatethe 4.(1 𝑟)S𝒓 𝟐. 𝟖𝟕%
SOLUTIONS (previous slide)
FINDING THE INTEREST RATETherefore, Henry’s investment grows at a compound interest rate of2.87% p.a. (to two decimal places).How can we test our answer to make sure it’s right?
APPRECIATIONAppreciation is the increase in value of goods.Example question:Flynn bought a block of land in 2014 for 319,000. If theland appreciates at 5% p.a., calculate it’s value after 5years. Answer to the nearest dollar.Solution:𝐹𝑉 𝑃𝑉(1 𝑟)*𝐹𝑉 319 000 (1 0.05)m𝐹𝑉 407 134𝑻𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒍𝒂𝒏𝒅 𝐚𝐟𝐭𝐞𝐫 𝟓 𝐲𝐞𝐚𝐫𝐬 𝐢𝐬 𝟒𝟎𝟕 𝟏𝟑𝟒, 𝒕𝒐 𝒕𝒉𝒆 𝒏𝒆𝒂𝒓𝒆𝒔𝒕 𝒅𝒐𝒍𝒍𝒂𝒓.
INFLATIONInflation is a steady increase in the price of goods and services inan economy over time.Example question:In 2010, the average cost of 1L ofmilk was 1.05. Accounting foran inflation rate of 6% p.a.,determine the cost of 1L of milkin 2020.Inflation affects boththe cost of living andthe wages earned byindividuals.Which formula do we use to calculateinflation?
INFLATIONLet’s use the compound interest formula.𝐹𝑉 𝑃𝑉(1 𝑟)*𝐹𝑉 1.05 (1 0.06)Ln𝐹𝑉 1.88Therefore, the cost of 1L of milk in 2020 is 1.88.
QUESTION TIME!!Question 10 (Multiple Choice)– 2017 HSC Exam:
QUESTION TIME!!Question 17 (Multiple Choice)– 2015 HSC Exam:
QUESTION TIME!!Question 26 (d) – 2015 HSC Exam: 369.17
DEPRECIATIONTo calculate depreciation, we use the declining-balance methodformula:𝑺 𝑽𝟎 (𝟏 𝒓)𝒏where: S is the salvage value of the asset after n periods V0 is the initial value of the asset r is the depreciation rate per period n is the number of periodThis formula is an application of the compound interest formula!
DEPRECIATIONExample question:Rodney purchased a Hyundai i30 in 2014 for 19 000. The cardepreciates at 16% per annum.Calculate the value of the asset after 10 years, correct to the nearestdollar.Solution:𝑆 𝑉n (1 𝑟)*𝑆 19 000(1 0.16)Ln𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒 𝟑𝟑𝟐𝟑After 10 years, Rodney’s Hyundai is worth 3323, to the nearestdollar.
SHARES AND DIVIDENDSA share is when an individual/company owns part of anothercompany. Shares are bought and sold on the stock market,including the ASX (Australian Securities Exchange).To buy and sell shares, we use a broker. A broker receives abrokerage fee for their work – a percentage of the transaction.
CALCULATING THE COST OF SHARESTo calculate the cost of shares, we must look at their originalmarket value and how many are bought. Additionally, we mustconsider the brokerage fees which apply.Example question:June purchased 100 shares at a market value of 12.64 eacheach. Sheincurred brokerage fees at 4% of the price of the shares. Calculatethe total cost of purchasing the shares.Solution:Cost of shares 12.64 (market value) 100 (quantity) 1264Brokerage fees 4% 1264 50.56Total cost of shares 1314.56
DIVIDENDSA shareholder is entitled to a share in the company’s profits –this is known as a dividend.To calculate the value of a dividend, we use the dividend yield.𝑎𝑛𝑛𝑢𝑎𝑙 ��𝑒𝑛𝑑 𝑦𝑖𝑒𝑙𝑑 100%𝑚𝑎𝑟𝑘𝑒𝑡 �� 𝑑𝑖𝑣𝑖𝑑𝑒𝑛𝑑 𝑦𝑖𝑒𝑙𝑑 𝑚𝑎𝑟𝑘𝑒𝑡 𝑝𝑟𝑖𝑐𝑒
DIVIDENDSExample question:The share price of Wesfarmers is currently 49.84.a) The predicted dividend yield is 3.7%. What is the dividend?(Correct to two decimal places)Dividend 3.7% (dividend yield) 49.84 (market price) 1.84b) Wesfarmers decide to pay a dividend of 4.98. What is thedividend yield? (Correct to two decimal places) S.wxDividend yield 100% 9.99% Sw.xS
HSC QUESTIONQuestion 26 (e)– 2017 HSC Exam:COST Shares: 500 3.20 1600Brokerage fees: 1.5% 1600 24Total cost: 1624INCOME Dividends: 500 0.26 130Selling shares: 500 4.80 2400Total income: 2530PROFITIncome - Expenses 2530 - 1624 906
NOT COVERED TODAY q Credit cardsq AnnuitiesThese harder topic areas will be covered innext year’s Head Start and Trial Revisionlectures J
Topic 2:Measurement
In Year 11, you would have learnedhow to:ü convert between metric units (length, capacity, area etc)ü calculate errors in measurementü use significant figures and scientific notationü calculate perimeter, area & volumeü use Pythagoras’ theorem & similar figuresü use Trapezoidal ruleü work with time (latitude, longitude)
Now in Year 12:The two BROAD topics we cover include:1. Non-right-angled Trigonometry2. Rates and ratios
You may already know:Trigonometric ratios – only apply to right-angled � 𝜃 { {} ‚ { *ƒ} (SOH)𝑐𝑜𝑠 𝜃 „ †„‡ * ‚ { *ƒ} (CAH)𝑡𝑎𝑛 𝜃 { {} „ †„‡ * (TOA)
FINDING UNKNOWN SIDESWhen finding unknown sides in right-angled triangles, thequestion will give you one angle other than the right angle, andone labelled side to help you answer the question.Example question:Find the length of the unknown side x in the triangle below. Answercorrect to two decimal places.𝟐𝟗 x20cm
FINDING UNKNOWN SIDESSolution:First, let’s label the sides of the triangle according to thetrigonometric ratios listed before. Remember, we label the sidesaccording to the other angle provided in the triangle (not the rightangle).𝟐𝟗 xA20cmHWe are given the adjacent and hypotenuse, therefore we will usethe cosine ratio.
FINDING UNKNOWN SIDESSolution continued:1.cos 𝜃 unknownlengthhypotenuselength𝒙𝐜𝐨𝐬 𝟐𝟗 �𝑜𝑡𝑒𝑛𝑢𝑠𝑒Now substitute our known values intothe formula.angle provided inthe question2.3. 𝐜𝐨𝐬 𝟐𝟗 𝒙𝟐𝟎𝒙 𝟐𝟎 𝐜𝐨𝐬 𝟐𝟗 To complete the equation, we need torearrange the formula so x is thesubject.With x as the subject, simplify theequation.𝒙 𝟏𝟕. 𝟒𝟗𝒄𝒎 (𝟐 𝒅𝒆𝒄𝒊𝒎𝒂𝒍 𝒑𝒍𝒂𝒄𝒆𝒔)
FINDING UNKNOWN ANGLESWhen finding unknown angles in right-angled triangles, thequestion will give you an unknown angle and two sides relevantto solving the question.Example question:Find the size of the unknown angle, to the nearest degree.24 cm𝜽17 cm
FINDING UNKNOWN ANGLESSolution:Again, let’s label the triangle according to the position of theunknown angle.24 cmA𝜽17 cmOWe are given the adjacent and opposite sides, therefore we willuse the tangent ratio.
FINDING UNKNOWN ANGLESSolution 𝑛 𝜃 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡Using our calculators, we can now solve this equation.CALCULATOR STEPS:SHIFT17ðð𝑡𝑎𝑛tan-1 will appearð24ð What answer appears on your calculator?
FINDING UNKNOWN ANGLESHopefully you got a decimal answer of 35.31121344.We must round our final answer to degrees. ′ ′′𝜽 𝟑𝟓 𝟏𝟖š 𝟒𝟎. 𝟑𝟕"𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 𝑡ℎ𝑒 𝑛𝑒𝑎𝑟𝑒𝑠𝑡 𝑑𝑒𝑔𝑟𝑒𝑒:𝜽 𝟑𝟓
PROPERTIES OFNON-RIGHT-ANGLED TRIANGLESCbaAcBThe sides of the triangle are named according tothe opposite angle.
AREA OF NON-RIGHT-ANGLEDTRIANGLES1𝑨 𝒂𝒃 𝐬𝐢𝐧 𝑪2sideanglesideThe area of a non-right-angled triangle is equal to half the product of two sidesmultiplied by the sine of the included angle.
AREA OF NON-RIGHT-ANGLEDTRIANGLESQuestion 27– 2009 HSC Exam:What is the area of this triangle? (Correct to two decimal places)2.34km2
SOLUTION (previous slide)
SINE RULEThe sine rule relates the sides and angles in a nonright-angled triangle.𝒂𝒃𝒄 𝐬𝐢𝐧 𝑨𝐬𝐢𝐧 𝑩𝐬𝐢𝐧 𝑪To use the sine rule, you must be given either:§ two angles and one side or§ two sides and an angle opposite one of the givensides
SINE RULEThe sine rule can be used to find both sides and angles.To find a side:𝒂𝒃𝒄 𝐬𝐢𝐧 𝑨𝐬𝐢𝐧 𝑩𝐬𝐢𝐧 𝑪To find an angle:𝐬𝐢𝐧 𝑨𝒂 𝐬𝐢𝐧 𝑩𝒃 𝐬𝐢𝐧 𝑪𝒄
SINE RULEExample question – Finding a sideFind the value of x correct to the nearest whole number.x33 12cm43 Do we have the correct information to use the sine rule?
SINE RULESolution:1.2.𝑎𝑏 sin 𝐴sin 𝐵𝑥12 sin 43sin 33𝑥12 sin 43 sin 43sin 43sin 33𝑥 3.Substitute the known values into the sinerule formula.Make x the subject of the formula.12 sin 43sin 33𝒙 𝟏𝟓𝒄𝒎Solve x to the nearest whole number.
SINE RULEExample question – Finding an angleFind the value of the missing angle, correct to the nearest degree.39cm98 42cm𝜽Do we have the correct information to use the sine rule?
SINE RULESolution:sin 𝐴sin 𝐵 𝑎𝑏Substitute the known values into the sinerule formula.sin 𝜃sin 98 3942sin 𝜃 39 sin(98)39 42sin 9842Make sin 𝜃 the subject of the formula.Insert equation into calculator.𝟎. 𝟗𝟏𝟗𝟓𝟑𝟒𝟔𝟑𝟓𝟑CALCULATOR STEPS:SHIFTðsinð Use calculator to find value of sin 𝜃.
SINE RULESolution continued:Convert to degrees and minutes. ′ ′′𝜃 66 51š 29.32”Round to required unit (nearest degree).𝜽 𝟔𝟕
COSINE RULEThe cosine rule can be used to solve problemsinvolving three sides and one angle.To find the third side given two sides and the included angle:𝒄𝟐 𝒂𝟐 𝒃𝟐 𝟐𝒂𝒃 𝐜𝐨𝐬 𝑪To find an unknown angle given three sides:𝒂𝟐 𝒃𝟐 𝒄𝟐𝐜𝐨𝐬 𝑪 𝟐𝒂𝒃
COSINE RULEQuestion 26 – 2013 HSC Exam:Which formula are we using here?aC𝑎 𝑏 𝑐 cos 𝐶 2𝑎𝑏53 66 98 cos 𝐶 2 53 66cbcos 𝐶 110
COSINE RULEExample question – Finding a sideFind the value of x correct to two decimal places.14cmx𝟑𝟐 13cmTry and solve this question and we’ll work through the solutionstogether!
SOLUTIONS (previous slide)
*Note: these questions are from the Cambridge General Maths textbook.SINE RULE or COSINE RULE?Cosine ruleSine ruleCould you solve these on your own?
NOT COVERED TODAY q Bearingsq Angles of elevation and depressionq Radial surveysThese harder topic areas will be covered innext year’s Head Start and Trial Revisionlectures J
Topic 3:Networks
NETWORK TERMINOLOGYTERMINOLOGYnetworkverticesDEFINITIONa set of objects/tasks connected together to create asequencethe objects of a network (normally drawn as points)edgesconnect the objects of a network togetherpatha sequence of events through a network connecting thestarting vertex to the finishing vertexdirected networkwhen the edges in a network point in only one directionundirected networkwhen the edges in a network are bidirectional (or undirected)degree of a vertexthe number of edges which protrude from each vertexweighted edgewhen the connection/edge have weights (numbers) assignedto them
WHAT IS A NETWORK DIAGRAM?A network diagram has several objects connected together to create a sequence.B318D1A4922022E7Cweighted edge this connection hasa value of twoassigned to itdegree of a vertex – asseen above, the vertex Chas two edges protrudingfrom it; therefore, itsdegree is 2edge –connectingpoints C and Etogethervertex –object Edenoted by apointThis network is directed – each edge only points in one direction (i.e. the edge connecting Ato B only travels from A to B).
DRAWING NETWORK DIAGRAMSLet’s begin this topic by learning how to draw simple network diagrams frominformation given in a table. This activity will help us understand the need forspecific sequences of tasks.In the table below, we have objects A, B, C and D and an explanation of theconnections between them.ABCDA-42-B4-5-C25-7D--7-
DRAWING NETWORK DIAGRAMSBefore drawing a network diagram from this table, we need to understand afew key points evident:§the same object (vertex) does not connectto each other i.e. object A does notconnect to itself§the values in the boxes represent theweighted edge between those twovertices i.e. the edge between A and B isassigned a value of 4§not all vertices connect to eachother i.e.vertex D does not connect to vertex B§the network is undirected – i.e. the edgeconnecting A to B is presented two times(A, B) (B, A) indicating the insignificanceof the edge
DRAWING NETWORK DIAGRAMSNow we can attempt to represent this information in a network diagram.There are, of course, many different ways to interpret and visualise thisinformation. If this explanation doesn’t work for you, take an approach whichbest suits your learning!If you haven’t already, quickly sketch down this information so we can drawthe network.ABCDA-42-B4-5-C25-7D--7-
DRAWING NETWORK DIAGRAMS1. Begin by creating a vertex for object A. Note which vertices also connect toA, and use edges to connect these vertices together. In this case, object Band C also connect to object A. Ensure you label the weighted edges ofthese connections!B4C2A
DRAWING NETWORK DIAGRAMS2. Continue to connect the vertices together according to the table.For example, C connects to B with a weighted edge of 5, and C connects to Dwith a weighted edge of 7.B452AC1D
DRAWING NETWORK DIAGRAMSRemember, there is no one way to draw this diagram. Have a look at the networkdiagram below – it is representing the same information!B4A52D7C
MORE TERMINOLOGY!!!WALK: A sequence of vertices and the edges between them.PATH: A path is a walk that doesn’t visit any vertex more thanonce.CYCLE: A cycle is a walk with the same start and end vertex,which doesn’t visit any vertex more than once.
PUTTING TERMINOLOGY INTOPRACTICEIdentify the following as a walk, path, cycle or other,giving reasons for your choice:a) ABCDEDb) DEFAc) ABCDEFAd) ADBCADFEWALK: Follows the edges of the network. Why can thisnot be a path?PATH: Follows the edges of the network and doesn'tvisit any vertex more than once.CYCLE: Follows the edges of the network and has thesame start and end vertex.NEITHER: There is no edge connecting A and D.
THE SHORTEST PATHThe shortest path is a path between two vertices in a network for whichthe sum of the weights of its edges is minimised.Example Question: Consider the following network.6CDa) What is the shortest path from A to B?41A65B124GACDB (11)FE6Hb) What is the shortest path from F to G?FBHG (9)
WHY IS THE MINIMUM SPANNING TREE NOTALWAYS THE SHORTEST PATH ?The minimum spanning tree does not always provide the shortest path betweentwo points. For example, the network diagram below shows the minimumspanning tree which has a path ABC, whereas the shortest path betweenvertices A and C is AC.CShortest path from A-C:AC54Minimum spanning tree:ABCA3B
MINIMUM SPANNING TREESUsing algorithmsA minimum spanning tree connects all the vertices of a network together withoutany cycles and with the minimum possible total edge weight. When solvingminimum spanning trees, there are two algorithms that can be used:Prim’s AlgorithmPrim’s algorithm suggests that to findthe minimum spanning tree, we firstselect the shortest edge, and thencontinue by selecting the shortestattached edge. This continues until allvertices are included, and helps avoidany possible loops.Kruskal’s AlgorithmKruskal’s algorithm suggests that tofind the minimum spanning tree, wefirst select the shortest edge, and thenselect the next shortest edge,regardless of whether it is attached tothe previous edge or not. We only addconnections that connect part of thenetwork that was not previously joined.
USING PRIM’S ALGORITHMLet’s use Prim’s Algorithm first to find the minimum spanning tree of this network:Total: 9
USING KRUSKAL’S ALGORITHMNow, let’s use Kruskal’s Algorithm to find the minimum spanning tree of thisnetwork. Note: It should produce the exact same weight!!
LET’S DO SOME PRACTICEQUESTIONS
LET’S DO SOME PRACTICEQUESTIONS
LET’S DO SOME PRACTICEQUESTIONS
SOLUTIONS (previous slide)
THANKYOU YEAR 12!See y
Question 10 (Multiple Choice)–2017 HSC Exam: QUESTION TIME!! Question 17 (Multiple Choice)–2015 HSC Exam: QUESTION TIME!! Question 26 (d) –2015 HSC Exam: . next year’s Head Start and Trial Revision lectures J. Topic 2: Measurement. In Year 11, you would have learned how to: üconvert between metric units (length, capacity, area etc)
Wed 19/10 HSC Mary Poppins Musical Capital Theatre Thur 20/10 HSC Yr 7 Maths Exam 2 Yr 8 Maths Exam 2 Case of Conspiracy – Selected Yr 8 Yr 7 visual Arts Assignment Due Fri 21/10 HSC WEEK 3 Mon 24/10 HSC Yr 8 History Incursion Group 1 Tues 25/10 HSC Yr 8 History Incursion Group 2 Wed 26/10 HSC Thur 27/10 HSC Fri 28/10 HSC WEEK 4 Mon 31/10 HSC
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Introduction of Chemical Reaction Engineering Introduction about Chemical Engineering 0:31:15 0:31:09. Lecture 14 Lecture 15 Lecture 16 Lecture 17 Lecture 18 Lecture 19 Lecture 20 Lecture 21 Lecture 22 Lecture 23 Lecture 24 Lecture 25 Lecture 26 Lecture 27 Lecture 28 Lecture
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HSC Assessment Schedule 2020 - 2021 2 Assessment Calendar 3 Contact List 4 . HSC Assessment Policy 5-14. Year 12 Curriculum 15 . HSC Assessment Policy English Faculty 16-20. HSC Assessment Policy Mathematics Faculty 21-25. HSC Assessment Policy Science Faculty 26-32.
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Target Project Target PtCo/HSC-f Ordered-PtCo/HSC-f PtCo/HSC-f Ordered-PtCo/KB 8 SOA Integration & DOE Validation Technical Accomplishment: Cathode: 30 wt.% Intermetallic ordered Pt 3 Co/HSC-f at 0.06 and 0.10 mg Pt /cm2, PFSA ionomer (D2020), 900 EW, I/C ratio of 0.8, Anode: Pt/HSC, 0.015 mg Pt /cm2 PEM: PFSA with reinforcement layer, 18 μmthick
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