Information of Racial Intolerance :: Racial Intolerance Obscenity Society Essays

Information of Racial Intolerance Intolerance: There are several forms of intolerance but in this case I am referring to that of racial intolerance. Ideas on the subject: †¢ Racial intolerance is an unfair act done by many in the present time. †¢ It shows ignorance and arrogance in the person. †¢ You would think that after so many years and so many advances that today’s society would be able to look past racial differences. †¢ Racial intolerance affects everyone in the world no matter what race or ethnicity they pertain to simply because it is a clear threat to justice and equality everywhere. Changes that need to be made: †¢ To minimize the racial intolerance of today’s world I believe we first have to educate our children of the inhumanity and unnecessary that it brings into society and people everywhere. †¢ I also believe that those who have deep-based angers towards other races should consider looking into some sort of medical attention to trace their hatred and rectify it. †¢ Lastly we should all work together and instead of magnifying the acts of a few or the bad characteristics of some we should look at every race as a whole, acknowledge the good in all people, and acknowledge the possibility that one or a few persons do not determine an entire race. Melissa Serna October 18, 2000 Period: #2 AP English III It’s everywhere you look†¦ Obscenity, according to Webster is anything that is offensive to decency or lewd. Our society is filled with obscenities everywhere you turn, whether it is on television, in a novel, or even in a school. An obscenity can be almost anything, depending on whom it is that sees it, and what their personal standards of decency are. Obscenities can be found anywhere now a days. For example, obscenities can be found in a gory scene from a movie, or even in a movies sexual content. Some CD’s are classified with an â€Å"explicit lyrics† label, meaning they have some sort of indecent song lyrics.

Bathsheba and Farmer Boldwood Essay

How does Hardy present the idea ‘the pain of love’ in his depiction of the relationships between Bathsheba Everdene and Sergeant Troy, and Bathsheba and Farmer Boldwood? Thomas Hardy, born in 1840, divided his works into character and environment, romance and fantasies and novels of ingenuity, in which case Far from the Madding Crowd is in the first category. The original works were in the form of a series in the Cornhill magazine, which was so successful he was able to give up his job (as an architect) and devote his time entirely to writing. Hardy is known for his controversial novels such as Jude the Obscure, but his best work is the world renowned Far From the Madding Crowd, which expresses the journey of Bathsheba Everdene and her loyal farmer, Gabriel Oak, who encounter love but at the terrible price of death and despair. This, perhaps reflecting the tragic loss of his own wife in 1912, sixteen years before his own death, in 1928. Bathsheba and Sergeant Troy’s first meeting outside the Fir Plantation, Troy flirts and compliments Bathsheba, by taking more time then is necessary to untie the knot that binds them. Bathsheba, however she appears to be quite uncivilized towards him, â€Å"Thank you for the sight of such a beautiful face! He said. ‘Twas unwillingly shown† she replied, stiffly. This unkind manner attracts Troy to her and ignites a relationship between them, where Troy’s flirtatious and dishonest ways, already expressed in their first meeting, are destined to destroy it. Troy’s display of swordsmanship in Chapter 28, indicates another example of the pain of love, in which Bathsheba must sacrifice her own safety to be with Sergeant Troy. Though Bathsheba, at first, oblivious to her near death, due to her believing Troy’s lie that the sword was blunt. Though Bathsheba does not object to him kissing her at the end of the chapter, after being made aware of his dishonesty. The relationship of Bathsheba and Troy, already beginning to fail due to Troy’s insensitive and lying persona, is made no easier by Fanny’s part in their lives. Ironically, it was her mistake, which brought together Bathsheba and Troy in the first place, unintentionally, and, unknowingly aided in the downfall of Bathsheba and Troy’s relationship. For example, when Fanny meets Troy, though by accident, Troy tries to convince Bathsheba she is unknown to him, however Bathsheba begins to have doubts, especially after seeing a lock of hair in his watch, that was not hers. The argument that followed the discovery of the hair also lead to Troy leaving Bathsheba, to secretly meet with Fanny, and the next time she hears of Troy he is assumed dead. This expresses Hardy’s pain of love theory exactly, for Bathsheba who fell so quickly in love with Troy, and married him soon after, now expresses little feeling towards him leaving. For example, in chapter 48, `Bathsheba underwent the enlargement of her husband’s absence, from hours to days, with a slight feeling of surprise and relief, yet neither sensation rose †¦ `. Soon after the discovery of Troy’s apparent death, Boldwood immediately steps onto the scene to offer his hand in marriage. The first time Bathsheba saw Boldwood was at the market, however Boldwood did not pay attention to her presence, unlike all the other men. Bathsheba’s vanity, encouraged by Liddy, caused her to send a Valentine to Boldwood in the attempt to get him to notice her and partly as a joke, with the accidental seal of â€Å"Marry Me†. Boldwood, having received the Valentine, become besotted with Bathsheba and soon called on her to accept the invitation of marriage. Though the answer of course was not what he expected. â€Å"I didn’t know†¦ I ought never to have dreamt of sending that valentine – forgive me, sir – it was a wanton thing which no woman with any self respect should have done. If you will only pardon my thoughtlessness†¦ † Farmer Boldwood replied†¦ â€Å"No, no, no. Don’t say thoughtlessness! †¦ You torture me to say it was done out of thoughtlessness†¦ This outburst again shows the reader that Hardy can express the theme `the pain of love` so vividly. However, Boldwood does not stop and continues to express his want to have her. For example, â€Å"I may think of you? Yes, I suppose you may And hope to obtain you? No – do not hope! Let us go on†. This obsessiveness driven by the act of Bathsheba’s valentine proposes another effect of love. At the end of chapter 19, Hardy tells us that Boldwood is almost spellbound by Bathsheba and by her leaving, he comes out of this stupor, â€Å"like the pain of a wound†¦ † Another of Hardy’s pain of love themes. Boldwood’s appearances from then on are stopped due to the arrival of Troy, who Boldwood at once is jealous of and wishes to extract him from their (Bathsheba and Boldwood’s) lives. The deal he tries to accomplish with Troy is of course a failure, however it proposes that Boldwood would do (or should I say ‘pay’) so much to expel him, and further expand his relationship with Bathsheba. His love for her, the reader could interpret as, is almost insane and compulsive. This is perfectly described in Chapter 53, where Boldwood begins to talk to himself. â€Å"I hope to God she comes, or this night will be nothing but misery to me! O my darling, my darling, why do you keep me in suspense like this? † However that is not the most important part of the chapter because of course this is the climax to the whole story and the final effects that love can have on somebody. ‘A strange voice came from the fireplace – a voice sounding far off and confined, as if from a dungeon. Hardly a soul recognized the thin tones to be those of Boldwood. Sudden despair had transformed him. ‘ Boldwood, at the realization of the return of Troy, and the conclusion that Bathsheba cannot be his, had the devastating affect of forcing him to take action. In the form of a bullet. Aimed at Troy. His love for Bathsheba perhaps deluded him into it, showing that the pain of love has lead one man to his death and another to the noose. However we know that sentence was not carried out, life imprisonment instead, though in a way just as bad as the loss of life some could argue. Hardy is clearly well adept into conveying the pain of love as a theme, though not the main theme of the story. Hardy presents the idea of ‘the pain of love’ clearly in the two relationships that Bathsheba experiences, even in Oak’s relationship with Bathsheba, his offer of marriage is turned down, and not accepted till the end of the book. Bathsheba’s relationship with Troy and Boldwood differ quite drastically. Boldwood is the calm farmer enticed by Bathsheba, and driven to killing another man. Troy is the dashing sergeant who seduces Bathsheba and causes their marriage to disintegrate due to his lies and deceit. Bathsheba also reacts differently to the two men. To Boldwood she thinks she is not good enough for him, and is almost scared of his persistence- ‘She was frightened as well as agitated by his vehemence’ She also taken aback by Troy, ‘Ah! There was a time Frank, when it would have taken a good many promises to other people to drag you away from me. ‘ She also is now quite sad to be married, preferring the unmarried version of him when it was dashing swordplay and romantic courting and love. However it is not the case with the married Troy, â€Å"What do you regret? † He asked. â€Å"That my romance has come to an end†, she relied. The pain of love that the characters encounter, i. e. Boldwood’s hopeless love for Bathsheba, and Bathsheba’s own foolish love for Troy, and the final act leading to Troy’s (perhaps deserved) death, are all excellently and elaborately depicted by the most English of English novelists, in one of the most English of great English novels.

Environmental Problematic in essays

Environmental Problematic in essays Environmental Problematic vs. Industrial revolution The common Japanese expression, mizu ni nagasuliterally, "let it be washed away," or to settle a dispute by letting bygones be bygonesoriginated in the belief that any and all troubles can be purified by nature. (nira.go.jp/publ/review/99summer/mori.html) Because Japan has always had a thriving and rich natural environment, Japanese people have for a long time taken this natural bounty for granted. In the late 1960s, however, environmental mayhem resulting from industrial pollution caused alarming damage to the people of Japan. For the first time, policy-makers and ordinary people alike began to recognize "the environment" as a problem... otherwise known as the environmental problematic. During the last thirty two years after the establishment of the Environment Agency in 1971, the environmental situation at the national and global levels has undergone substantial changes in Japan. At the national level, notable achievements have been made in combating severe pollution during the period of high economic growth. However, air pollution in major urban areas and water pollution caused by industrial expansion are some of the dynamics that continue to impose themselves as heavy factors in the environmental problematic. Increased severity of global issues such as depletion of the ozone layer and global warming are playing a national and political roll in defining state policy within Japan. This paper will describe some major environmental problems in Japan, several Japanese environmental policies, the process of implementing these policies, and how they are maintained through the Japanese political system. After the Second World War, Japan emerged as one of the three dominant players influencing the global environment and the direction of international environment protection. The structure of Japans industry has drastically changed over the last few decades w...

USS Ranger CV-4 World War II Aircraft Carrier

USS Ranger CV-4 World War II Aircraft Carrier USS Ranger (CV-4) Overview Nation: United StatesType: Aircraft CarrierShipyard: Newport News Shipbuilding Drydock CompanyLaid Down: September 26, 1931Launched: February 25, 1933Commissioned: June 4, 1934Fate: Scrapped Specifications Displacement: 14,576 tonsLength: 730 ft.Beam: 109 ft., 5 in.Draft: 22 ft., 4.875 in.Propulsion: 6 Ãâ€" boilers, 2 Ãâ€" Westinghouse geared steam turbines, 2 Ãâ€" shaftsSpeed: 29.3 knotsRange: 12,000 nautical miles at 15 knotsComplement: 2,461 men Armament 8 Ãâ€" 5 in./25 cal anti-aircraft guns40 Ãâ€" .50 in. machine guns Aircraft 76-86 aircraft Design Development In the 1920s, the US Navy commenced the construction of its first three aircraft carriers. These efforts, which produced USS Langley (CV-1), USS Lexington (CV-2), and USS Saratoga (CV-3), all involved the conversion of existing hulls into carriers. As work on these ships progressed, the US Navy began designing its first purpose-built carrier. These efforts were constrained by the limits imposed by the Washington Naval Treaty which capped both the size of individual ships and the total tonnage. With the completion of Lexington and Saratoga, the US Navy had 69,000 tons remaining which could be assigned to aircraft carriers.   As such, the US Navy intended for the new design to displace 13,800 tons per ship so that five carriers could be constructed. Despite these intentions, only one ship of the new class would actually be built.    Dubbed USS Ranger (CV-4), the new carriers name hearkened back to the sloop of war commanded by Commodore John Paul Jones during the American Revolution. Laid down at the Newport News Shipbuilding and Drydock Company on September 26, 1931, the carriers initial design called for an unobstructed flight deck with no island and six funnels, three to side, that were hinged to fold horizontally during air operations. Aircraft were housed below on a semi-open hangar deck and brought to the flight deck via three elevators. Though smaller than Lexington and Saratoga, Rangers purpose-built design led to an aircraft capacity that was only marginally less than its predecessors. The carriers reduced size did present certain challenges as its narrow hull required the use of geared turbines for propulsion.   As work on Ranger progressed, alterations to the design occurred including the addition of an island superstructure on the starboard side of the flight deck. The ships defensive armament consisted of eight 5-inch guns and forty .50-inch machine guns. Sliding down the ways on February 25, 1933, Ranger was sponsored by First Lady Lou H. Hoover. Over the next year, work continued and the carrier was completed. Commissioned on June 4, 1934 at the Norfolk Navy Yard with Captain Arthur L. Bristol in command, Ranger commenced shakedown exercises off the Virginia Capes before beginning air operations on June 21. The first landing on the new carrier was conducted by Lieutenant Commander A.C. Davis flying a Vought SBU-1. Further training for Rangers air group was conducted in August. Interwar Years Later in August, Ranger departed on an extended shakedown cruise to South America which included port calls at Rio de Janeiro, Buenos Aires, and Montevideo. Returning to Norfolk, VA, the carrier conducted operations locally before receiving orders for the Pacific in April 1935. Passing through the Panama Canal, Ranger arrived at San Diego, CA on the 15th. Remaining in the Pacific for the next four years, the carrier took part in fleet maneuvers and war games as far west as Hawaii and as far south as Callao, Peru while also experimenting with cold weather operations off Alaska. In January 1939, Ranger departed California and sailed for Guantanamo Bay, Cuba to participate in winter fleet maneuvers. With the completion of these exercises, it steamed to Norfolk where it arrived in late April. Operating along the East Coast through the summer of 1939, Ranger was assigned to the Neutrality Patrol that fall following the outbreak of World War II in Europe. The initial responsibility of this force was to track warlike operations of combatant forces in the Western Hemisphere. Patrolling between Bermuda and Argentia, Newfoundland, Rangers seakeeping ability was found lacking as it proved difficult to conduct operations in heavy weather. This issue had been identified earlier and helped contribute to the design of the later Yorktown-class carriers. Continuing with the Neutrality Patrol through 1940, the carriers air group was one of the first to receive the new Grumman F4F Wildcat fighter that December. In In late 1941, Ranger was returning to Norfolk from a patrol to Port-of-Spain, Trinidad when the Japanese attacked Pearl Harbor on December 7. World War II Begins Departing Norfolk two weeks later, Ranger conducted a patrol of the South Atlantic before entering drydock in March 1942. Undergoing repairs, the carrier also received the new RCA CXAM-1 radar. Deemed too slow to keep up with newer carriers, such as USS Yorktown (CV-5) and USS Enterprise (CV-6), in Pacific, Ranger remained in the Atlantic to support operations against Germany. With the completion of repairs, Ranger sailed on April 22 to deliver a force of sixty-eight P-40 Warhawks to Accra, Gold Coast. Returning to Quonset Point, RI in late May, the carrier conducted a patrol to Argentia before delivering a second cargo of P-40s to Accra in July. Both shipments of P-40s were destined for China where they were to serve with the American Volunteer Group (Flying Tigers). With the completion of this mission, Ranger operated off Norfolk before joining four new Sangamon-class escort carriers (Sangamon, Suwannee, Chenango, and Santee) at Bermuda. Operation Torch Leading this carrier force, Ranger provided air superiority for the Operation Torch landings in Vichy-ruled French Morocco in November 1942. Early on November 8, Ranger began launching aircraft from a position approximately 30 miles northwest of Casablanca. While F4F Wildcats strafed Vichy airfields, SBD Dauntless dive bombers struck at Vichy naval vessels. In three days of operations, Ranger launched 496 sorties which resulted in the destruction of around 85 enemy aircraft (15 in the air, approx. 70 on the ground), the sinking of the battleship Jean Bart, severe damage to the destroyer leader Albatros, and attacks on the cruiser Primaugut. With the fall of Casablanca to American forces on November 11, the carrier departed for Norfolk the next day. Arriving, Ranger underwent an overhaul from December 16, 1942 to February 7, 1943. With the Home Fleet Departing the yard, Ranger carried a load of P-40s to Africa for use by the 58th Fighter Group before spending much of the summer of 1943 conducting pilot training off the New England coast. Crossing the Atlantic in late August, the carrier joined the British Home Fleet at Scapa Flow in the Orkney Islands. Putting out on October 2 as part of Operation Leader, Ranger and a combined Anglo-American force moved towards Norway with the goal of attacking German shipping around Vestfjorden. Avoiding detection, Ranger began launching aircraft on October 4. Striking a short time later, the aircraft sank two merchant vessels in Bodo roadstead and damaged several more. Though located by three German aircraft, the carriers combat air patrol downed two and chased off the third. A second strike succeeded in sinking a freighter and a smaller coastal vessel. Returning to Scapa Flow, Ranger commenced patrols to Iceland with the British Second Battle Squadron. These continued until late November when the carrier detached and sailed for Boston, MA. Later Career Too slow to operate with the fast carrier forces in the Pacific, Ranger was designated as a training carrier and ordered to operate out of Quonset Point on January 3, 1944. These duties were interrupted in April when it transported a cargo of P-38 Lightning to Casablanca. While in Morocco, it embarked several damaged aircraft as well as numerous passengers for transport to New York. After arriving in New York, Ranger steamed to Norfolk for an overhaul. Though Chief of Naval Operations Admiral Ernest King favored a massive overhaul to bring the carrier on par with its contemporaries, he was discouraged in following through by his staff who pointed out that the project would draw resources away from new construction. As a result, the project was limited to strengthening the flight deck, installation of new catapults, and improving the ships radar systems. With the completion of the overhaul, Ranger sailed for San Diego where it embarked Night Fighting Squadron 102 before pressing on to Pearl Harbor. From August to October, it conducted night carrier flight training operations in Hawaiian waters before returning to California to serve as a training carrier. Operating from San Diego, Ranger spent the remainder of the war training naval aviators off the California coast. With the end of the war in September, it transited the Panama Canal and made stops at New Orleans, LA, Pensacola, FL, and Norfolk before reaching the Philadelphia Naval Shipyard on November 19. After a brief overhaul, Ranger resumed operations on the East Coast until being decommissioned on October 18, 1946. The carrier was sold for scrap the following January. Selected Sources DANFS: USS Ranger (CV-4)Air Group 4: USS RangerMilitary Factory: USS Ranger (CV-4)

Models of Health Behavior Essays

Models of Health Behavior Essays Models of Health Behavior Paper Models of Health Behavior Paper Note: If candidate uses incorrect formula: maximum 1/4 marks (for standard form) substitution into correct formula Note: If an error in subs and 4  ± ? 80 gets: and 6 states â€Å"no solution†: maximum 3/4 marks 4  ± 16 + 96 6 4  ± 112 = 6 2 ±2 7 = 3 = 2,43 or ? 1,10 112 If doesn’t conclude with â€Å"no solution†: maximum 2/4 marks 4  ± 112 or 6 decimal answer (4) Copyright reserved Mathematics/PI 3 NSC – Memorandum DBE/November 2011 OR 3x 2 ? 4 x = 8 3x 2 ? 4 x ? 8 = 0 x= = ? b  ± b ? 4ac 2a ? (? 4)  ± 2 (? 4)2 ? 4(3)(? 8) 2(3) Note: Penalise 1 mark for inaccurate rounding off to ANY number of decimal places if candidate gives decimal answers tandard form substitution into correct formula answer answer (4) = 2,43 or ? 1,10 1. 1. 3 4 x 2 + 1 ? 5x (4 x ? 1)(x ? 1) ? 0 + 0 1 4 x? OR 4 x 2 ? 5x + 1 ? 0 factors ? 0 1 + 1 4 1? ? OR ? ? ? ; ? ? [1; ? ) 4? ? 1 both critical 1 values of and 1 4 or OR ? answer (4) 1 or x ? 1 4 1 4 1 x OR 1 4 1 xNote: If candidate gives either of these correct graphical solutions but writes down the incorrect intervals or uses AND: max 3/4 marks NOTES: If a candidate gives an answer of 1 ? x ? If a candidate gives an answer of 1 then max 3/4 marks. 4 1 ? x ? 1 then max 2/4 marks. 4 1 If a candidate gives an answer of x ? nd x ? 1 then max 3/4 marks. 4 If the candidate leaves out the equality of the notation then penalty of 1 mark. 1 If a candidate gives an answer of x ? ; x ? 1 then max 3/4 marks. 4 1 If candidate gives x ? and/or x ? 1 , BREAKDOWN: max 2/4 marks. 4 If candidate gives : 0 ? 0 + + award 3/4 marks 1 1 4 Copyright reserved Please turn over Mathematics/PI 4 NSC – Memorandum DBE/November 2011 1. 2. 1 x 2 + 5 xy + 6 y 2 = 0 (x + 3 y )(x + 2 y ) = 0 x + 3y = 0 x = ? 3 y OR x = ? 3 y OR x + 2y = 0 x = ? 2 y x = ? 2 y Note: If a candidate gives x x ? = 3 or ? = 2 y y award 2/3 marks factors answers (3) Let k = y 2 x + 5 xy + 6 y 2 = 0 2 ?x? ?x? ? ? + 5? ? + 6 = 0 ? y? ? y? ? ? ? ? 2 k + 5k + 6 = 0 (k + 3)(k + 2) = 0 k = – 3 or k = – 2 x x = ? 3 or = ? 2 y y OR factors answers (3) x 2 + 5 xy + 6 y 2 = 0 x= x= ? 5 y  ± (5 y ) 2 ? 4(1)(6 y 2 ) 2(1) ? 5y  ± y2 2 ? 5y  ± y x= 2 x = ? 3 y x = ? 2 y or x x = ? 3 = ? 2 y y substitutes correctly into correct formula answers (3) OR x 2 + 5 xy + 6 y 2 = 0 ?5 x 2 + 5 xy + ? ?2 ? 5 ? y ? = ? 6 y 2 + ? ?2 ? 2 2 ? y? ? 2 5 ? 1 2 ? ? x + y? = y 2 ? 4 ? 5 1 x+ y= ± y 2 2 5 1 x=? y ± y 2 2 completing the square Copyright reservedPlease turn over Mathematics/PI x = ? 3 y x = ? 3 y x = ? y 5 NSC – Memorandum DBE/November 2011 answers (3) or x = ? 2 y OR Let k = x = ky x y x 2 + 5 xy + 6 y 2 = 0 (ky )2 + 5 y(ky ) + 6 y 2 = 0 k 2 y 2 + 5 y 2k + 6 y 2 = 0 y 2 k 2 + 5k + 6 = 0 ( (k ) 2 + 5k + 6 = 0 ) factors (k + 3)(k + 2) = 0 k = – 3 or k = – 2 x x = ? 3 or = ? 2 y y answers (3) Note: (x;y) = (0;0) is also a solution, but in this case OR x is undefined y Let y = 1 , x 2 + 5x + 6 = 0 (x + 2)(x + 3) = 0 x = ? 2 or x = ? 3 x x = ? 2 or = ? 3 y y x+ y =8 ? 3y + y = 8 ? 2y = 8 y = ? 4 x = 12 factors answers (3) x+ y =8 ? 2y + y = 8 ? y =8 y = ? 8 x = 16 1. 2. 2 OR substitution x = – 3y subs x = ? 2 y values both x values correct (5) OR 8? y = ? 3 y 8 ? y = ? 3 y 8 = ? 2 y y = ? 4 x = 12 Copyright reserved 8? y = ? 2 OR y 8 ? y = ? 2 y 8 = ? y y = ? 8 x = 16 x=8–y substitution y values both correct x values (5) Please turn over Mathematics/PI 6 NSC – Memorandum DBE/November 2011 OR x+ y =8 y =8? x x = ? 3 OR 8? x x = ? 3(8 ? x ) x = ? 24 + 3 x ? 2 x = ? 24 x = 12 y = ? 4 OR y =8? x x = ? 2 8? x x = ? 2(8 ? x ) x = ? 16 + 2 x ? x = ? 16 x = 16 y = ? 8 substitution x values correct both y values correct (5) (x + 2 y )(x + 3 y ) = 0 x+ y =8 x =8? y ( y + 8)(2 y + 8) = 0 y = ? 8 or y = ? 4 x = 16 x = 12 x =8? y ubstitution y values correct both x values correct (5) OR x = 8? y x = 8? y 2 (8 ? y ) 2 + 5(8 ? y ) y + 6 y = 0 64 ? 16 y + y 2 + 40 y ? 5 y 2 + 6 y 2 = 0 2 y 2 + 24 y + 64 = 0 y 2 + 12 y + 32 = 0 ( y + 8)( y + 4) = 0 y = ? 8 or y = ? 4 x = 16 x = 12 OR substitution factors both y values correct both x values correct (5) Copyright reserved Please turn over Mathematics/PI 7 NSC – Memorandum DBE/November 2011 OR x =8? y (8 ? y ) 2 + 5(8 ? y ) y + 6 y = 0 2 2 2 2 x = 8? y substitution 64 ? 16 y + y + 40 y ? 5 y + 6 y = 0 2 y 2 + 24 y + 64 = 0 y 2 + 12 y + 32 = 0 ? 12  ± 12 ? 4(1)(32) y= 2(1) 2 ? 12  ± 16 2 y = ? 8 or y = ? x = 16 x = 12 = Note: If a candidate uses the formula and replaces x for y and then answers are swapped: maximum 4/5 marks substitutes into correct formula both y values correct both x values correct (5) OR y =8? x x 2 + 5 x(8 ? x ) + 6(8 ? x )2 = 0 x + 40 x ? 5 x + 6 64 ? 16 x + x 2 x 2 ? 56 x + 384 = 0 x 2 ? 28 x + 192 = 0 (x ? 16)(x ? 12) = 0 x = 16 x = 12 or y = ? 8 y = ? 4 2 2 y =8? x 2 ( )= 0 substitution factors both x values correct both y values correct (5) OR y =8? x x + 5 x(8 ? x ) + 6 (8 ? x ) = 0 2 2 y =8? x substitution x + 40 x ? 5 x 2 2 + 6(64 ? 16 x + x ) = 0 2 2 x 2 ? 56 x + 384 = 0 x 2 ? 28 x + 192 = 0 x= = ? ? 28)  ± 28  ± 416 2 (? 28)2 ? 4(1)(192) 2(1) x = 12 x = 16 or y = ? 4 y = ? 8 substitutes into correct formula both x values correct both correct y values (5) [19] Please turn over Copyright reserved Mathematics/PI 8 NSC – Memorandum DBE/November 2011 QUESTION 2 2. 1. 1 x ? 4 = 32 ? x 2 x = 36 x = 18 T2 ? T1 = T3 ? T2 Note: If answer only: award 2/2 marks answer (2) a + 2d = 32 and a = 4 OR a=4 a + 2d = 32 2d = 28 d = 14 x = 14 + 4 x = 18 OR Note: If candidate writes x? 4 32 ? x only (i. e. omits equality) : 0/2 marks answer (2) substitutes correctly into arithmetic mean 4 + 32 formula i. e. 2 answers (2) T2 T3 = T1 T2 x= + 32 = 18 2 2. 1. 2 x 32 = 4 x x 2 = 128 x =  ± 128 x =  ±8 2 OR a=4 x r= 4 ? x? ar 2 = 4? ? ? 4? ? x? 32 = 4? ? ? 4? x 2 = 128 OR x =  ±11,31 OR x =  ± 2 2 7 Note: If candidate 32 x only writes 4 x (i. e. omits e quality) : 0/2 marks x 2 = 128 both answers (surd or decimal or exponential form) (3) Note: If only x = 128 then penalty 1 mark 2 2 2 ? x? 32 = 4? ? ? 4? 2 x = 128 x =  ± 128 x =  ±8 2 or x =  ±11,31 or x =  ± 2 2 7 both answers (surd or decimal or exponential form) (3) substitutes correctly into geometric mean formula i. e.  ± 4? 32 both answers (surd or decimal or exponential form) (3) Please turn over OR x =  ± 4 ? 32 =  ± 128 or x =  ±8 2 or x =  ±11,31 or x =  ± 2 2 7 Copyright reserved Mathematics/PI 9 NSC – Memorandum DBE/November 2011 2. 2 13 P = ? 3k ? 5 a = 3? 4 or +3 3? 5 =3 k =1 1? 5 +3 2? 5 + + 3 13 ? 5 = 3 ? 4 + 3 ? 3 + 3- 2 + + 38 3 ? 4 313 ? 1 = 3 ? 1 = 9841,49 Note: Correct answer only: 1/4 marks only 1 81 ( ) r =3 subs into correct formula or 9841 40 797161 or 81 81 answer (4) OR 13 P = ? 3k ? 5 k =1 1? 5 =3 + 32 ? 5 + 33 ? 5 + + 313 ? 5 2. 3 2 S n = [2a + (n ? 1)d ] + [2a + (n ? 1)d ] + + [2a + (n ? 1)d ] + [2a + (n ? 1)d ] = n[2a + (n ? 1)d ] Sn = n [2a + (n ? 1)d ] 2 S n = [a + (n ? 1)d ] + [a + (n ? 2)d ] + [a + d ] + a = 3 ? 4 + 3? 3 + 3- 2 + + 38 1 1 1 = + + + + 6561 81 27 9 40 797161 or = 9841,49 or 9841 81 81 S n = a + [a + d ] + [a + 2d ] + + [a + (n ? 2 )d ] + [a + (n ? 1)d ] Note: If the candidate rounds off and gets 9841,46 (i. e. correct to one decimal place): DO NOT penalise for the rounding off. expand the sum 13 terms in expansion answer (4) writing out Sn â€Å"reversing† Sn expressing 2Sn grouping to get 2 S n = n[2a + (n ? 1)d ] (4) writing out Sn â€Å"reversing† Sn expressing 2Sn grouping to get 2S n = n[a + a + (n ? 1)d ] (4) OR S n = a + [a + d ] + [a + 2d ] + + (Tn ? d ) + Tn S n = Tn + (Tn ? d ) + [a + d ] + a = n[a + a + (n ? 1)d ] = [2a + (n ? 1)d ] Sn = n [2a + (n ? 1)d ] 2 2 S n = a + Tn + a + Tn + a + Tn + + a + Tn Note: If a candidate uses a circular argument (eg S n +1 = S n + Tn ): max 1/4 marks (for writing out Sn) Note: If a candidate uses a specific linear sequence, then NO marks. [13] Copyright reserved Please turn over Mathematics/PI 10 NSC – Memorandum DBE/November 2011 QUESTION 3 3. 1 21; 24 Note: If candidate writes T8 = 21 T7 = 24 : award 1/2 marks 21 24 (2) 3. 2 T2 k = 3. 2 k ? 1 and so T52 = 3. 2 26 ? 1 = 100663296 Note: If candidate writes out all 52 terms and gets correct answer: award 5/5 marks . 2 k ? 1 T52 6k ? 3 T51 T2k ? 1 = 3 + 6(k ? 1) = 6k ? 3 and so T51 = 6(26) ? 3 = 153 T52 ? T51 = 100663296 ? 153 = 100663143 answer Note: If candidate used k = 52: max 2/5 Note: if candidate interchanges order i. e. does T51 ? T52 : max 4/5 marks Note: writes out all 52 terms and subtracts T51 ? T52 : max 4/5 marks (5) OR Consider sequence P: 3 ; 6 ; 12 †¦ Pn = 3. 2 n? 1 P26 = 3. 2 26 ? 1 = 100663296 Consider sequence Q: 3 ; 9 ; 15 †¦ Qn = 6n ? 3 Q26 = 6(26) ? 3 = 153 T52 ? T51 = P26 ? Q26 = 100663296 ? 153 = 100663143 Pn = 3. 2 n? 1 P26 Qn = 6n ? 3 Q26 answer (5) Copyright reserved Please turn over Mathematics/PI 1 NSC – Memorandum DBE/November 2011 3. 3 For all n ? N , n = 2k or n = 2k ? 1 for some k ? N If n = 2k : Tn = T2k = 3. 2 k ? 1 If n = 2k ? 1 : Tn = T2k ? 1 = 6k ? 3 = 3(2k ? 1) factors 3. 2 k ? 1 Note: If a candidate only illustrates divisibility by 3 with a specific finite part of the sequence, not the general term: 0/2 marks factors 3(2k ? 1) (2) In either case, Tn has a factor of 3, so is divisible by 3. OR Pn = 3. 2 n ? 1 Which is a multiple of 3 Qn = 6 n ? 3 = 3(2n ? 1) Which is also a multiple of 3 Since Tn = Q2 k ? 1 or Tn = P2 k for all n ? N , Tn is always divisible by 3 OR factors 3. 2 n ? 1 factors 3(2n ? 1) (2)The odd terms are odd multiples of 3 and the even terms are 3 times a power of 2. This means that all the terms are multiples of 3 and are therefore divisible by 3. odd multiples of 3 3 times a power of 2 (2) [9] Copyright reserved Please turn over Mathematics/PI 12 NSC – Memorandum DBE/November 2011 QUESTION 4 4. 1 The second, third, fourth and fift h terms are 1 ; – 6 ; T4 and – 14 First differences are: – 7 ; T4 + 6 ; – 14 – T4 So T4 + 6 + 7= – 14 – 2T4 – 6 T4 = – 11 d = – 11 + 6 + 7 = 2 or – 14 + 22 – 6 = 2 Note: Answer only (i. e. d = 2) with no working: 3 marks Note: Candidate gives T4 = ? 11 and d = 2 only: award 5/5 marks 7 T4 + 6 – 14 – T4 setting up equation T5 ? T2 = (T5 ? T4 ) + (T4 ? T3 ) + (T3 ? T2 ) answer (5) –7 –7+d – 7 + 2d setting up equation T5 ? T2 = (T5 ? T4 ) + (T4 ? T3 ) + (T3 ? T2 ) OR T2 1 -7 T3 -6 -7+d d T4 -7+2d d T5 -14 T5 ? T2 = (T5 ? T4 ) + (T4 ? T3 ) + (T3 ? T2 ) ? 15 = (? 7 + 2d ) + (? 7 + d ) + ? 7 ? 15 = ? 21 + 3d 6 = 3d d =2 Note: Candidate uses trial and error and shows this: award 5/5 marks answer (5) OR 4a + 2b + c = 1 9a + 3b + c = ? 6 5a + b = ? 7 25a + 5b + c = ? 14 16a + 2b = ? 8 10a + 2b = ? 14 6a = 6 a =1 d = 2a = 2 4a + 2b + c = 1 9a + 3b + c = ? 6 25a + 5b + c = ? 1 4 solved simultaneously answer (5) ORT1 1 – T1 T1 8 1 -7 T4+13 -6 T4+6 -20-2 T4 -14 T4 -14 T4 –7 T4 + 6 ? 14 ? T4 setting up equation answer (5) Please turn over T4 + 13 = ? 20 ? 2T4 3T4 = ? 33 T4 = ? 11 d = ? 11 + 13 d =2 Copyright reserved Mathematics/PI 13 NSC – Memorandum DBE/November 2011 OR T1 x 1–x -8+x T2 1 -7 y +13 T3 -6 y+6 20 – 2y T4 y -14 y T5 -14 –7 y+6 ? 14 ? y y + 13 = ? 20 ? 2 y 3 y = ? 33 y = ? 11 Second difference = y + 13 = ? 11 + 13 = 2 4. 2 T1 1 –6 -9 2 -7 Note: Answer only: award 2/2 marks Note: If incorrect d in 4. 1, 2/2 CA marks for T1 = d + 8 (since 1 ? T1 = ? 7 ? d ) setting up equation answer (5) method T1 = 10 (2) T1 = 10 OR =1 5a + b = ? 7 5(1) + b = ? 7 b = ? 12 a +b+c =1 4(1) + 2(? 12) + c = 1 c = 21 Tn = n 2 ? 12n + 21 T1 = (1) 2 ? 12(1) + 21 = 10 OR method T1 = 10 (2) T4 + 13 = ? 8 + T1 ? 11 + 13 = ? 8 + T1 T1 = 10 y + 13 = ? 8 + x method T1 = 10 (2) [7] OR ? 11 + 13 = ? 8 + x x = 10 Copyright res erved Please turn over Mathematics/PI 14 NSC – Memorandum DBE/November 2011 QUESTION 5 5. 1. 1 y = f (0) ? 6 = ? 1 0? 3 =1 (0 ; 1) OR ?6 ? 1 x? 3 ? 6 1= x? 3 x ? 3 = ? 6 0= x = ? 3 (? 3 ; 0) x = 0 and y = 1 5. 1. 2 Note: Mark 5. 1. 1 and 5. 1. 2 as a single question. If the intercepts are interchanged: max 3/5 marks y =1 x=0 (2) y=0 x ? 3 = ? 6 nswer (3) shape y 5. 1. 3 Note: The graph must tend towards the asymptotes in order to be awarded the shape mark (? 3; 0) (0 ; 1) 0 3 x y = ? 1 ?1 x=3 both intercepts correct horizontal asymptote vertical asymptote (4) Note: A candidate who draws only one ‘arm’ of the hyperbola loses the ‘shape’ mark i. e. max 3/4 marks 5. 1. 4 ? 3 lt; x lt; 3 OR (? 3; 3) OR ? 3 lt; x and x lt; 3 Note: if candidate writes ? 3 lt; x only: 1/2 marks Note: if candidate writes x lt; 3 only: 1/2 marks ?3 and 3 inequality OR interval notation (2) Copyright reserved Please turn over Mathematics/PI 15 NSC – Memorandum DBE/Novem ber 2011 5. 1. y= ?6 ? 1 ? 2? 3 1 = 5 1? 1 5 1 5 m= 0 ? (? 2) 2 = 5 formula substitution answer (4) OR m= = f (0) ? f (? 2) 0 ? (? 2) 1? 1 5 formula f (? 2) = 1 5 0+2 2 = 5 b lt; 0 since b lt; 0 and a lt; 0 2a y x substitution answer (4) y-intercept negative turning point on the x axis turning point on the left of the y axis maximum TP and quadratic shape 5. 2 x=? 0 (4) [19] Copyright reserved Please turn over Mathematics/PI 16 NSC – Memorandum y DBE/November 2011 QUESTION 6 f C(0 ; 4,5) g x O A B 6. 1 0 = 2x ? 8 8 = 2x 23 = 2 x x=3 A(3 ; 0) f (0) = 2 0 ? 8 = 1? 8 = ? 7 B(0 ; –7) Note: no CA marks Note: answer only: award 2/2 marks =0 answer for A x=0 answer for B (4) answer (1) 6. 2 6. 3 y = ? 8 OR y + 8 = 0 h( x ) = f ( 2 x ) + 8 = 22x ? 8 + 8 ( ) (2 2 x ? 8) answer of h( x ) = 4 x or 2 2 x (2) = 4 x or 2 2 x 6. 4 x = 4y y = log 4 x OR x = 22 y 2 y = log 2 x 1 y = log 2 x OR y = log 2 x 2 Note: answer only award 2/2 marks Note: candidate works out f -1 and gets y = l og 2 ( x + 8) award 1/2 marks log x OR y = log 4 switch x and y answer in the form y =†¦ (2) 6. 5 p ( x) = ? log 4 x OR p( x) = log 1 x 4 answer (1) OR p ( x) = log 4 OR 1 x OR 1 p( x) = ? log 2 x 2 y = ? log 2 x Copyright reserved Please turn over Mathematics/PI 17 NSC – Memorandum 5 DBE/November 2011 . 6 ? g (k ) ? ? g (k ) = g (0) + g (1) + g (2) + g (3) ? g (4) ? g (5) x = 3 is the axis of symmetry of g ? by symmetry g (2) = g (4) and g (1) = g (5) Answer = g (0) + g (3) = 4,5 + 0 = 4,5 OR k =0 k =4 3 = g (0) + g (1) + g (2) + g (3) ? g (4) ? g (5) g (2) = g (4) and g (1) = g (5) g (0) + g (3) answer (4) ? g (k ) ? ? g (k ) ? g (k ) = g (0) + g (1) + g (2) + g (3) ? g (k ) = g (4) + g (5) k =4 k =0 5 3 5 expansion k =0 3 k =4 x = 3 is the axis of symmetry of g ? by symmetry g (4) = g (2) g (5) = g (1) k =0 g (2) = g (4) and g (1) = g (5) ? g (k ) ? ? g (k ) k =4 3 5 = g (0) + g (3) = 4,5 + 0 = 4,5 OR g (0) + g (3) answer (4) ( x) = a( x ? 3) + 0 2 4,5 = a(0 ? 3) 2 + 0 4,5 = 9a 1 a= 2 1 2 g ( x) = ( x ? 3) 2 k =0 3 k =0 g ( x) = 1 (x ? 3)2 2 ? g (k ) ? ? g (k ) k =4 3 5 ? g (k ) = g (0) + g (1) + g (2) + g (3) = 4,5 + 2 + 0,5 + 0 =7 expansion Copyright reserved Please turn over Mathematics/PI 18 NSC – Memorandum DBE/November 2011 k =4 ? g (k ) = g (4) + g (5) = 0,5 + 2 = 2,5 ? g (k ) ? ? g (k ) k =4 3 5 5 k =0 = 7 ? 2,5 = 4,5 7 ? 2,5 answer (4) OR g ( x) = ax 2 + bx + c g (k ) = ak 2 + bk + c g (0) = c g (1) = a + b + c g (2) = 4a + 2b + c g (3) = 9a + 3b + c k =0 ? g (k ) = 14a + 6b + 4c 3 g (4) = 16a + 4b + c ? g (k ) = 41a + 9b + 2c 5 5 (5) = 25a + 9b + c k =4 3 k =0 ? g (k ) ? ? g (k ) = ? 27a ? 3b + 2c k =4 ? 27 a ? 3b + 2c g ( x) = a ( x ? 3)2 + 0 4,5 = a (0 ? 3) 2 + 0 4,5 = 9a 1 2 1 g ( x) = ( x ? 3)2 2 1 2 9 = x ? 3x + 2 2 a= k =0 ? g (k ) ? ? g (k ) = ? 27 a ? 3b + 2c k =4 3 5 g ( x) = 1 (x ? 3)2 2 ?9? ?1? = ? 27? ? ? 3(? 3) + 2? ? ? 2? ?2? = 4,5 answer (4) [14] Copyright reserved Please turn over Mathematics/PI 19 NSC – M emorandum DBE/November 2011 QUESTION 7 7. 1 A = P(1 ? i ) P n = P(1 ? 0,07 ) 2 1 = 0,93n 2 1 log = n log 0,93 2 1 log 2 n= log 0,93 = 9,55 years n OR A = P(1 ? i )n P = P(1 ? 0,07 )n 2 1 = 0,93 n 2 1 log 0,93 = n 2 n = 9,55 yearsP 2 subs into correct formula A= log answer (4) Note: If candidate uses incorrect formula: max 1/4 marks P for A = 2 Note: If candidate interchanges A and P A i. e. uses P = : max 2/4 marks 2 Copyright reserved Please turn over Mathematics/PI 20 NSC – Memorandum DBE/November 2011 7. 2 Radesh: A = P(1 + in ) = 8 550 Bonus = 0,05 ? 6 000 = 300 = 6 000(1 + 0,085 ? 5) A = 6 000 + 8,5% of 6000 ? 5 OR = 6000 + 510 ? 5 = 6000 + 2550 = 8 550 8 550 Received = 8 550 + 300 = R 8 850 Thandi: n A = P(1 + i ) R8 850 ? 0,08 ? = 6 000? 1 + ? 4 ? ? = R 8 915,68 20 n = 20 0,08 i= 4 answer choice made (6) 0,15 1 or or 0,0125 12 80 n = 18 n = 18 7. 3Thandis investment is bigger. Fv = initial deposit with interest + annuity ? ? ? 0,15 ? 18 ? ?1 + ? ? 1? 18 ? 12 ? ? 0,15 ? = 1 000? 1 + ? + 700? ? 0,15 12 ? ? ? ? ? ? 12 ? ? = 1 250,58 + 14 032,33 = R15 282,91 i= OR ? 0,15 ? 1 000? 1 + ? 12 ? ? ? ? 0,15 ? 18 ? ? ? 1 + ? ?1? ? 12 ? 700? ? 0,15 ? ? ? ? 12 ? ? answer (6) 18 Fv = initial deposit with interest + annuity ?18 ? ? ? ? 1 ? ?1 + 0,15 ? ? ? 18 18 ? ? 0. 15 ? 0,15 ? 12 ? ? = 1 000? 1 + ? ? + 700? 1 + 0,15 12 ? 12 ? ? ? ? ? 12 ? ? 0,15 1 or or 0,0125 12 80 n = 18 n = 18 i= 0. 15 ? ? = 1 250,58 + 11220,68? 1 + ? 12 ? ? = 1 250,58 + 14 032,33 = R15 282,91 Copyright reserved 8 ? 0,15 ? 1 000? 1 + ? 12 ? ? ?18 ? ? ? 1 ? ?1 + 0,15 ? ? ? ? 12 ? 700? 0,15 ? ? 12 ? 18 ? ? 18 0. 15 ? ? 1 + 12 ? ? ? answer (6) Please turn over Mathematics/PI 21 NSC – Memorandum DBE/November 2011 OR ? ? 0,15 ? 19 ? ? ? 1 + ? ?1? 18 ? 12 ? ? 0,15 ? Fv = 300? 1 + ? + 700? ? 0,15 12 ? ? ? ? ? ? 12 ? ? = 375,17 + 14 907,74 = R15 282,91 0,15 1 or or 0,0125 80 12 n = 19 (corresponding to 700) n = 18 (corresponding to 300) i= ? 0,15 ? 300? 1 + ? 12 ? ? ? ? 0,15 ? 19 ? ? ? 1 + ? ?1? 12 ? ? 700? ? 0,15 ? ? ? ? 12 ? ? answer (6) [16] 18 QUESTION 8 8. 1 f ? ( x ) = lim = lim f (x + h ) ? f (x ) h h;0 ? 4( x + h )2 ? 4 x 2 h h;0 ( ) Note: Incorrect notation: formula substitution expansion = lim = lim ? 4 x 2 + 2 xh + h 2 + 4 x 2 h h;0 ? 4 x 2 ? 8 xh ? 4h 2 + 4 x 2 h h;0 ( ) no lim written: penalty 2 marks lim written before equals sign: penalty 1 mark Note: A candidate who gives –8x only: 0/5 marks Note: A candidate who omits brackets in the line lim (? 8 x ? 4h ) : h ;0 ? 8 xh ? 4h 2 = lim h h;0 h(? 8 x ? 4h ) = lim h h;0 = lim (? 8 x ? 4h ) h;0 ? 8 x ? 4h answer (5) = ? 8 x NO penalty OR Copyright reserved Please turn over Mathematics/PI 22 NSC – Memorandum DBE/November 2011 f ( x ) = ? 4 x 2 f ( x + h) = ? 4( x + h) 2 = ? 4 x 2 ? xh ? 4h 2 f ( x + h) ? f ( x) = ? 8 xh ? 4h 2 ? 8 xh ? 4h 2 f ? ( x) = lim h h;0 h ( ? 8 x ? 4 h ) = lim h h;0 = lim (? 8 x ? 4h) h;0 substitution expansion formula ? 8 x ? 4h = ? 8 x 8. 2. 1 y= 3 x2 ? 2x 2 3 1 = x ? 1 ? x 2 2 2 answer (5) 3 ? 1 x 2 ? Note: Incorrect notation in 8. 2. 1 and/or 8. 2. 2: Penalise 1 mark 3 dy = ? x ? 2 ? x 2 dx 3 =? 2 ? x 2x 8. 2. 2 f ( x) = (7 x + 1) 2 3 ? 2 x 2 (3) ?x = 49 x 2 + 14 x + 1 f ? ( x) = 98 x + 14 f ? (1) = 98(1) + 14 = 112 multiplication 98 x 14 answer (4) OR f ( x) = (7 x + 1) 2 f ? ( x) = 2(7 x + 1)(7 ) By the chain rule . f ? ( x) = 98 x + 14 f ? (1) = 98(1) + 14 = 112 hain rule answer (4) [12] Copyright reserved Please turn over Mathematics/PI 23 NSC – Memorandum DBE/November 2011 QUESTION 9 9. 1 f ( x ) = ? 2 x 3 + ax 2 + bx + c f ? ( x ) = ? 6 x 2 + 2ax + b = ? 6( x ? 5)( x ? 2) = ? 6 x 2 ? 7 x + 10 ( ) = ? 6 x 2 + 42 x ? 60 2a = 42 a = 21 b = ? 60 f (5) = ? 2(5) + 21(5) ? 60(5) + c 3 2 Note: A candidate who substitutes the values of a, b and c and then checks (by substitution) that T (2; ? 9 ) and S (5;18) lie on the curve: award max 2/7 marks f ? ( x ) = ? 6 x 2 + 2ax + b ? 6( x ? 5)( x ? 2 ) b= –60 2a = 42 f (2) = ? 2(2) + 21(2) ? 60(2) + c OR ? 9 = ? 52 + c c = 43 3 2 18 = ? 5 + c c = 43 subs (5 ; 18) or (2 ; -9) c = 43 (7) Note: A candidate who substitutes the values of a, b and c into the a = 21 ; b = ? 60 ; c = 43 function i. e. gets f ( x) = ? 2 x 3 ? 21x 2 ? 60 x + 43 and then shows by substitution that T (2; ? 9 ) and S (5;18) are on the curve and works out the derivative i. e. gets f ? ( x ) = ? 6 x 2 ? 42 x ? 60 and shows (by substitution into the derivative) that the turning points are at x = 2 and x = 5 (assuming what s/he sets out to prove and proving what is given): award max 4/7 marks as follows: x = 2 from f ? ( x ) = 0 OR subs x = 2 into the derivative and gets 0 x = 5 from f ? x ) = 0 OR subs x = 5 into the derivative and gets 0 substitution of x = 2 in f and gets – 9 substitution of x = 5 in f and gets 18 f ? ( x ) = ? 6 x 2 + 2ax + b f ? (2) = 0 f ? (5) = 0 6a = 126 OR f ? ( x ) = ? 6 x 2 + 2ax + b f ? (2) = ? 6(2) + 2a (2) + b 0 = ? 24 + 4a + b 2 b = 24 ? 4a f ? (5) = ? 6(5) 2 + 2a (5) + b 0 = ? 150 + 10a + b 0 = ? 150 + 10a + (24 ? 4a) 0 = ? 126 + 6a 6a = 126 a = 21 b = ? 60 f (5) = ? 2(5) + 21(5) ? 60(5) + c 3 2 Note: If derivative equal to zero is not written: penalize once only b = – 60 f (2) = ? 2(2) + 21(2) ? 60(2) + c 3 2 subs (5 ; 18) or (2 ; -9) c = 43 (7) Please turn over 18 = ? 25 + c c = 43Copyright reserved a = 21 ; b = ? 60 ; c = 43 OR ? 9 = ? 52 + c c = 43 Mathematics/PI 24 NSC – Memorandum DBE/November 2011 OR f (2) = ? 9 i. e. ? 16 + 4a + 2b + c = ? 9 4a + 2b + c = 7 f (5) = 18 i. e. ? 250 + 25a + 5b + c = 18 25a + 5b + c = 268 21a + 3b = 261 f ? (5) = 0 f ? ( x ) = ? 6 x 2 + 2ax + b and f ? (2 ) = 0 OR 4a + b = 24 10a + b = 150 ? 16 + 4a + 2b + c = ? 9 and ? 250 + 25a + 5b + c = 18 f ? ( x ) = ? 6 x 2 + 2ax + b f ? (2) = 0 or f ? (5) = 0 12a + 3b = 72 9a = 189 189 9 a = 21 a= 12(21) + 3b = 72 3b = ? 180 b = ? 60 4(21) + 2(? 60 ) + c = 7 c = 43 4a + 2b + c = 7 OR 30a + 3b = 450 9a = 189 189 a= 9 a = 21 9a = 189 b = – 60 5a + 5b + c = 268 c = 43 subs (5 ; 18) or (2 ; -9) c = 43 (7) subs f ? (1) m tan = ? 24 f(1) = 2 f ? ( x) = ? 6 x + 42 x ? 60 2 OR 25(21) + 5(? 60 ) + c = 268 9. 2 f ? ( x) = ? 6 x + 42 x ? 60 2 m tan = ? 6(1) + 42(1) ? 60 2 = ? 24 3 2 f (1) = ? 2(1) + 21(1) ? 60(1) + 43 =2 Point of contact is (1 ; 2) y ? 2 = ? 24( x ? 1) y = ? 24 x + 26 9. 3 f ? ( x) = ? 6 x + 42 x ? 60 f ( x) = ? 12 x + 42 2 OR y = ? 24 x + c 2 = ? 24(1) + c c = 26 y = ? 24 x + 26 y ? 2 = ? 24( x ? 1) OR y = ? 24 x + 26 f ( x ) = ? 12 x + 42 (5) 0 = ? 12 x + 42 x= 7 2 x= 7 2 (2) OR 2+5 x= 2 Please turn over Copyright reserved Mathematics/PI 25 NSC – MemorandumDBE/November 2011 2+5 2 7 x= 2 OR ? 21 x= 3(? 2 ) 7 = 2 x= QUESTION 10 y x= 7 2 (2) x= ? 21 3(? 2 ) 7 x= 2 (2) [14] ?4 0 1 x y = f /(x) 10. 1 x-value of turning point: ? 4 +1 x= 2 3 =? 2 3 ? 3 ? ?x ;gt; ? OR ? x ? ? ? ; ? ? 2 ? 2 ? f has a local minimum at x = ? 4 because: (1; y) f ? 4 xgt;? 3 ? 3 ? OR ? ? ; ? ? 2 ? 2 ? (1) 10. 2 x=†“4 graph (3) f 1 –4 OR f ( x) lt; 0 for x lt; ? 4 , so f is decreasing for x lt; ? 4 . f / ( x) gt; 0 for ? 4 lt; x lt; 1 , so f is increasing for ? 4 lt; x lt; 1 . / i. e. –4 OR Copyright reserved ? f has a local minimum at x = ? 4 x=–4 f / ( x) lt; 0 for x lt; ? 4 f / ( x) gt; 0 for ? lt; x 0 so graph is concave up at x = – 4, so f has a local minimum at x = – 4. x=–4 gradient negative for x lt; ? 4 gradient positive for ? 4 lt; x lt; 1 (3) f ? (? 4) = 0 f (? 4) gt; 0 x=–4 (3) [4] QUESTION 11 11. 1 11. 2 V (0) = 100 ? 4(0) = 100 litres Rate in – rate out = 5 – k l / min answer (1) 5–k –4 units stated once 5 ? k = ? 4 k=9 V ? (t ) = ? 4 l / min 11. 3 5 ? k = ? 4 k = 9 l / min OR (3) (2) Note: Answer only: award 2/2 marks Volume at any time t = initial volume + incoming total – outgoing total 100 + 5t ? kt = 100 ? 4t 5t ? kt = ? 4t 9t ? kt = 0 t (9 ? k ) = 0 At 1 minute from start, t = 1, 9 â€⠀œ k = 0, so k = 9 OR 00 + 5t ? kt = 100 ? 4t k=9 (2) dV = ? 4 , the volume of water in the tank is decreasing by 4 dt litres every minute. So k is greater than 5 by 4, that is, k = 9. Since k=9 (2) [6] Copyright reserved Please turn over Mathematics/PI 27 NSC – Memorandum DBE/November 2011 QUESTION 12 Note: If the wrong inequality 50x + 25y ? 500 is used, candidate wrongly says that there are more learners than available seats. Maximum of 10 marks. 12. 1 x, y ? N x + y ? 15 50 x + 25 y ? 500 y? 8 y ? ? x + 15 OR y ? ?2 x + 20 y? 8 Note: for the inequality’s marks to be awarded, the LHS and the RHS must be correctNote: If candidate gives 50 x + 25 y = 500 : max 5/6 marks x + y ? 15 y ? 8 50 x + 25 y ? 500 (6) 12. 2 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 y x + y ? 15 50 x + 25 y ? 500 y? 8 feasible region (4) Blue buses x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Red buses 12. 3 12. 4. 1 C = 600 x + 300 y (6 ; 8) ; (7 ; 6) ; (8 ; 4) ; (9 ; 2) and (10 ; 0) NOTE: The gradient of the search line is m = ? answer 2 1 (1) 3 marks for all correct solutions 2 marks if only 3 or 4 correct solutions 1 mark if only 1 or 2 correct solutions (3) subs answer (2) 2. 4. 2 12. 5 C = 6(600) + 8(300) = R 6 000 or C = 7(600) + 6(300) = R 6 000 or C = 8(600) + 4(300) = R 6 000 or C = 9(600) + 2(300) = R 6 000 or C = 10(600) + 0(300) = R 6 000 8 red ; 4 blue answer (1) [17] TOTAL: 150 Copyright reserved Please turn over Mathematics/P1 28 NSC – Memorandum DBE/November 2011 QUESTION 12. 2 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 y Blue Buses Red Buses 1 2 3 4 5 6 7 8 x 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Copyright reserved

Who you are what defines who you are Essay Example | Topics and Well Written Essays - 500 words

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Use of Force Policy in New Jersey Essay Example | Topics and Well Written Essays - 1000 words

Use of Force Policy in New Jersey - Essay Example It will then evaluate the policy by presenting the advantages and the disadvantages of the policy, and end by submitting some recommendations on the measures which can be used to improve the policy for the law enforcement officers. For the use of force in general, be it lethal or non-lethal, the policy states certain ground points that need to be followed by the police department. These four points include the use of force only by authorized and certified police officials, the use of force only when appropriate and required, the appropriate and reasonable use of force both in intensity and type, and the use of force only to attain lawful or legal ends (Use of force, 2000). Furthermore, the policy defines these lawful ends, in general, as any of the following four situations: when the officer is confronted with resistance directed at himself or others; when there is application of force against the officer or a third party, which could be another officer or an innocent bystander; when there is a threat of damage to property; or when the officer needs to make other legal ends possible, such as making an arrest (Use of force, 2000). For the use of deadly force, the parameters are further defined. Deadly force can be used only when the officer believes the criminal will cause death or serious bodily harm to the officer or a third party and when there are no other methods of stopping him, or in cases of an escape of such a criminal who has been charged with causing serious bodily harm or killing someone, and who the officer has reasonable belief will cause such harm if he succeeds in escaping (Use of force, 2000). Again, deadly force should only be used if there are no other options available and no innocent bystanders are in the danger of being hurt (Use of force, 2000). In 2009 (Hester, 2010), the Attorney General, on the basis of the recommendations and findings of his advisory committee on the use of force policy (Use of force, 2000), introduced the concept and the use of non-lethal weapons or force in police action in the state of New Jersey. This policy concerned mainly with the use of stun guns by police officers (Hester, 2009), although others measures of non-lethal force such as the use of rubber bullets, baton, foam, pressurized water, and other such devices does constitute non-lethal force (Capstick, 2001). There are two versions o f the policy; the original policy, issued in 2009, and a revised version, issued in 2010 (Hester, 2010). The revised version is currently in place (Hester, 2010). The use of non-lethal force does not replace the use of lethal force; it helps merely in reducing the number of incidents in which the police had to resort to lethal force in the absence of any alternative measures (Hester, 2009). Similarly, there are laws defining the use of such force just as there are for lethal force. The use of stun guns is the most common non-lethal force employed by the police (Hester, 2009). According to the old version of the policy, the police could only use this force against those deemed mentally ill or disturbed who were posing imminent danger either to themselves or to the public (Hester, 2010). Such people could be unarmed forcefully and arrested by the use of stun guns (Hester, 2010). The police had to take into account the opinion of an expert on the scene for such matters (Hester, 2010). Stun guns could not be used to forcefully make