Technical Mathematics Grade 11 P2 2023 Memo (Eng)

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 12 NOVEMBER 2023 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 16 pages./ Hierdie nasienriglyn bestaan uit 16 bladsye.

Downloaded from hlayiso.com 2 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/NOVEMBER 2023) NOTE: • Continuous accuracy (CA) applies only where indicated in this marking guideline. • Assuming values/answers in order to solve a problem is unacceptable. LET WEL: • Volgehoue akkuraatheid (CA) is slegs van toepassing soos aangedui in hierdie nasienriglyn. • Aanvaarding van waardes/antwoorde om ʼn probleem op te los, is onaanvaarbaar. MARKING CODES / NASIENKODES M Method / Metode A Accuracy / Akkuraatheid AO Answer only / Slegs antwoord CA Consistent accuracy / Deurlopende akkuraatheid F Formula / Formule I Identity / Identiteit R Rounding / Afronding S Simplification / Vereenvoudiging ST Statement / Bewering RE Reason / Rede ST RE Statement and correct reason / Bewering en korrekte rede SF Substitution correctly in correct formula / Korrekte vervanging in die korrekte formule NPU No penalty for omitting units / Geen penalisering vir eenhede uitgelaat Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 3 QUESTION/VRAAG 1 1.1 product / produk ✓ A (1) 1.2 𝑦2 − 𝑦1 ✓F A 𝑚𝑃𝑄 = 𝑥2 − 𝑥1 ✓ SF A 6−2 𝑚𝑃𝑄 = 2−(−1) 4 ✓ Ans. / Antw. CA 𝑚𝑃𝑄 = 3 (3) 1.3 𝑚𝑃𝑄 × 𝑚𝑃𝑅 = −1 ✓M A 4 6−𝑘 ✓ SF A × 2 − 6 = −1 3 4 6−𝑘 × −4 = −1 ✓ Simpl. / Vereenv. CA 3 6−𝑘 = −1 −3 6−𝑘 =3 −𝑘 = −3 𝑘=3 (3) 1.4 𝑀𝑄𝑅 ( 𝑥2 + 𝑥1 𝑦2 + 𝑦1 ; ) ✓F A 2 2 −1 + 6 2 + 3 ✓ SF A 𝑀𝑄𝑅 ( ; ) 2 2 5 5 𝑀𝑄𝑅 (2 ; 2) ✓ Ans. / Antw.CA (3) Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com 4 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/NOVEMBER 2023) 1.5 𝑀𝑄𝑅 = 𝑀𝑃𝑆 5 5 2 + 𝑥𝑆 6 + 𝑦 𝑆 ( ; )=( ; ) 2 2 2 2 5 2 + 𝑥𝑆 5 6+𝑦 ∴2= 2 and/en ∴ 2 = 2 𝑆 ✓2= 5 2 + 𝑥𝑆 CA 2 5 = 2 + 𝑥𝑆 5 = 6 + 𝑦𝑆 5 6 + 𝑦𝑆 ✓2= 2 CA 3 = 𝑥𝑆 −1 = 𝑦𝑆 ∴ 𝑆(3 ; −1) ✓ value of/waarde van x CA ✓ value of/waarde van y CA OR/OF 𝑃 → 𝑅 = 𝑃(2 ; 6) → 𝑅(6 ; 3) ∴ 𝑃 → 𝑅 = (𝑥 + 4 ; 𝑦 − 3) ✓M A ∴ 𝑄 → 𝑆 = 𝑄(−1 ; 2) → 𝑆(−1 + 4 ; 2 − 3) ✓S CA ∴ 𝑆(3 ; −1) ✓ value of/waarde van x CA ✓ value of/waarde van y CA (4) 1.6 𝑚𝑃𝑅 = − 𝟒 𝟑 𝑚𝑃𝑅 = − 𝟒 𝟑 ✓ Gradient/gradient 𝟑 𝟑 CA ∴ 𝑦 = −𝟒𝑥 + 𝑐 ∴ 𝑦 − 𝑦1 = − 𝟒 (𝑥 − 𝑥1 ) 𝟑 𝟑 (2 ; 6): 6 = − (2) + 𝑐 (2; 6): 𝑦 − 6 = − (𝑥 − 2) 𝟒 𝟒 𝟑 𝟑 𝟑 ✓ SF A ∴ 6 = −𝟐 + 𝑐 OR/OF ∴ 𝑦 − 6 = −𝟒𝑥 + 𝟐 10 𝟑 𝟑 ∴ 3 =𝑐 ∴ 𝑦 = −𝟒𝑥 + 𝟐 + 6 𝟑 𝟏𝟓 𝟑 𝟏𝟓 ✓ c-value / -waarde ∴ 𝑦 = −𝟒𝑥 + 𝟐 ∴ 𝑦 = −𝟒𝑥 + 𝟐 CA ✓ Ans. /Antw. CA (4) 1.7 𝑡𝑎𝑛𝜃 = 𝑚𝑃𝑅 3 𝑡𝑎𝑛𝜃 = −4 ✓ SF CA 3 ∴ 𝑟𝑒𝑓/𝑣𝑒𝑟𝑤. ∠ = 𝑡𝑎𝑛−1 (4) ≈ 36,87° ✓ ref. ∠ / verw. ∠ ∴ 𝜃 = 180° − 36,87° CA ∴ 𝜃 = 143,13° ✓ Ans. /Antw. CA (3) Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 5 1.8 RQ = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 ✓F A ✓ SF A = √(6 + 1)2 + (3 − 2)2 ✓ length/lengte PQ = √50 CA 5 ✓ cos ratio/verh. CA ∴ cos Q = √50 ✓ size of / grootte van ̂ ∴ Q = 45° Q CA OR/OF OR/OF PR = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 ✓F A ✓ SF A = √(6 − 2)2 + (3 − 6)2 ✓ length/lengte PR =5 CA ✓ tan ratio/verh. CA 5 ∴ tan 𝑄 = ✓ size of / grootte van 5 Q CA ∴̂ Q = 45° (5) [26] Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com 6 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/NOVEMBER 2023) QUESTION/VRAAG 2 2.1.1 tan(30,5˚ + 130,5˚) ≈ −0,34 ✓ SF A ✓ −0,34 A AO: Full marks / Volpunte (2) 2.1.2 cosec(130,5° − 30,5°) ✓ SF A 1 = sin(100°) ✓ reciprocal ratio /resiprook verh. A ≈ 1,02 ✓ 1,02 A AO: Full marks / Volpunte (3) 2.2.1 54° (1)2 = 𝑘 2 + 𝑥 2 ✓ Pythagoras A 1 1 − 𝑘2 = 𝑥2 𝑘 ✓ Diagram A ±√1 − 𝑘 2 = 𝑥 36° ∴ 𝑥 = √1 − 𝑘 2 √1 − 𝑘 2 ✓𝑥 = √1 − 𝑘2 A ∴ cos36° = √1 − k2 ✓ √1 − 𝑘 2 CA (4) 2.2.2 sin(216 °) = 𝑠𝑖𝑛(180° + 36°) ✓ Reduction/reduksie sin(216 °) = −𝑠𝑖𝑛36° A sin(216 °) = −𝑘 ✓Ans. / Antw. CA (2) 2.3 𝑡𝑎𝑛𝜃 = 2𝑠𝑖𝑛38,1° 𝑡𝑎𝑛𝜃 = 1,234 … ✓ 1,234 … A 𝑟𝑒𝑓/𝑣𝑒𝑟𝑤. ∠ ≈ 50,98° ✓ ref/verw. ∠ A ∴ 𝑄1: 𝜃 = 50,98° ✓ Quadrants/kwadrante AND/EN A ✓ Both answers / beide ∴ 𝑄3: 𝜃 = 180° + 50,98° = 230,98° antwoorde CA (4) [15] Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 7 QUESTION/VRAAG 3 3.1 1 ✓ 𝑐𝑜𝑠𝜃 A cos(360° − 𝜃). . tan (360° + 𝜃) cot(180° + 𝜃) 1 ✓ 𝑐𝑜𝑡𝜃 A cos(180° + 𝜃). 𝑡𝑎𝑛(180° − 𝜃) ✓ 𝑡𝑎𝑛𝜃 A 1 𝑐𝑜𝑠𝜃. 𝑐𝑜𝑡𝜃 . 𝑡𝑎𝑛𝜃 ✓ −𝑐𝑜𝑠𝜃 A = ✓ −𝑡𝑎𝑛𝜃 A −cos𝜃. −𝑡𝑎𝑛𝜃 1 ✓ 𝑡𝑎𝑛𝜃 CA = cotθ = 𝑡𝑎𝑛𝜃 (6) 3.2 1 2 1+ 𝑠𝑖𝑛𝑥 (𝑡𝑎𝑛𝑥 + ) = 𝑐𝑜𝑠𝑥 1−𝑠𝑖𝑛𝑥 1 2 ∴ 𝐿𝐻𝑆 = (𝑡𝑎𝑛𝑥 + ) 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 ✓ 𝑐𝑜𝑠𝑥 A 𝑠𝑖𝑛𝑥 1 2 ∴ 𝐿𝐻𝑆 = (𝑐𝑜𝑠𝑥 + 𝑐𝑜𝑠𝑥) 𝑠𝑖𝑛𝑥+1 2 (𝒔𝒊𝒏𝒙+𝟏)𝟐 ∴ 𝐿𝐻𝑆 = ( 𝑐𝑜𝑠𝑥 ) ✓ CA 𝒄𝒐𝒔𝟐 𝒙 (𝑠𝑖𝑛𝑥 + 1)2 ∴ 𝐿𝐻𝑆 = 𝑐𝑜𝑠2 𝑥 ✓ 1 − 𝑠𝑖𝑛2 𝑥 A (𝑠𝑖𝑛𝑥 + 1)(𝑠𝑖𝑛𝑥 + 1) ∴ 𝐿𝐻𝑆 = 1−𝑠𝑖𝑛2 𝑥 ✓ LHS = RHS (𝑠𝑖𝑛𝑥 + 1)(𝑠𝑖𝑛𝑥 + 1) ∴ 𝐿𝐻𝑆 = (1−𝑠𝑖𝑛𝑥)(1 + 𝑠𝑖𝑛𝑥) 1+𝑠𝑖𝑛𝑥 ∴ 𝐿𝐻𝑆 = = 𝑅𝐻𝑆 1−𝑠𝑖𝑛𝑥 (4) [10] Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com 8 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/NOVEMBER 2023) QUESTION/VRAAG 4 4.1 𝒇(𝒙): ✓ both x-int. / albei x-afsnitte A ✓ y-int. / y-afsnit ✓ 𝑔(𝑥) Shape / vorm A 𝒈(𝒙): 𝑓(𝑥) ✓ both x-int. / albei x-afsnitte A ✓ y-int. / y-afsnit A ✓ Shape / vorm A ✓ Turning points / draaipunte A (7) 4.2 𝑦 ∈ [0 ; 2 ] ✓ Notation / notasie A OR/OF ✓ start- and endpoints / 0≤𝑦≤2 begin- en eindpunte CA (2) 4.3 360 ✓Ans. / Antw. A (1) 4.4 90° ≤ 𝑥 ≤ 270° ✓ 90° ≤ CA ✓ ≤ 270° CA (2) [12] Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 9 QUESTION/VRAAG 5 C B 120° 1 600 m 40° A 2 400 m D 5.1 BĈA = 180° − 120° − 40° ✓ ST ✓ RE BĈA = 20° (int. ∠of ∆ / binne ∠ van ∆) (2) 5.2 AC AB = ̂ sinABC sinBĈA ✓ SF CA AC 1600 = ✓S CA sin120° sin20° 1600 ✓ Rounded Ans. / AC = × sin120° sin20° Afgeronde Antw. CA AC = 4 051 m NPU (3) 5.3 1 Area ∆ABC = 2 . a. b. sinC OR/OF 1 ✓F A Area ∆ABC = . AB. AC. sinBA ̂C 2 1 Area ∆ABC = 2 (1600)(4051)sin40° ✓ SF CA Area ∆ABC = 2 083 146,085 m2 ✓Ans. / Antw. A ̂ D = 50° (compl. ∠’s / kompl. ∠’e) CA 1 Area ∆ABC = 2 . a. b. sinC OR /OF 1 Area ∆ABC = . AC. AD. sinCA ̂D 2 1 Area ∆ABC = 2 (2400)(4051)sin50° Area ∆ABC = 3 723 895,247 m2 ✓Ans. / Antw. A ∴ Total Area = 2 083 146,085 + 3 723 895,247 ✓M A ∴ Totale Opperv. = 5 807 041,33 m2 ✓ Ans. / Antw. A NPU (6) [11] Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com 10 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/NOVEMBER 2023) QUESTION/VRAAG 6 6.1 Double or Twice/Dubbel of Twee keer ✓Ans. / Antw. (1) 6.2 A 48° O 𝑥 1 2 𝑦 1 C 2 B 6.2.1 𝑥 = 48° × 2 = 96° ✓ ST ✓ RE (∠ at centre = 2 × ∠ at circumf. / middelpts ∠ = 2 × omtreks ∠) (2) 6.2.2 OB = OC (radii) Ĉ2 = ̂ B2 = 𝑦 (∠’s opp = sides / ∠’e teenoor = sye) ✓ ST ✓RE 180°−96° ∴𝑦= = 42° (int. ∠of ∆ / binne ∠ van ∆) 2 (2) 6.3 Equal / gelyk ✓Ans. / Antw. (1) 6.4 S 𝑧 𝑥 R O 120° 1 𝑦 T Q P 6.4.1 𝑥 = 120° (ext. ∠ of cq / buite ∠ van kvh) ✓ ST ✓ RE (2) 6.4.2 𝑦 = 90° (∠ in semi-circle / ∠ in semi-sirkel) ✓ ST ✓ RE (2) 6.4.3 𝑧 = 30° (ext. ∠ of ∆/ buite ∠ van ∆) ✓ ST ✓ RE (2) [12] Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 11 QUESTION/VRAAG 7 7.1 equal / gelyk ✓Ans. / Antw. (1) 7.2 7.2.1 𝐶̂4 = 25° (tan-chord / raaklyn koord) ✓ ST ✓ RE (2) 7.2.2 𝐶̂3 = 42° (∠’s in same seg. / ∠’e in dieselfde seg.) ✓ ST ✓ RE (2) 7.2.3 𝐵𝐶̂ 𝐷 = 90° (∠ in semi-circle / sirkel) ✓ ST ✓ RE ̂1 = 65° (int. ∠of ∆ / binne ∠ van ∆) 𝐷 ✓ ST ✓ RE (4) 7.3 outside / buite ✓Ans. / Antw. (1) 7.4 S P K 1 4 23 2 3 M 4 O1 T 2 100° 1 3 10° R 7.4.1 OQ = OR (radii) 𝑆̂3 = 10° (∠’s opp = sides / ∠’e teenoor = sye) ✓ ST ✓RE ∴ 𝑆̂2 = 40° ✓ ST (∠ at centre = 2 × ∠ at circumf. / middelpts ∠ = 2 × omtreks ∠) ✓ RE (4) 7.4.2 𝑆̂4 + 𝑆̂3 = 90° ✓ ST ∴ 𝑆̂4 = 80° (tan ⊥radius) ✓ RE (2) 7.4.3 𝑅̂1 = 80° tan from same pt. / 𝑟𝑎𝑎𝑘𝑙𝑦𝑛 𝑢𝑖𝑡 𝑑𝑖𝑒𝑠𝑒𝑙𝑓𝑑𝑒 𝑝𝑡.) ✓ ST ✓ RE 𝑃̂ = 20° (int. ∠of ∆ / binne ∠ van ∆) ✓ ST ✓ RE (4) [20] Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com 12 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/NOVEMBER 2023) QUESTION/VRAAG 8 8.1 Perpendicular / loodreg ✓Ans. / Antw. (1) 8.2 B E 8 D 𝑥 12 O A 8.2.1 𝑂𝐸 = 𝑥 + 8 ✓ Ans. / Antw. (1) 8.2.2 𝑂𝐴2 = 𝑥 2 + (12)2 (Pyth) ✓ ST ✓RE 𝑂𝐴2 = 𝑥 2 + 144 𝑂𝐴 = ±√𝑥 2 + 144 ✓ Answer / Antw.CA ∴ 𝑂𝐴 = √𝑥 2 + 144 (3) 8.2.3 𝑥 + 8 = √𝑥 2 + 144 ✓M A (𝑥 + 8)2 = 𝑥 2 + 144 ✓S A 𝑥 2 + 16𝑥 + 64 = 𝑥 2 + 144 16𝑥 = 80 ✓S CA 𝑥=5 ✓Ans. / Antw. CA (4) 8.2.4 𝑟 = 13 ✓Ans. / Antw. (1) [10] Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 13 QUESTION/VRAAG 9 9.1 𝜔 = 2𝜋𝑛 ✓F A 𝜔 = 2𝜋(12) ✓ SF CA 𝜔 = 24𝜋 ✓Answer/Antw.CA 𝜔 ≈ 75,40 rad. / sec NPU (3) 9.2 𝐷 = 80 mm ✓diameter/middellyn 𝑉 = 𝜋𝐷𝑛 A 𝑉 = 𝜋(80)(20) ✓F A 𝑉 = 1 600𝜋 ✓ SF CA 𝑉 ≈ 5 026,55 mm/min ✓Answer/Antw. CA NPU (4) 9.3 500 mm = 50 cm ✓conv. /herleid. A 4ℎ2 − 4𝑑ℎ + 𝑥 2 = 0 ✓F A 4ℎ2 − 4(56,6)ℎ + (50)2 = 0 ✓ SF CA 4ℎ2 − 226,4ℎ + 2500 = 0 ÷ 4: ℎ2 − 56,6ℎ + 625 = 0 −𝑏 ± √𝑏 2 − 4𝑎𝑐 ℎ= 2𝑎 ✓ SF CA −(−56,6) ± √(−56,6)2 − 4(1)(625) ℎ= 2(1) ✓ Both answers/ ∴ ℎ ≈ 41,56 or ℎ ≈ 15,04 beide antwoorde CA ∴ ℎ = 15,04 cm ✓ Answer/Antw. CA NPU (6) Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com 14 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/NOVEMBER 2023) 9.4 2m 𝜃 2,5 m 9.4.1 𝑠 = 𝑟𝜃 ✓F A 2,5 = 2𝜃 ✓ SF CA 1,25 rad = 𝜃 ✓Answer/Antw.CA (3) 9.4.2 Area = r2 θ ✓F A 2 ✓ SF CA (2)2 (1,25) Area = 2 ✓Ans./Antw. CA Area = 2,5 cm2 NPU (3) OR/OF rs Area = 2 ✓F A (2)(2,5) ✓ SF CA Area = 2 ✓Ans./Antw. CA Area = 2,5 cm2 NPU (3) 9.4.3 𝐶 = 2𝜋𝑟 ✓F A 2,5 = 2𝜋𝑟 ✓ SF A 0,40 = 𝑟 ✓Answer/Antw.CA (2)2 = ℎ2 + (0,40)2 ✓M A (2)2 − (0,40)2 = ℎ2 3,84 = ℎ2 1,96 m = ℎ ✓Answer/Antw.CA NPU (5) 9.5 VolShell = Volouter − Volinner ✓M A 4 4 VolShell = 3 π(5,5)3 − 3 π(3,5)3 ✓ SF A 1331 343 VolShell = π− 6 π ✓Answer/Antw.CA 6 494 3 VolShell = 3 π cm 494 ✓M A ∴ Mass/𝐺𝑒𝑤𝑖𝑔 = 3 π × 30 ∴ Mass/𝐺𝑒𝑤𝑖𝑔 = 4940π ∴ Mass/𝐺𝑒𝑤𝑖𝑔 = 15519,47 grams ✓Ans./Antw. CA NPU (5) [29] Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 15 QUESTION/VRAAG 10 12 cm 𝑥 cm 13 cm 11 cm 11,8 cm 18 cm 25 cm 𝑂 +𝑂 𝐴𝑟𝑒𝑎 = 𝑎 ( 1 2 𝑛 + 𝑂2 + 𝑂3 +. . . +𝑂𝑛−1 ) ✓F A 25 12+18 ✓𝑎=5 A 329 = 5 ( 2 + 𝑥 + 13 + 11 + 11,8) ✓ SF CA 329 = 5(𝑥 + 50,8) ✓S CA 65,8 = 𝑥 + 50,8 ✓ 𝑥 = 15 CA 15 = 𝑥 OR / OF OR / OF 𝐴𝑟𝑒𝑎 = 𝑎(𝑚1 + 𝑚2 + 𝑚3 +. . . +𝑚𝑛−1 ) ✓F A 25 12 + 𝑥 𝑥 + 13 13 + 11 11 + 11,8 11,8 + 18 ✓𝑎=5 A 329 = 5 ( + + + + ) ✓ SF CA 2 2 2 2 2 25 12 + 𝑥 𝑥 + 13 𝑎= ✓S CA 329 = 5 ( + + 12 + 11,4 + 14,9) 5 2 2 𝑎=5 ✓ 𝑥 = 15 CA 25 + 2𝑥 329 = 5 ( + 38,3) 2 25 + 2𝑥 65,8 = + 38,3 2 25 + 2𝑥 27,5 = 2 55 = 25 + 2𝑥 30 = 2𝑥 15 = 𝑥 (5) [5] TOTAL/TOTAAL: 150 Copyright reserved /Kopiereg voorbehou Please turn over / Blaai om asseblief