Grade 12 Answers

Grade Twelve Answers

1)
Answer:
T= 10 years Therefore T/4 = Maxima and (3T)/4 = Minima 3(10)/4 = 7.5 years

2)
Answer:
m1 = 1 m2 = -1 y = x is perpendicular to y = -x+5 because m1 = -1/m2 y = x intersects y = -x+5 x + x = 5 x = 2.5 Check: y =x y = -x+5 y = 2.5 y = -2.5 + 5 y = 2.5 Use length formula: a = [(x1-x2)² + (y1-y2)²]^1/2 where (x1-x2) = 2.5-0 = 2.5 (y1-y2) = 2.5-0 = 2.5 a = [2(2.5)²]^1/2 a = (12.50)^1/2 b = [(x1-x2)² + (y1-y2)²]^1/2 where (x1-x2) = 5-2.5 = 2.5 (y1-y2) = 0-2.5 = -2.5 b = [2(2.5)²]^1/2 b = (12.50)^1/2 AREA: 0.5*b*h = 0.5*a*b = 0.5*[(12.50)^1/2]* [(12.50)^1/2] = 6.25

3)
Answer:
a) circumference = 2Pi*r arc length = (x/360 degrees) * 2Pi*r 20 = (72/360)*2Pi*r r = 15.915 cm b) circumference = 2Pi*r base of cone = circumference 20 cm = 2Pi*r r = 10/Pi r = 3.18 cm

4)
Answer:
xy = (x-3)*(y+2) xy = (x+3)(y-1) xy = xy-3y+2x-6 xy = xy+3y-x-3 3y = 2x-6 3y = x+3 -3y = -x-3 0 = x-9 x = 9 ------------> 3y = (9)+3 12/3 = y 4 = y Therefore, area of (i) = x*y = (9)*(4) = 36 units² (ii) = (x-3)*(y+2) = (9-3)*(4+2) = 36 units² (iii) = (x+3)*(y-1) = (9+3)*(4-1) = 36 units²

5)
Answer:
f(x) = 5 sin x = 0 g(x) = 5 cos x = 0 x = sin^-1(0) = 0, Pi, 2Pi x = cos^-1(0) = Pi/2, Pi, 3Pi/2 (0,0) B= (Pi,0) (2Pi,0) A= (Pi/2,0) (3Pi/2, 0) Maxima of f(x) = 5 sin x = 5 1 = sin x x = sin^-1 (1) x = Pi/2 C = (Pi/2, 5) Therefore, A = (Pi/2,0) B = (Pi,0) C = (Pi/2,5) AREA = 0.5(length AB)(length AC) NOTE: Use length formula = 0.5[(Pi-Pi/2)²]^1/2[(5-0)²]^1/2 = 0.5[(Pi/2)²]^1/2[(5²)]^1/2 = 0.5*Pi/2*5 = 5Pi/4 AB is perpendicular to AC AB: (y1-y2)/(x1-x2) = AC: -(x1-x2)/(y1-y2) 0/(Pi-Pi/2) = -(Pi/2-Pi/2)/(5-0) 0 = 0 Therefore, slope AB is negative reciprocal of slope AC

6)
Answer:
Let x = side opposite the barn wall (<= 100m) x + 2y = 160 y = (160 - x)/2 y = 80 - x/2 Area = xy = x(80-x/2) = 80x-(x²/2) Area’ = 80-x = 0 x = 80 y = 80 - 80/2 = 80-40 = 40 Maximum area is (x)(y) = (80)(40) = 3200

7)
Answer:
y = ax²+bx+c -b/(2a) = 2 0 = a(2)²-4a(2)+c <------- b = -4a 0 = 4a-8a+c 4a = c y = ax²-4ax+4a 1 = a(3)2-4a(3) +4a 1 = 9a-12a+4a 1=a Therefore, c = 4(1) = 4 b = -4(1) = -4 y = x²-4x+4 t = 9-4(-3)+4 t = 9+12+4 t = 25

8)
Answer:
Step (i) b = AC = BC because triangle ABC is isosceles (ii) sin 70 degrees = h/4 h = 3.759 Area = 0.5*b*h = 0.5*(4)(3.759) = 7.518 cm²

9)
Answer:
BC/ED = AB/AE 4/ED = 3/1 3 ED = 4 length ED = 4/3

10)
Answer:
After Quake: v = 50(1/2)(3/4)(30)(7) v = 3937.5 m³ Before Quake: v = (50m)(30m)(d) = 3937.5 m³ where d = depth Therefore, d = 2.625 m

11)
Answer:
100 - 3t = 88 100 - 88 = 3t t = 12/3 t = 4 km/h reduction in speed per person When six persons are on board, the van travels at 100 - 6t = 100 - 6(4) = 76 km/h

12)
Answer:
0.5*b*h = 11 b = length AC b*h = 22 h = length CB b = 22/h c = length AB c² = b² + h² 100 = b² + h² 100 = (22/h)² + h² 100 = 22²/h² +h² 100h² = 22² + h,sup>4 h⁴-100h²+284 = 0 To solve, use quadratic formula: h² = 100+[(100)²-4(484)] = 9.74 2 h² = 100- [(100)²-4(484)] = 2.258 2 Since b = 22/h, then b = 9.74 <------ Therefore, use h= 2.258

13)
Answer:
Let V= volume of water in pond Let P = volume drained by one large pipe in one hour Let p = volume drianed by one small pipe in one hour Let t = time V - 9P*8 = 0 V-6*p*(16) = 0 +(-V+6p*16 = 0) V = 6*p*16 -9P*8+6p*16 = 0 9P*8 = 6p*16 V-3(P)(t)-5(p)(t) = 0 P = 1.33p --------------> 6p*16-3(1.33p)(t)-5(p)(t) = 0 96p-4pt-5pt = 0 96p = 9pt t = (96p)/(9p) t = 10.667 hrs = 10 hrs 40 min

14)
Answer:
Let x = rate of car 1 (m/s) Let y = rate of car 2 (m/s) (1) 1800m = 30x+30y 1800-30x = 30y 7200-120x = 120y (2) 1800 + 120x = 120y Solve for Car 1: (1)=(2) 7200-120x = 1800+120x 5400 = 240x x = 22.5 m/s Convert to km/hr x = 22.5m/sec *60sec/min*60min/hr = 81 km/hr 1000 m/km Solve for Car 2: Substitute x = 22.5 m/s 1800+120x = 120y y = 1800+120x 120 y = 1800+120(22.5) 120 y = 37.5 m/s Convert to km/hr x =37.5m/sec *60sec/min*60min/hr = 135 km/hr 1000 m/km

15)
Answer:
Area A = Pi * r1² = Pi*1²=Pi r1 = 1 circumference = 2 pi r = 2 pi A C is the center of circle B it is equidistant from points P and Q. Thales says that delta PCQ will be a right isosceles triangle as PQ = diameter of A Find length PC = r of B r2 = PC = PQ (right isosale triangle) r1 = 1 2 * r2 ² = (2r1)² 2 * r2 ² = 4 r1 ² r2 ² = 2 (1) ² r2 = (2)^1/2 Therefore: Area shaded = 1/2 Area A - (90 degrees/360degrees * Pi * r2² - 1/2 r2 ²) = 1/2 pi - (0.25 * pi * (2) - 0.5 (2)) = 0.5pi - 0.5pi + 1 = 1

16)
Answer:
y = ax ² + bx + c -4 = c [1 - (17)^1/2] / 2 = a[(1-(17)^1/2) / 2]^1/2 + b(1-(17)^1/2) / 2 [1 - (17)^1/2] / 2 = a[(1- 2(17)^1/2 + 17)] / 4 + b(1-(17)^1/2) / 2 1/2 - [(17)^1/2 / 2] + 4 = a[(1- 2(17)^1/2 + 17) / 4] + b[(1- (17)^1/2 / 2)] 2 - 2(17)^1/2 + 16 = a[(1 - 2(17)^1/2 + 17) / 4] + 2b (1 - (17)^1/2) [1 + (17)^1/2 / 2] = a[(1 + 2(17)^1/2 + 17)] / 4 + b[(1 + (17)^1/2) / 2] - 4 4 [(1 + (17)^1/2 / 2) + 4] = a (1+ 2(17)^1/2 +17) + 2b (1+(17)^1/2) 2 + 2 (17)^1/2 + 16 = a (1 + 2(17)^1/2 + 17 ) + 2b (1 + (17)^1/2) 18 = a (1+2(17)^1/2 +17) + 2b(1 + (17)^1/2) 18 = a (1-2(17)^1/2 +17) + 2b(1 - (17)^1/2) ----------------------------------------------- 0 = a(4(17)^1/2) + 2b(2(17)^1/2) - 4(17)^1/2 4(17)^1/2 - 4a(17)^1/2) = 4b(17)^1/2 1 - a = b y = ax² + (1 - a)x - 4 1 + ((17)^1/2 / 2) = a[(1+(17)^1/2) / 2]² + (1 - a)(1 + ((17)^1/2 / 2 )) -4 1 + ((17)^1/2 / 2) + 4 - (1 + ((17)^1/2 / 2)) = a[(1 + 2(17)^1/2 +17 - (1 + (17)^1/2)2) / 4] y = x² - 4 1 - a = b a = 1 b = 0 4 = a(-1 + 17)/4 c = 4

17)
Answer:
y = ax² + bx + c =0 a = 5 b = d -3 b² - 4ac =0 c = (-9d² + 15d + 30) (d - 3)² - 4(5)(-9d² + 15d + 30) = 0 d² - 6d + 9 + 20(9)d² - 20(30) = 0 181d² - 306d - 591 = 0 d = 306 +/- [(306)² - 4(181)(-591)]^1/2 ----------------------------------- 2(181) = 2.84, -1.15

18)
Answer:
x + 1 = 0 4 10 k b -2 x = -1 + -4 -6 6 - k -b - 8 + k ------------------------------------------------- 4 6 k - 6 b + 6 - k -b - 8 + k x - 1 = 0 4 10 k b -2 x = 1 + 4 14 k + 14 b + k + 14 -------------------------------------------------- 4 14 k + 14 b + k + 14 b + k + 12 -b - 8 + k = -3 b + k + 12 = 25 -b + k = 5 b + k = 13 b + k = 13 b = 13.9 2k = 18 = 4 k = 9 f(x) = 4x⁴ + 10x³ + 9x² + 4x - 2 x + 2 = 0 4 10 9 4 -2 x = -2 + -8 -4 -10 12 ------------------------------------- 4 2 5 -6 10 R = 10

19)
Answer:
m² 5 k² 5mk 6mk -m² m² -5 5 - k² - m² k² + m² - 5 - 5mk ----------------------------------------------------------------------- m² (5 - m²) k² + m² - 5 5mk + 5 - k² - m² R = 6mk - k2 + m2 - 5 - 5mk 32 = mk + k2 + m2 - 5 6mk = 72 37 = m2 + mk + k2 m = 72 / 64 37 = (72 / 6k)² + (72 / 6k )k + k² 37 = (144 / k²) + (72 / 6) + k² 25k² = 144+ k 4 144 - 25k² + k⁴ k² = 25 +/- ((25² - (144))^1/2 -------------------------- 2 = 25 +/- 7 = 18 , 32 -------- --- --- 2 2 2 if k (+) m(+) k² = 9 , 16 (sign dependent) k = +/- 3 , +/- 4 try m=3 k=4 therefore: k = +/- 3, +/- 4 k = +/- 4 m = +/- 4 9x⁴ + 5x³ + 16x² + 60 + 72 x + 1 = 0 9 5 16 60 72 x = -1 -9 4 -20 -40 -------------------- 9 -4 20 40 32 m = 4 k = 3 --------------> 16 5 9 60 72 -16 11 -20 -40 -------------------- 16 -11 20 40 32 therefore: k = 3 m = 4 k = 4 m = 3 k = -3 m = -4 k = -4 k = -3

20)
Answer:
P(5) = 30 + 6 log(5 + 2) -------------- log 2 = 46.84 million dollars 80 = 30 + 6 log2 (x + 2) 50 / 6 = log2 (x + 2) 2^50/6 = (x + 2) x = 320.54 million dollars. therefore: increase in spending = 320.54 - 5 = 215.54 million

21)
Answer:
Slope OP = delta y = (3)^1/2 / 2 - 0 = (3)^1/2 ------- --------------- delta x 1/2 - 0 Therefore: m PR = delta y -1 -1 ------- = --- = ---- delta x mOP 3^1/2 slope OQ = delta y -3^1/2 - 0 -(3)^1/2 ------- = ---------- = delta x 1/2 - 0 Therefore: mQR -1 -1 = ------- = ------ -(3)^1/2 (3) y1 = -1 / (3)^1/2 x + 2 / (3)^1/2 y2 = 1 / (3)^1/2 x - 2 / (3)^1/2 PR: y1 = -1x + b ------- -(3)^1/2 (3)^1/2 / 2 = -1 / (3)^1/2 (1 / 2) + b (3)^1/2 / 2 + 1/2 (3)^1/2 = b 1/(3)^1/2x - 2/(3)^1/2 3/2(3)^1/2 + 1/2(3)^1/2 = b 2/(3)^1/2 = b y1 = -1x / (3)^1/2 + 2/(3)^1/2 y1 = y2 -1x / (3)^1/2 + 2/(3)^1/2 = 1x / (3)^1/2 - 2/(3)^1/2 4 / (3)^1/2 = 2x / (3^1/2) x = 2 length OR = 2-0 = 2 delta POQ = cos^-1 (1/2) + cos^-1 (1/2) = 120 degrees QR: y2 = 1x / (3)^1/2 + b -(3)^1/2 / 2 - 1 / 2(3)^1/2 = b -4 / 2(3)^1/2 = b -2 / (3)^1/2 = b

22)
Answer:
1. 2x + y + z = 155 2. x + 2y + 2z = 235 2x +2y - 2z = 0 3. 3x+ 4y + 0z = 235 z = x+y 3x + 4y = 235 3x = 253 - 4y 3x = 235 - 4(40) x = 25 z = x + y = 25 + 40 2x + y (x + y) = 155 z = 65 3x + 2y = 155 (235 - 4y) + 2y = 155 235 - 155 = 2y 80 = 2y y = 40

23)
Answer:
A = Pi r² r = x tan 30 degrees = x / 3^1/2 = Pi x² / 3 dx / dt = 3 m/s dA / dt = dA / dx * dx / dt = (2Pi * x / 3) * 3m/s = 2Pi * x m/s at x = 6 dA/dt = 2 Pi (6) m²/s = 12 Pi m²/s

24)
Answer:
find points of intersection 1/4 x² = 2 - 1/4 x² 1/2 x² = 2 x² = 4 x = +/- 2 y = (1/4)(4) y = 1 (-2, 1) and (2, 1) extreme values x (0, 0) and (0, 2) extreme y A = 2x (y,sup>2 (x) - y1 (x)) =2x (2-(1/4 x²) - 1/4 x²) = 4x - 2x (1/2 x²) = 4x - x³ to find max A tangent line = 0 when dA/dx = 4 - 3x² x is an element of (-2, 2) y is an element of (0, 2) 0 = 4 - 3x² A = 4((4/3)^1/2) - ((4/3)^1/2)³ 4 = 3x² = 3.08 x² = 4/3 x = +/-(4/3)^1/2 = 1.155 use (+) root

25)
Answer:
y1 = y2 1/4 x² = 2 - 1/4 x² x = 2 y1 = 1/4 (2)² 1/2 x² = 2 y1 = 1 x² = 4 x = +/- 2 (2, 1) = A find tangents at (2, 1) m1 = y1 = 1/2x y2 = -1/2 x = 1/2 (2) m2 = -1 m1 = 1 y = m1x + b y = m2x + b = x + b = -x + b 1 = 2 + b 1 = -2 + b -1 = b 3 = b y1 = x - 1 y2 = -x + 3 y1 is perpendicular to y2 therefore: curve is 90 degrees

26)
Answer:
P(0) = 2572.55 = 2000 + 600 log (a) 572.55 / 600 = log (a) 10^0.95425 = a a = 9.000 P(7) = 2000 + 600 log (7 + 9) = 2722.47 really 2722 bacteria

27)
Answer:
l = (64² + 128y + y² + 12288)^1/2 = ( 16384 + 128y + y²)^1/2 = y + 128 l = ((-64 - y)² + (110.85)²)^1/2 cos 30 degrees = x / 128 x = 110.85 sin 30 degrees = y / 128 y = 64 y / 80 + y + 128 = 4 hours -------- 80 2y + 128 = 320 2y = 192 y = 96 km that can be safely traveled.

28)
Answer:
know two points (-3 + (19)^1/2, 0) (-3 - (19)^1/2, 0) ax² + 3x - 5 = 0 x = -3 + (19)^1/2 a (9 - 6(19)^1/2 + 19) + 3(-3 + (19)^1/2) -5 = 0 a (28 - 6(19)^1/2) - 9 + 3(19)^1/2 - 5 = 0 a (28 - 6(19)^1/2) = 14 - 3(19)^1/2 a = 1/2 1/2 x² + 3x - 5 = y y = -1/2 x - 6 y = y 1/2 x² + 3x - 5 = - 1/2 x - 6 x² + 7x + 2 = 0 x = -7 +/- (41)^1/2 --------------- 2 y = - 1/2 (-7 + (41)^1/2) / 2 - 6 third (-7 + (41)^1/2 / 2, -5.85) y = -5.85 y = -1/2 (-7 -(41)^.5) / 2 -6 third ((-7 - (41)^.5) / 2, -2.65) y= -2.65 therefore: a = 1/2

29)
Answer:
-1/3 x² + 4x - 2 >= (1/4)x² - 1 - x² + 12x - 6 >= 3/4 x² - 3 -4x² + 48x - 24 >= 3x² - 12 0 >= 7x² - 48x + 12 x = 48 +/- ((48)² - 4(7)(12))^1/2 ------------------------------ 14 = 6.597, 0.260 use test values on graph x = 6 -1/3 (6) squared + 4(6) - 2 >= (1/4)(6) squared - 1 10 >= 8 true x=7 -1/3 (7) squared + 4(7) - 2 >= (1/4)(7) squared -1 9 2/3 is not greater or equal to 11.25 x >= 6.597 x = 0 -2 is not greater or equal to -1 x = 1 -1/3 + 4 - 2 >= (1/4) (1) -1 x >= 0.260 1 2/3 >= -3/4 true x is an element of [0.260, 6.597]

30)
Answer:
We know * F to be 90 degrees (radius perpendicular to tangent) * CED 90 degrees delta DGF is similar to delta DCE by AAA DG/DC = GF/CE = D/DEF (DE)²2 + (EC)² = (DC)² FG/5 = 6(5)/DE (DE)² = (DC)² - (EC)² = (25)² - (5)² = 600 FG = (30)(5) / 10(6^1/2) DE = 10 (6)^1/2. = 15/(6)^1/2 FG = 15 (6)^1/2 / 6

31)
Answer:
mg = T1 sin theta 1, + sin theta 2 100N = T1 sin 20š, + T2 sin 50š T1 cos 20š = T2 cos 50š T1 = T2 cos 50š / cos 20š 100 N = T2 cos 50š * sin 20š + T2 sin 50š / cos 20š 100 N = T2 (cos 50š sin 20š + sin 50š cos 20š) / cos 20š 100N (cos (20š)) / sin 70š = T2 T2 = 100 N T1 = 100 cos 50š/ cos 20š= 68.4N

32)
Answer:
1 b 59 c 60 -1 1-b b-60 60-c-b ------------------------------------ 1 b-1 60-b c+b-60 120-c-b x² + 4x +3 = 0 R = 0 = 120 - c - b (x + 1)(x + 3) = 0 c + b = 120 x + 1 = 0 x + 3 = 0 1 b 59 c 60 -3 -3b-9 9b-204 -3(c + 96 - 204) -------------------------------------------- 1 (b - 3) 68 - 3b (c + 9b - 204) R = 0 = 60 - 3(c - 27b + 612) = 60 - 3c - 27b + 612 = 672 - 3c - 27b c = 672 - 27b / 3 (672 - 27b + b) / 3= 120 672 - 27b + 3b = 360 672 - 360 = 24b b = 13 c + b = 120 c = 120 - b c = 107 check: 1 13 59 107 60 1 13 59 107 60 -1 -12 -47 -60 -3 -30 -87 -60 ------------------------- -------------------------- 1 10 29 20 0 1 12 47 60 0

33)
Answer:
y = 5/2x + 3 x = 1 (1, 1) y = mx - 3 = 5/5 1 = m - 3 = 1 4 = m y = 4x - 3 5/(2x + 3) = 4x - 3 5 = (4x - 3)(2x + 3) 5 = 8x² + 6x - 9 0 = 8x² + 6x - 14 x = -6 +/- (36 + 4 (8)(4))^1/2 --------------------------- 16 = 1 , - 1.75 Therefore: a = -1.75

34)
Answer:
t0 = 2s for speeder ds = 40 m/s to + 40 m/s t dp = 3.0 m/s² (t)² ds = dp 40 (2) + 40t = 3.0t² 3.0t² - 40t - 80 = 0 t = 40 +/- (1600 + 4(3)(80))^1/2 = 15.1, -1.77 time always (t) It will take the police 15.1 s to catch the speeder.

35)
Answer:
-20 = - 1/2 gt² + 0 40/g = t² t = 2.02s time to reach ground y = - 1/2 gt² + y_o = 0 y₁ = v = (- 1/2)(2) gt at t = 1 v = -g (1) = -g = - 9.81 m/s

36)
Answer:
R = 0.500 (1/2)^(t/25) = 0.500 (1/2)^(5/25) = 0.435 kg 0.700 = Ro (1/2)^(10/25) 0.700 --------------- = Ro = 0.924 kg (1/2)^10/25

37)
Answer:
R = Ro (1/2)^20/n 0.85 Ro = Ro (1/2)^20/n 0.85 = (1/2)^20/n log 0.85 20 --------- = ---- log (1/2) n n = 85.3 years

38)
Answer:
p (t) = A sin ((g/l)^1/2 t ) = A sin ((1/4)^1/2 t) = A sin (1/2 t) T = 2 pi / (1/4)^1/2 = 4 pi p(5) = A sin (5/2) = 0.5985 A

39)
Answer:
dA/dt = dA/dl * dl/dt r = tan 30 degrees / l = d(pi (tan 30)² (1/l)²)/dl * dl/dt = 2 pi (1/(3)^1/2)² * 1 * (-2) --------------------- l = -4 pi/3l l = 3 = -4 pi/9

40)
Answer:
y - x - 2 = y - 3x + 2 y - x - 2 = 0 2x = 4 y - 2 - 2 = 0 x = 2 y = 4 Point of intersection = (2, 4). 3y - kx - 4 = 0 3(4) - k(2) - 4 = 0 12 - 2k - 4 = 0 8 = -2k k = 4

41)
Answer:
42)
Answer:
log₆ (x - 3)(x - 7) = 1 (x - 3)(x - 7) = 6 x² - 10x + 21 = 6 x² - 10x + 15 = 0 x = 10 +/- (100-4(1)(15))^1/2 = 8.16, 1.84

43)
Answer:
4/b = b/1 4 = b² b = 2 sin theta/2 = sin 90/4 sin theta = 2/4 theta = 30š

44)
Answer: