M ATH 31 0 Self-Test M EASURE Answers (NOTE– In pdf format, some images do not display correctly) 1. a. 2 C B(2u)2 + [(4u)2 ! B(2u)2] = (4 B+ 16)u2(see diagram `) 2.b. approximately 11+ 18/2 u2 = 20u2 (count th e unit s & halves)c. 5@7 ! (2@2 + 1@1 + ½(1@7 + 1@5 + 1@3 + 1@1 + 1@2)) u2= 21u 2(work from outside in...)d. The perimeter is the same as that of tw o circles: 8B units 1a.3. a. h@(a + b)/2 b. Br2c. (xo/360o)@ Br2d. C = 2Br e. V = (4/3)Br3SA = 4Br2 ))))))) )4. a. r 42 + 82 u = 4r 5 units. b. D[(!1,4),(2,0)] = 5; D[(2,0),(!2,!3)] = 5; D[(!1,4),(!2,!3)] = 501/2 . Yes: d12 + d22 = d32.5. (12 cm)2 ! B(6 cm)2 = (144 ! 36B)cm26. V o f p rism o r cy linder = (Area of base)@(Height) = 8000 m2 7. a. (1/3)@(30 cm)220 cm = 6000 cm3Volume of pyramid or cone = (1/3) (area of base)(height)b. (1/3)@(20 cm)@(30 cm)@20cm = 4000 cc.8. a. V = (1/3)@B(5 cm)2@25cm = 625 B/3 cm3lateral SA = (1/2)@B(10 cm )(5r &2& &6& & cm) = 25r26Bcm2b. V = B(5 cm)2@25cm = 625B cc (cc is abbreviation for cm3)þpart of Clateral SA = B(10cm)@25cm = 250Bcm2 radial faceþ9a. SA = top+ bottom + radial faces + strip of C V = one-six th of th e V o f t he origin al w heelSA = 2@(15cm@3cm) + 2@(1/6)@B(15cm)2 + 3@30Bcm2/6 V = (1/6)@B(15cm)2@ 3cm = 225B/2cm39b. SA = 2@(15cm@3cm) + 2@(30/360)@B(15cm)2 + (1/12 )3@30Bcm2V = (30/360)@B(15cm)2@ 3cm = 225B/4 cm330 ° is 1/12 of the c ircle, so V is half V of part a10. SA = (1/2)@4B(5cm)2 + (1/2)@B(10cm)@13cm = 115Bcm2V = (1/2)@(4/3)B(5c m)3 + (1/3)@B(5cm)2@ (12cm) = (550B/3)cm3 11. The relationship between oC and oF is a line going from (0,32) to (100,212 ); that makes a rise of 100oC equal to a rise of 180oF: i.e. 100oC = 180oF, so ( 100oC/180oF ) = 1! .100oC/180oF reduc es to 5oC/9oF. An d, y es, 5oC/9oF = 1 also. So oC = 5oC/9oF@(oF ! 32oF).(Not ice oF must be adjust ed dow n t o 0 bef ore m ult iplyin g, so th at 3 2oF will end up being 0oC.).C = (5 /9 )(F ! 32) a. 72oF = 2 2oC (18 °Reaumur) c. 98.6oF = 3 7oC F = (9/5 )C + 32 b. 30oC = 86oF12. a. cm b. cm c. m d. km e. kg f. g g. g (act ually! mg , bu t t hat ' s not in t he list )h. t i. mL or cc j. L k. mL or cc l. kL13. 1 kL = 1000 L = 1000@1000 mL = 106mL = 106cc =j 106g = 100 0 kg = 1 metric ton. OR:1 kL = 1 kL @ 1000L/kL = 1000 L = 1000L @ 100 0 mL/L = 100 000 0 mL = 100 000 0 cc = 100 000 0 g =100 000 0g @ 1kg/1 000 g = 100 0 kg = 100 0 kg @ 1 metric ton/1 000 kg = 1 metric tonj valid ON LY f or w at er at 4°C14. a. Carpet cost s $32/yd2; wood costs $4/ft2 = $4/ft2@3ft/yd@3ft/yd = $36/yd2; carpet is ch eaper.b. 1 ft2 = (12 in)2 = 144 in2; 1 yd2 = (3 ft)2 = 9 ft2c. (1 ft)3 = (12 in)3 = 1728 in3; (1 yd)3 = (3 ft)3 = 27 ft3; 1 yd3 = 27 ft3 = 27@1728 in3d. 1 m3 = (100 cm)3 = 1000 000 cm3; (see #13 )e. (see #13; t he conversion at * is valid for water only!)f. 300 dam g. 200 cm2 = 200 cm2 @ 1m @ 1 m. = .02m2h. 3 28 dl = 32 8d l @100ml@1cc = 3000 m 100cm.@100 cm dl ml = 30000 cm = 32800 cc15. Just barely n ot. Th e longest dimen sion in a 2 4" x7 " x7 " box is the ext reme diagonal, (67 4)½ “ Ñ 25.96"M AT H 3 1 0 Self -T est M EAS URE ad dit iona l not es reg ardin g solut ions . . . . . . . . . . . . . . 1. a. Find b. approxim ate a. b. . . . . . . . c. . . . . . . . th e area enclosed by t he 8 4B u2 . . . . . . . . . . . . . . simple closed curve sketched. 6u . . . . . . . . . . . . . . In part a, all curv y part s are 9 A2. . . . . . . . . . . . . . ¼ or ½ circles w ith equal radii. . . . . . . . . . . . . . . One Solution: 78u6 . . . . . . . . . . . . . . For the Area bounded by all those arcs, . . . . . . . . . . . . . . I see tw o full circular regions, plus the region marked A2. . . The radius of each arc is 2 units, so the area of each . . = 1 sq unitcircular regio n is 4B u2 .The region marked A2 is similar to the shaded region described and illustrated in #5. Each of th e fo ur qu arters o f t his region is the area betw een the vert ex of a 4-un it square, and t he inscribedcircle... Area shaded is area of square ! area of circle... (4 u nit s)2 ! B (2 unit s)2 16 u2 ! 4 B u2So t he t ot al area enclosed by t he heart -shaped f igure c om posed o f circular arc s is: 2( 4B u2 ) + ( 16 ! 4B ) u2 ( 4B + 16 ) u2b. W ith in the bou ndary , w e see 11 fu lly enc losed squares, and 18 part ial squares.We assume t he part ially enc losed squares w ould av erage out to being 5 0% con tained w ithin the boundary , so w e estim ate t he area inside t he bou ndary to be appro xim ately ( 11 + 18 /2 ) u2.c. T o f ind t he area of th e region enclosed by t he poly gon joining th e respect ive vertic es in the Cart esianpl an e, m ak e s ur e t o “ con ne c t t he do t s” IN TH E ORD ER GI V EN:(0, 0) (2 ,0 ) (1,1 ) (3,3 ) (4,6 ) (!1,7) (0,0)5@7 ! (3@1 + 1@ 1 + ½(1@ 7 + 1@ 5 + 1@ 3 + 1@ 1 + 2@ 2)) u24. a. Sket ch t he poin ts (!4,5) and (4,1 ) in th e plane. Find the dist ance …