:Q :SB :SG210112174272055151 :SP183064267136267136267136 :SP182136238088238088238088 :SH1135A :SH1827B :SH0827C :SH1838D :SF 1. Which of the following equations correctly identifies the line CD in the diagram? (a) 3x - 4y = 7 (b) 3x + 4y = 12 (c) 4x + 3y = 12 (d) y = 4x + 3 (e) 4y - 3x = 7 :RCB 1. (b) 3x + 4y = 12 Ans. Two points on the graph are (-4,6) and (8,-3). Although we have selected points with integral values to avoid inaccuracies, we have purposely avoided the points at which x or y is 0 in order to make a more general case. :RA subtract to eliminate b 6 = (-4)m + b -3 = (8)m + b - --------- 9 = -12m 3 m = - - 4 :RA substitute for m 3 6 = (- -)(-4) + b 4 b = 3 Therefore: y = -3/4x + 3 4y = -3x + 12 3x + 4y = 12 Ans. :RA :SD :Q 2. From the graph find the value of x on line CD when y = 3/4x. (a) 2 (b) 4 (c) -2 (d) -4 (e) 3 :RCA 2. (a) 2 Ans. If you know the equation, you substitute 3/4x for y and solve for x. If you know only the graph, then, on the same axis, draw the graph of y = 3/4x. One point is (0,0). Another is (4,3). :RA Where this line meets the original line, read the value of x. x = 2 Ans. Note: In this problem you found graphically the solution to the simultaneous set of equations, y = 3/4x and 3x + 4y = 12. :RA :SD :Q 3. Find the linear relationship between x and y as shown by the table: x -1 1 3 4 ------------------- y 4 -2 -8 -11 (a) y = -3x + 1 (b) y = 3x + 1 (c) y = x + 1 (d) x = -3y + 1 (e) y = 3x - 1 :RCA 3. (a) y = -3x + 1 Ans. Treat the pairs of values as coordinates of points on the graph. Selecting (-1,4) and (1,-2), substitute each pair in y = mx + b. 4 = m(-1) + b, and -2 = m(1) + b -------------- 6 = -2m subtract :RA m = -3 b = 1 y = -3x + 1 Ans. :RA :SD :Q :SB :SG175112174272039111 :SP189104252104266048189104 :SH1427R :SH1437S :SH0639T :SF 4. Find in square units the area of {RST. (a) 30 sq. units (b) 31.5 sq. units (c) 35 sq. units (d) 33.5 sq. units (e) 32 sq. units :RCB 4. (b) 31.5 sq units Ans. 1 Note: A = - bh 2 To find b, count the number of units in the base RS: 9. To find h, count the number of units in the perpendicular from T to RS: 7. :RA Therefore: 1 A = - x 9 x 7 2 63 = -- 2 = 31.5 sq. units Ans. :RA Note: If each x-division were equal to 3 units, then RS would equal 9 x 3 = 27 units. If each y-space were equal to 2 units, we would multiply each y-dimension by 2 to obtain the number of vertical units. :RA :SD :Q :SB :SG210088181251031111 :SP189096245048245048245048 :SH1327A :SH0636B :SF 5. The coordinates of point A are (-3,-1) and of B, (5,5). Find the number of units in the length AB. (a) 5 (b) 2 (c) 10 (d) 29 (e) 20 :RCC 5. (c) 10 units Ans. Use the distance formula. ___________________ / 2 2 d=\/(x - x') + (y - y') _______________ / 2 2 AB = \/(-3-5) + (-1-5) ___________ / 2 2 AB = \/(-8) + (-6) :RA _______ AB = \/64 + 36 ___ AB = \/100 AB = 10 Ans. :RA :SD :Q :SB :SG189112188258031111 :SP196072252096217040196072 :SH0930A :SH1337B :SH0532C :SF 6. Find the number of square units in the area of {ABC. (a) 20.5 (b) 10.5 (c) 20 (d) 15.5 (e) 0 :RCA :SD :SB :SG189112188258031111 :SP196072252096217040196072 :SH0930A :SH1337B :SH0532C :SP196040196096252096252040 :SP252040196040196040196040 :SH0729I :SH0734II :SH1229III :SF 6. (a) 20.5 Ans. In this case, neither the base nor the altitude of the triangle can be found directly by counting units. Through the vertices of the triangle, draw horizontal and vertical lines, as shown, to form a rectangle. This rectangle is composed of 3 right triangles and {ABC. :RA {I = 1/2 bh = 1/2 x 3 x 4 = 6 {II = 1/2 x 5 x 7 = 35/2 = 17.5 {III = 1/2 x 8 x 3 = 12 The sum of the areas of triangles is 6 + 17.5 + 12 = 35.5 sq unit :RA area of rectangle is bh = 8 x 7 = 56 sq units Therefore: area of {ABC is 56 - 35.5 = 20.5 Ans. :ET :ET Copyright ARROW INSTRUCTIONAL SYSTEMS July 1983 The sum of the areas of triangles is 6 + 17.5 + 12 = 35.5 sq unit :RA area of rectangle is bh = 8 x 7 = 56 sq units Therefore: area of {ABC is 56 - 35.5 = 20.5 Ans. :ET :ET Copyright ARROW INSTRUCTIONAL SYSTEMS July 1983 '!m.. '!DISK1 j!nBASICA EXE `qåÇ√EDLIN COM `qÜáSAMPLE BAK ¿Γ2|WDISK2 m!oDISK3 o!pDISK4 p!rDISK5 r!sDISK6 t!tDISK7 u!uDISK8 w!vDISK9 x!wDISK10 z!xDISK11 }!ynown examples of such coevolution are the obligatory relations between figs and fig wasps and between yuccas and yucca moths. In both instances the plants have come to need the services of the insects as pollen carriers and the insects in turn have come to call on the plant to sacrifice some of its ovules as larval feeding sites to promote the insect's reproduction. In such instances plants and animals have taken turns acting as agents of natural selection with respect to each other. Not all coevolution, however, is mutualistic. In many instances one of the interacting organisms is parasitic on the other. One of the most :RA remarkable interactions for the study of animal-plant coevolution of the host-parasite type is the interaction between certain brightly colored butterflies of the New World Tropics and certain vines. The butterflies are members of the genus Heliconius; the vines are passion-flower vines, members of the genus Passiflora. The passion-flower vines have evolved effective chemical defenses against insects, but a few insects, the Heliconius butterflies among them, have evolved the ability to circumvent these defenses. That ability apparently precludes the Heliconius butterflies' being parasitic on other plants. Heliconius butterflies thus deposit their eggs only on Passiflora :RA vines, where the eggs hatch into larvae that feed voraciously on the leaves of the vine. The remarkable thing is that some species of the vine have features that appear to mimic the distinctive bright yellow eggs of the butterflies. What accounts for this mimicry, if that is what it is? One possibility is the coevolution of Passiflora with Heliconius. How could any one trait of a plant be causally attributed to natural selection imposed by one species or genus of insects among so many? The answer is that such a selective effect is almost impossible. That is why plants such as the passion-flower vines, with their :RA simplified, specialized populations of animal parasites, are of such interest. With only a few major herbivores such as Heliconius to account for, interpreting the defensive traits of passion-flower vi┌ΦΩφ)SΦM■ç┌Y[XQ╣▌╙Q╣QPRΦA■ZXèΦ■╚è╚:░r├ï≥è
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1130 PB=M1:PS=R0:ER=R0:J=R0:SK=R0:T2=R1:GOSUB 10200:GOSUB 10400:GOSUB 2010:GOSUB 42:T2=R0:IF IN$=M$ THEN 1170
1150 IF MI=R0 THEN 1162
1160 GOTO 1040
1162 CLS:T$="TIME IS UP":VT=R4:GOSUB 10000:GOSUB 1980:HT=R1:VT=12:GOSUB 10000:PRINT"PRESS 'C' TO CONTINUE":VT=14:GOSUB 10000:PRINT"PRESS 'S' TO STOP THE TEST"
1164 PRINT CHR$(7);:GOSUB 830:IF IN$="S" THEN 1097