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## College Algebra

by: Vernie McCullough

35

0

15

# College Algebra MATH 101

Vernie McCullough

GPA 3.58

Bruce Ikenaga

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COURSE
PROF.
Bruce Ikenaga
TYPE
Class Notes
PAGES
15
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 15 page Class Notes was uploaded by Vernie McCullough on Thursday October 15, 2015. The Class Notes belongs to MATH 101 at Millersville University of Pennsylvania taught by Bruce Ikenaga in Fall. Since its upload, it has received 35 views. For similar materials see /class/223530/math-101-millersville-university-of-pennsylvania in Mathematics (M) at Millersville University of Pennsylvania.

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Date Created: 10/15/15
Math 101 672 772009 Review Sheet for Test 2 These problems are provided to help you study The presence of a problem on this handout does not imply that there will be a similar problem on the test And the absence of a topic does not imply that it won t appear on the test 1 a Divide I4 13 1 by 12 7 1 using long divisioni b Compute the quotient and remainder When 214 312 41 is divided by 212 c Compute the quotient and remainder When 212 51y 7 3y2 is divided by z By d Factor 13 7 212 7 111 12 completely given that z 7 4 is one of the factors 4 1 4 2 a Show that it is not valid to cancel 17s in I l to get ll 5 by giving a speci c value of z for z 4 Which I l is not equal to 5 z a b Show that is not always the same as g by giving speci c values of a b and c for Which is not a 2 2 equal to i c 3 Simplify cancelling any common factors 127172 127516 12312 1274z4i a 13741 I27516 1013 b i 512 1274z4 6112 4i Simplify cancelling any common factors 13761291 I276z8 a12781172 I74 M12717 12514 514 71513 13 4x2 12 7 21y 13 y 0 12 7 41y 7 5y2 3 1 171 11 g 1 5 Combine the fractions into a single fraction and simplify 2 3x 2 a I 7 2 1 6 b 7 7 z2i5z6 1279 1 z 1 12721 1 c27 I y d 7 i 90er 1 9 2y 12732 172 e 1275 11 171 3 1277z1 13 5 z3i 31 12 bSolve16I74zi4i 3 1 1 S1 777 7 00V612721 12741221 fzii 61 a Solve 12 25 dSl 7 i ovezi5 175 6 71 a Solvez57lt z z 311 1 b Solve z11272173173 81 Pipe A can ll a tank in 3 hours If pipes A and B work together they can ll 5 tanks in 6 hours How long would it take for pipe B to ll a tank by itself 91 Calvin can eat 800 Brussels sprouts in 4 hours Phoebe can eat 600 Brussels sprouts in 5 hours It takes Bonzo 3 hours to eat 240 Brussels sprouts How long Will it take them to eat 600 Brussels sprouts if they work together 10 Calvin can eat 396 tacos in 6 hours Calvin and Phoebe eating together can eat 324 tacos in 3 hours Phoebe and Bonzo eating together can eat 456 tacos in 4 hours How long Will it take Bonzo to eat 576 tacos by himself 11 The numerator of a fraction is 8 less than the denominator If 7 is added to the top and the bottom of the fraction7 the new fraction is equal to Find the original fraction 12 A river ows at a constant rate of 16 miles per hour Calvin rows 32 miles downstream With the current in 5 hours less than it takes him to row the same distance upstream against the current What is Calvin7s rowing speed in still water 6 13 Given that 2 2 z and 91 12 nd a f3 b f1 0 f1 1 01 f91 and 90 e we 14 Graph the function 1 1 if I S 0 f12izifzgt0 by computing for z 73 72 71 07 17 2 and 3 11 15 a Find the domain of z 7 z b Find the domain ofgz xz 7 ll 16 Find the range of the function Whose entire graph is pictured belowi Solutions to the Review Sheet for Test 2 3 1 a Divide I4 13 1 by 12 7 1 using long divisioni x2x1 x21lx4x31 x4 x x3x2 x3 x x2 x 1 x2 1 x2 z4zgl D 1271 12 1271 12 z 1 b Compute the quotient and remainder When 214 312 41 is divided by 212 214 312 4x 2x4 312 4x 2 3 2 22221 t tquot 2x 2x 2x 2x 2 I Since there7s only one term on the bottom its easier to break the fraction up into pieces than to do the long divisioni D c Compute the quotient and remainder When 212 51y 7 3y2 is divided by z By 2xy x 3y 12x2 5xy3y2 2x2 6xy xy 3y2 xy 3y2 21251y73y2 2 7 i D z3y I y d Factor 13 7 2x2 7 111 12 completely given that z 7 4 is one of the factors Divide 13 7 2x2 7111 12 by z 7 4 x2 2X 3 X 4 1x3 2x2 11X 12 x3 4x2 2X2 11X 2x2 8X 3X12 3X12 Thus 13 7212 711112 I74z22173174z3171i a 4 1 4 2 a Show that it is not valid to cancel 17s in I l to get l 5 by giving a speci c value of z for z 4 Which I l is not equal to 5 z 4 2 4 6 4 For I 2 i i i 3 f 5 Thus you can7t cancel 17s in I l to get 5 D I z a a 7 a b Show that Z is not always the same as A by giving speci c values of a b and c for which 3 is not c a c c E equal to c lfalblandc2then Thus is not in general equal to b i D c E 3i Simplify cancelling any common factors 127172 127516 312312391274z4 127172 1275167172z139172173173 7 i D z2312 z2i4z4 zlz2 1722 12 M13741 1275I6 1013 512 1274z4 6112 13741 1275I6 1013 711274 172x173 1013 7 512 1274z4 6z12 512 35722 6z2 1172z2 172x173 1013 712173 D 512 35722 6z2 3 4i Simplify cancelling any common factors 13761291 276z8 021 z 78zl72 174 13761291 1 32 1I 32 1I 32 z2iazs Imam remain ltz72gtltz74gt 12781172 1278117 174 12781172zi4 1276z9 174 174 174 174 174 11732 172x174 II732 I 174 i U I732 172x174 I732 I72 174 b 127172 I25I4 51471513 I34I2 127172 127172 125I475I4715I3 127172 I34I2 51471513 I34I2 7 I25I4 75147151339124r514r4 I34I2 I72I1 I2I4 172 5353173 39 11z4 535173 127213 I3 I2 c 12 I274Iy75y2 127213 m 127213 I274Iy75y2 1172y I75yIy 1753 I2 74y2 1312y39 I2 74y2 12Iyl172y12y IltI2y I274Iy75y2 2 3 CD1 I 9I2 1 2 3 2 3 ligip ligip39IQ712721737173I17I1 D 17 17 I2 1279 I73I3 13 I I 1 1 e 171I1 1I2 1 1 1 1 1 1 1 FEET 1 7 1 Imam 171 I 1 H1711 1x71z1 1x71x1 1 zilgtltz1gtltz71gtltz1gt 171 lt 311M7311111111137221gt1 3 D 1 7 I 1 7 I I 7 I I I1I1171I1 137512 2 f 321151 1274 137512 W 137512 1274 12175 39I72I2II72 U I22I 12721715 I22I I75I3 II2 13 1274 1 1 I g1717171 z 1 1 1 71 1 z 7z1 z 7z1z7 1 D 171 11717l z 11iz1 117 11 711 1 z 5 Combine the fractions into a single fraction and simplify a 2 31 2 4 12 z22zi 2 31 27 2 31 275 2 31 12 2 12 122z ziz2 1z2 171 12 1z2 12 1 2z312z2 7 314 1z2 1z2 D 1 6 b 71 z2516 12 1 6 7 1 6 7 13 6 1727 125167129 71721737173z3 172z7339m773z339 13 7 6x 2 7 136z2 7 136z12 7 1 2I3I3 1 2I I3 7 1 2I3I3 7 1 2I3I3 7 5115 7 5z3 7 5 ltz72gtltz73gtltz3gt ltz72gtltz73gtltz3gt ltz72gtltz3gt 1 1 027m1 1 1 7 zz1 1 11 1 z 21z1 11 1 7 27gz17239zz1739z1 1139 zz17zz1zz17 21z1z1z72122zz1z72122z1 D zz1 7 zz1 7 zz1 I y 1me I y Iiy yIy2 Iy Iiy Iy Iiy Iiy Iy IiyIy IiyIy I2IyyIy27 I2y2 IinEer IinEery 12y 2y 7 12y 2y 7 12y zy 2y I7 I27Iy I27y2 IIy IiyIy IIy Iy IiyIy I 7 12 71y7 2y2 21y 7 12 7 1y 7 2y2 21y 12 1y 7 2y2 11 7 MI y 11 7 MI y zltz7ygtltzygt zltz7ygtltzygt 12yr7y 12y zltz7ygtltzygt zltzygt D 7 7 2 7 7 17111zl11 171I1 1 11391711 1 1 1 1 171 1171 777 M 1 171 171171171 11711171 121717171111137127111217 1311112711 D 11 71 11 71 11 71 11 71 3 12 711 6 S1 7 1 a 0V6135 13 3 27711 13 5 13 3 12 711 5ltI3gt39ltm35lt13gt39 13 3 12 711 513I35137513 13 151213 5711 15121363515 511213515 46 121 351 46231 21 Check 3 127 3 127312715 73 711772171573 13 5723 5 5 5 5 13723757 The solution is 1 2 El 31 12 bSl 16 71 ove I7 174 31 12 16 77 174 174 31 12 174lt16I74gt174zi4 31 12 17416174zi4174 161743112 1617643112 19176412 19176 14 3 12 Plugging z 4 into 16 14 4 results in division by zero Therefore7 there are no solutions I 7 z 7 D 3 1 1 S l 7 7 7 7 00V612721 12741221 12721 1274 12217 3 1 1 1372 172I2zz2 0 1172z2 II72z20 31172z2 7 1172z2 zz72z2 1172 I72z2 zz2 3z27zz720 3167Iz720 0 3140 3174 4 I 3 Check When I 7 3 9 0 12721 I274 1221740 20 8 7 4 The solution is z 77 El 3 12 25 d S1 7 i ovez75 175 25 7 175 12 25 17517571757z75I75 12717525 127173525 12735710 1277z100 I72z750 12 or 15 Check When I 2 z 4 25 25 4 7 d 7 77777i 175 3 an 175 3 3 However7 z 5 causes division by 0 in the original equationi Therefore7 the only solution is z 2 7 a Solve z 5 7 9 Ir lt 6 zzz 577 z 12 5x 7 6 12 7 5x 6 0 I72z730 The possible solutions are I 2 and z 3 Check For I 2 6 57 5775732 z 2 I Forz37 6 6 57 57 5723z The solutions arez2andz31 El 1 311 1 bSOlvez112721737173 z 311 7 1 11 12721737173 1 311 1 73 1 7 73 1 ac gtltzgtltz1mggtltz1gtgt z gtltzgt173 173z1 I I73z1SI iz73x1 1 11 I3I17 I73 1173311z1 I2731311z1 z21z1 127z0 11710 The possible solutions are I 0 and z 1 Check lfz07 z 311 1 1 1 0 i d i 11 1272173 1 3 3 an 3 73 lfz17 I 3I1 1 4 1 1 1 1 7 777 cl 777 z1x272173 2 74 2 an 173 72 The solutions are I 0 and z 1 El 8 Pipe A can ll a tank in 3 hours If pipes A and B work together7 they can ll 5 tanks in 6 hours How long would it take for pipe B to ll a tank by itself Let I be pipe A s rate in tanks per hour7 and let y be pipe B s rate in tanks per hour hours tanks per hour tanks A 3 z 1 B t y 1 together 6 z y 5 1 The rst equation says 31 1 so I 7 3 1 The third equation says 61 y 5 Substitute z g and solve for y 1 6 7 5 w 26y5 6y3 1 PE 1 1 The second equation says yt 1 Substitute y i into yt 1 this gives it 1 so t 2 hours D 9 Calvin can eat 800 Brussels sprouts in 4 hours Phoebe can eat 600 Brussels sprouts in 5 hours It takes Bonzo 3 hours to eat 240 Brussels sprouts How long Will it take them to eat 600 Brussels sprouts if they work together Let I be Calvin s rate in sprouts per hour let y be Phoebels rate in sprouts per hour and let 2 be Bonzo7s rate in sprouts per hour hours sprouts per hour sprouts Calvin 4 z 800 Phoebe 5 y 600 Bonzo 3 2 240 together t z y 2 600 The rst equation says 41 800 so I 200 The second equation says 5y 600 so y 120 The third equation says 32 240 so 2 80 The last equation says tzyz 600 Substitute z 200 y 120 and 2 80 into tzyz 600 t200 120 80 600 400t 600 t 115 It takes 15 hours D 10 Calvin can eat 396 tacos in 6 hours Calvin and Phoebe eating together can eat 324 tacos in 3 hours Phoebe and Bonzo eating together can eat 456 tacos in 4 hours How long Will it take Bonzo to eat 576 tacos by himself Let I be Calvin s rate in tacos per hour let y be Phoebe s rate in tacos per hour and let 2 be Bonzols rate in tacos per hour hours tacos per hour tacos Calvin 6 z 396 Calvin and Phoebe 3 z y 324 Phoebe and Bonzo 4 y 2 456 Bonzo t 2 576 The rst row says 61 396 so I 66 The second row says 31 y 324i Plug in z 66 and solve for y 366 7 y 324 66 y 108 y 42 The third row says 49 2 456i Plug in y 42 and solve for 2 442 7 2 456 42 2 114 2 72 The fourth row says t2 576i Plug in 2 72 and solve for t 72t 576 t 8 It takes Bonzo 8 hours D 11 The numerator of a fraction is 8 less than the denominator If 7 is added to the top and the bottom of the fraction the new fraction is equal to Find the original fraction n Let 7 be the original fraction The numerator is 8 less than the denominator so n d 7 8 If 7 is added to the top and the bottom the result equals 2 Plug n d7 8 into d78773 d71 3 d 7 7 5 d 7 5 Multiply by 5d 7 to clear denominators d71 3 5d77d75d73 5d713d7i Solveford 5d753d21 2d26 d13i Hence n d7 8 13 7 8 5 The fraction is 153 El 12 A river ows at a constant rate of 16 miles per hour Calvin rows 32 miles downstream With the current in 5 hours less than it takes him to row the same distance upstream against the current What is Calvin7s rowing speed in still water 12 Let I be Calvin7s rowing speed in still water His downstream rate is 1 167 While his downstream rate is z 7 1161 Let t be the time it takes Calvin to row upstreami Then it takes him t7 5 that is7 5 hours less to row time speed distance downstream t 7 5 z 16 32 upstream t z 7 116 32 The equations are t 7 5z 116 32 and tz 7116 321 Solve the second equation for t 15 7116 7 32 l l 1711639t171 6 17116 3932 32 t z 7116 Plug this into t 7 5 1 116 32 and solve for z t 7 5z 116 32 32 I716 75gt 147116 732 716 32 75 16771632 z 1 1716 I 1 7 z 1 32 7116 lt 7 5z 7116 1 116 7 32z 7116 I7 1 32 7 5z 71161 116 7 32z 7116 32 7 51 8z 116 7 321 7 512 40 7 5zz 116 7 321 7 512 7552 32x 7 64 7 321 7 512 7552 7 71152 12 7 23104 I 7 i418 Since Calvin s speed can t be negative7 I must have I 418 miles per hour El 13 Given that 126 2 and 91 12 nd a f3 6 6 3 m E D 0 164 6 D f1m2 Cfr1 6 6 f11z12212213 13 01 MM and 901 6 14 gltfltzgtgt g fltgltzgtgt f I e f f 1 6 6 6 6 I2 22 10061 f 6 2 36 2 36 2 I 1222 1242 2 WW2 MN 612 22 U W 1411 Graph the function 7 11 ifzgo Il lwz ifzgt0 by computing for z 737 72 717 07 17 2 and 3 1117317217110 w 172171 10 1 z 1 151 a Find the doma1n of 7 1 7 z 1 I27312 7171z72 consists of all real numbers except for 1 and 2 El b Find the domain ofgz xz 7 1 Since 7 f Will be unde ned if z 1 or if z 2 The domain 9 Will be unde ned if the stuff inside the square root 1 7 1 is negative So 1 must have I 7 1 2 07 or z 2 1 The domain consists of all real numbers greater than or equal to 1 16 Find the range of the function Whose entire graph is pictured belowi The range of a function is the set of its outputs7 ie the set of y Values it attainsi The function attains every yValue height from 72 to 37 including 3 but not including 72 the open circle at 727 72 indicates that the point is not on the graph Therefore7 the range is 72 lt y S 3 D One hears only those questions for which one is able to nd answers FRIEDRICH NIETZSCHE 2009 by Bruce Ikenaga 15

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