Tutor:

New! Get regular updates about newly solved problems via algebra.com's RSS system.

---

Recent problems solved by 'jim_thompson5910'

jim_thompson5910 answered: 28583 problems
Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 4740..4769 , 4770..4799 , 4800..4829 , 4830..4859 , 4860..4889 , 4890..4919 , 4920..4949 , 4950..4979 , 4980..5009 , 5010..5039 , 5040..5069 , 5070..5099 , 5100..5129 , 5130..5159 , 5160..5189 , 5190..5219 , 5220..5249 , 5250..5279 , 5280..5309 , 5310..5339 , 5340..5369 , 5370..5399 , 5400..5429 , 5430..5459 , 5460..5489 , 5490..5519 , 5520..5549 , 5550..5579 , 5580..5609 , 5610..5639 , 5640..5669 , 5670..5699 , 5700..5729 , 5730..5759 , 5760..5789 , 5790..5819 , 5820..5849 , 5850..5879 , 5880..5909 , 5910..5939 , 5940..5969 , 5970..5999 , 6000..6029 , 6030..6059 , 6060..6089 , 6090..6119 , 6120..6149 , 6150..6179 , 6180..6209 , 6210..6239 , 6240..6269 , 6270..6299 , 6300..6329 , 6330..6359 , 6360..6389 , 6390..6419 , 6420..6449 , 6450..6479 , 6480..6509 , 6510..6539 , 6540..6569 , 6570..6599 , 6600..6629 , 6630..6659 , 6660..6689 , 6690..6719 , 6720..6749 , 6750..6779 , 6780..6809 , 6810..6839 , 6840..6869 , 6870..6899 , 6900..6929 , 6930..6959 , 6960..6989 , 6990..7019 , 7020..7049 , 7050..7079 , 7080..7109 , 7110..7139 , 7140..7169 , 7170..7199 , 7200..7229 , 7230..7259 , 7260..7289 , 7290..7319 , 7320..7349 , 7350..7379 , 7380..7409 , 7410..7439 , 7440..7469 , 7470..7499 , 7500..7529 , 7530..7559 , 7560..7589 , 7590..7619 , 7620..7649 , 7650..7679 , 7680..7709 , 7710..7739 , 7740..7769 , 7770..7799 , 7800..7829 , 7830..7859 , 7860..7889 , 7890..7919 , 7920..7949 , 7950..7979 , 7980..8009 , 8010..8039 , 8040..8069 , 8070..8099 , 8100..8129 , 8130..8159 , 8160..8189 , 8190..8219 , 8220..8249 , 8250..8279 , 8280..8309 , 8310..8339 , 8340..8369 , 8370..8399 , 8400..8429 , 8430..8459 , 8460..8489 , 8490..8519 , 8520..8549 , 8550..8579 , 8580..8609 , 8610..8639 , 8640..8669 , 8670..8699 , 8700..8729 , 8730..8759 , 8760..8789 , 8790..8819 , 8820..8849 , 8850..8879 , 8880..8909 , 8910..8939 , 8940..8969 , 8970..8999 , 9000..9029 , 9030..9059 , 9060..9089 , 9090..9119 , 9120..9149 , 9150..9179 , 9180..9209 , 9210..9239 , 9240..9269 , 9270..9299 , 9300..9329 , 9330..9359 , 9360..9389 , 9390..9419 , 9420..9449 , 9450..9479 , 9480..9509 , 9510..9539 , 9540..9569 , 9570..9599 , 9600..9629 , 9630..9659 , 9660..9689 , 9690..9719 , 9720..9749 , 9750..9779 , 9780..9809 , 9810..9839 , 9840..9869 , 9870..9899 , 9900..9929 , 9930..9959 , 9960..9989 , 9990..10019 , 10020..10049 , 10050..10079 , 10080..10109 , 10110..10139 , 10140..10169 , 10170..10199 , 10200..10229 , 10230..10259 , 10260..10289 , 10290..10319 , 10320..10349 , 10350..10379 , 10380..10409 , 10410..10439 , 10440..10469 , 10470..10499 , 10500..10529 , 10530..10559 , 10560..10589 , 10590..10619 , 10620..10649 , 10650..10679 , 10680..10709 , 10710..10739 , 10740..10769 , 10770..10799 , 10800..10829 , 10830..10859 , 10860..10889 , 10890..10919 , 10920..10949 , 10950..10979 , 10980..11009 , 11010..11039 , 11040..11069 , 11070..11099 , 11100..11129 , 11130..11159 , 11160..11189 , 11190..11219 , 11220..11249 , 11250..11279 , 11280..11309 , 11310..11339 , 11340..11369 , 11370..11399 , 11400..11429 , 11430..11459 , 11460..11489 , 11490..11519 , 11520..11549 , 11550..11579 , 11580..11609 , 11610..11639 , 11640..11669 , 11670..11699 , 11700..11729 , 11730..11759 , 11760..11789 , 11790..11819 , 11820..11849 , 11850..11879 , 11880..11909 , 11910..11939 , 11940..11969 , 11970..11999 , 12000..12029 , 12030..12059 , 12060..12089 , 12090..12119 , 12120..12149 , 12150..12179 , 12180..12209 , 12210..12239 , 12240..12269 , 12270..12299 , 12300..12329 , 12330..12359 , 12360..12389 , 12390..12419 , 12420..12449 , 12450..12479 , 12480..12509 , 12510..12539 , 12540..12569 , 12570..12599 , 12600..12629 , 12630..12659 , 12660..12689 , 12690..12719 , 12720..12749 , 12750..12779 , 12780..12809 , 12810..12839 , 12840..12869 , 12870..12899 , 12900..12929 , 12930..12959 , 12960..12989 , 12990..13019 , 13020..13049 , 13050..13079 , 13080..13109 , 13110..13139 , 13140..13169 , 13170..13199 , 13200..13229 , 13230..13259 , 13260..13289 , 13290..13319 , 13320..13349 , 13350..13379 , 13380..13409 , 13410..13439 , 13440..13469 , 13470..13499 , 13500..13529 , 13530..13559 , 13560..13589 , 13590..13619 , 13620..13649 , 13650..13679 , 13680..13709 , 13710..13739 , 13740..13769 , 13770..13799 , 13800..13829 , 13830..13859 , 13860..13889 , 13890..13919 , 13920..13949 , 13950..13979 , 13980..14009 , 14010..14039 , 14040..14069 , 14070..14099 , 14100..14129 , 14130..14159 , 14160..14189 , 14190..14219 , 14220..14249 , 14250..14279 , 14280..14309 , 14310..14339 , 14340..14369 , 14370..14399 , 14400..14429 , 14430..14459 , 14460..14489 , 14490..14519 , 14520..14549 , 14550..14579 , 14580..14609 , 14610..14639 , 14640..14669 , 14670..14699 , 14700..14729 , 14730..14759 , 14760..14789 , 14790..14819 , 14820..14849 , 14850..14879 , 14880..14909 , 14910..14939 , 14940..14969 , 14970..14999 , 15000..15029 , 15030..15059 , 15060..15089 , 15090..15119 , 15120..15149 , 15150..15179 , 15180..15209 , 15210..15239 , 15240..15269 , 15270..15299 , 15300..15329 , 15330..15359 , 15360..15389 , 15390..15419 , 15420..15449 , 15450..15479 , 15480..15509 , 15510..15539 , 15540..15569 , 15570..15599 , 15600..15629 , 15630..15659 , 15660..15689 , 15690..15719 , 15720..15749 , 15750..15779 , 15780..15809 , 15810..15839 , 15840..15869 , 15870..15899 , 15900..15929 , 15930..15959 , 15960..15989 , 15990..16019 , 16020..16049 , 16050..16079 , 16080..16109 , 16110..16139 , 16140..16169 , 16170..16199 , 16200..16229 , 16230..16259 , 16260..16289 , 16290..16319 , 16320..16349 , 16350..16379 , 16380..16409 , 16410..16439 , 16440..16469 , 16470..16499 , 16500..16529 , 16530..16559 , 16560..16589 , 16590..16619 , 16620..16649 , 16650..16679 , 16680..16709 , 16710..16739 , 16740..16769 , 16770..16799 , 16800..16829 , 16830..16859 , 16860..16889 , 16890..16919 , 16920..16949 , 16950..16979 , 16980..17009 , 17010..17039 , 17040..17069 , 17070..17099 , 17100..17129 , 17130..17159 , 17160..17189 , 17190..17219 , 17220..17249 , 17250..17279 , 17280..17309 , 17310..17339 , 17340..17369 , 17370..17399 , 17400..17429 , 17430..17459 , 17460..17489 , 17490..17519 , 17520..17549 , 17550..17579 , 17580..17609 , 17610..17639 , 17640..17669 , 17670..17699 , 17700..17729 , 17730..17759 , 17760..17789 , 17790..17819 , 17820..17849 , 17850..17879 , 17880..17909 , 17910..17939 , 17940..17969 , 17970..17999 , 18000..18029 , 18030..18059 , 18060..18089 , 18090..18119 , 18120..18149 , 18150..18179 , 18180..18209 , 18210..18239 , 18240..18269 , 18270..18299 , 18300..18329 , 18330..18359 , 18360..18389 , 18390..18419 , 18420..18449 , 18450..18479 , 18480..18509 , 18510..18539 , 18540..18569 , 18570..18599 , 18600..18629 , 18630..18659 , 18660..18689 , 18690..18719 , 18720..18749 , 18750..18779 , 18780..18809 , 18810..18839 , 18840..18869 , 18870..18899 , 18900..18929 , 18930..18959 , 18960..18989 , 18990..19019 , 19020..19049 , 19050..19079 , 19080..19109 , 19110..19139 , 19140..19169 , 19170..19199 , 19200..19229 , 19230..19259 , 19260..19289 , 19290..19319 , 19320..19349 , 19350..19379 , 19380..19409 , 19410..19439 , 19440..19469 , 19470..19499 , 19500..19529 , 19530..19559 , 19560..19589 , 19590..19619 , 19620..19649 , 19650..19679 , 19680..19709 , 19710..19739 , 19740..19769 , 19770..19799 , 19800..19829 , 19830..19859 , 19860..19889 , 19890..19919 , 19920..19949 , 19950..19979 , 19980..20009 , 20010..20039 , 20040..20069 , 20070..20099 , 20100..20129 , 20130..20159 , 20160..20189 , 20190..20219 , 20220..20249 , 20250..20279 , 20280..20309 , 20310..20339 , 20340..20369 , 20370..20399 , 20400..20429 , 20430..20459 , 20460..20489 , 20490..20519 , 20520..20549 , 20550..20579 , 20580..20609 , 20610..20639 , 20640..20669 , 20670..20699 , 20700..20729 , 20730..20759 , 20760..20789 , 20790..20819 , 20820..20849 , 20850..20879 , 20880..20909 , 20910..20939 , 20940..20969 , 20970..20999 , 21000..21029 , 21030..21059 , 21060..21089 , 21090..21119 , 21120..21149 , 21150..21179 , 21180..21209 , 21210..21239 , 21240..21269 , 21270..21299 , 21300..21329 , 21330..21359 , 21360..21389 , 21390..21419 , 21420..21449 , 21450..21479 , 21480..21509 , 21510..21539 , 21540..21569 , 21570..21599 , 21600..21629 , 21630..21659 , 21660..21689 , 21690..21719 , 21720..21749 , 21750..21779 , 21780..21809 , 21810..21839 , 21840..21869 , 21870..21899 , 21900..21929 , 21930..21959 , 21960..21989 , 21990..22019 , 22020..22049 , 22050..22079 , 22080..22109 , 22110..22139 , 22140..22169 , 22170..22199 , 22200..22229 , 22230..22259 , 22260..22289 , 22290..22319 , 22320..22349 , 22350..22379 , 22380..22409 , 22410..22439 , 22440..22469 , 22470..22499 , 22500..22529 , 22530..22559 , 22560..22589 , 22590..22619 , 22620..22649 , 22650..22679 , 22680..22709 , 22710..22739 , 22740..22769 , 22770..22799 , 22800..22829 , 22830..22859 , 22860..22889 , 22890..22919 , 22920..22949 , 22950..22979 , 22980..23009 , 23010..23039 , 23040..23069 , 23070..23099 , 23100..23129 , 23130..23159 , 23160..23189 , 23190..23219 , 23220..23249 , 23250..23279 , 23280..23309 , 23310..23339 , 23340..23369 , 23370..23399 , 23400..23429 , 23430..23459 , 23460..23489 , 23490..23519 , 23520..23549 , 23550..23579 , 23580..23609 , 23610..23639 , 23640..23669 , 23670..23699 , 23700..23729 , 23730..23759 , 23760..23789 , 23790..23819 , 23820..23849 , 23850..23879 , 23880..23909 , 23910..23939 , 23940..23969 , 23970..23999 , 24000..24029 , 24030..24059 , 24060..24089 , 24090..24119 , 24120..24149 , 24150..24179 , 24180..24209 , 24210..24239 , 24240..24269 , 24270..24299 , 24300..24329 , 24330..24359 , 24360..24389 , 24390..24419 , 24420..24449 , 24450..24479 , 24480..24509 , 24510..24539 , 24540..24569 , 24570..24599 , 24600..24629 , 24630..24659 , 24660..24689 , 24690..24719 , 24720..24749 , 24750..24779 , 24780..24809 , 24810..24839 , 24840..24869 , 24870..24899 , 24900..24929 , 24930..24959 , 24960..24989 , 24990..25019 , 25020..25049 , 25050..25079 , 25080..25109 , 25110..25139 , 25140..25169 , 25170..25199 , 25200..25229 , 25230..25259 , 25260..25289 , 25290..25319 , 25320..25349 , 25350..25379 , 25380..25409 , 25410..25439 , 25440..25469 , 25470..25499 , 25500..25529 , 25530..25559 , 25560..25589 , 25590..25619 , 25620..25649 , 25650..25679 , 25680..25709 , 25710..25739 , 25740..25769 , 25770..25799 , 25800..25829 , 25830..25859 , 25860..25889 , 25890..25919 , 25920..25949 , 25950..25979 , 25980..26009 , 26010..26039 , 26040..26069 , 26070..26099 , 26100..26129 , 26130..26159 , 26160..26189 , 26190..26219 , 26220..26249 , 26250..26279 , 26280..26309 , 26310..26339 , 26340..26369 , 26370..26399 , 26400..26429 , 26430..26459 , 26460..26489 , 26490..26519 , 26520..26549 , 26550..26579 , 26580..26609 , 26610..26639 , 26640..26669 , 26670..26699 , 26700..26729 , 26730..26759 , 26760..26789 , 26790..26819 , 26820..26849 , 26850..26879 , 26880..26909 , 26910..26939 , 26940..26969 , 26970..26999 , 27000..27029 , 27030..27059 , 27060..27089 , 27090..27119 , 27120..27149 , 27150..27179 , 27180..27209 , 27210..27239 , 27240..27269 , 27270..27299 , 27300..27329 , 27330..27359 , 27360..27389 , 27390..27419 , 27420..27449 , 27450..27479 , 27480..27509 , 27510..27539 , 27540..27569 , 27570..27599 , 27600..27629 , 27630..27659 , 27660..27689 , 27690..27719 , 27720..27749 , 27750..27779 , 27780..27809 , 27810..27839 , 27840..27869 , 27870..27899 , 27900..27929 , 27930..27959 , 27960..27989 , 27990..28019 , 28020..28049 , 28050..28079 , 28080..28109 , 28110..28139 , 28140..28169 , 28170..28199 , 28200..28229 , 28230..28259 , 28260..28289 , 28290..28319 , 28320..28349 , 28350..28379 , 28380..28409 , 28410..28439 , 28440..28469 , 28470..28499 , 28500..28529 , 28530..28559 , 28560..28589, >>Next

Miscellaneous_Word_Problems/203188: How many points in the first quadrant satisfy both of the following conditions: (i) the x and y coordinates are positive integers and (ii) x + y ≤ 5? 1 solutions Answer 153301 by jim_thompson5910(28595)   on 2009-07-15 17:32:35 (Show Source): You can put this solution on YOUR website! Since "x" and "y" are both positive integers (ie positive whole numbers 1, 2, 3, 4, etc...) just ask yourself: "what two numbers add to another number that is either less than or equal to 5?" So for instance, 2+2=4 which is less than or equal to 5. Also, 1+2=3 which is also less than or equal to 5. Furthermore, 2+3=5 which is less than or equal to 5. These values can be thought of as the ordered pairs (2,2), (1,2), and (2,3) Here are all of the possible ordered pairs (0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (2, 0), (2, 1), (2, 2), (2, 3), (3, 0), (3, 1), (3, 2), (4, 0), (4, 1), (5, 0) So for example, the ordered pair (1,4) means that 1+4=5 which is less than or equal to 5.

Square-cubic-other-roots/203195: I need help rationalizing.... 9 times the cubic root 2 over/divided by the cubic root 5 minus cubic root 2. Please help! Soory my keyboard will not allow me to type the equation. 1 solutions Answer 153300 by jim_thompson5910(28595)   on 2009-07-15 17:23:25 (Show Source): You can put this solution on YOUR website! Start with the given expression. Let and Replace each with "x" and each with "y" Now this is where you have to get a bit creative. It's very helpful to remember that . Why is this important? Because if you cube a cube root, they will both cancel out (leaving the stuff inside as a remainder). So if we multiply both numerator and denominator by , we get: Use the identity given above Plug in and Cube the cube roots (to eliminate them) Subtract Reduce Combine the roots. Multiply Use the identity Square 5 to get 25 and square 2 to get 4 Distribute Combine the roots. Multiply ================================================================== Answer: So This can be verified (as I have verified it myself) with a calculator.

Trigonometry-basics/203109: i was wondering how you do these problems: The directions are to find the exact value of each expression. 6 cos 3π/4+ 2tan(-π/3) and sec ( -π/3)- cot (-5π/4) please i am struggling to do this work. I would appreciate if you would explain in detail. Thank you in advance. 1 solutions Answer 153249 by jim_thompson5910(28595)   on 2009-07-14 23:48:19 (Show Source): You can put this solution on YOUR website! Ok, first off let's bring up the unit circle This is a very handy tool. For each given angle (located on the spokes of the wheel), there is a point that lies on the circle. It turns out that the x coordinate of the point refers to the cosine of that given angle. Likewise, the y coordinate of the point refers to the sine of that angle. For example, the cosine of is since the x coordinate of this point is . Also, the sine of is since the y coordinate of this point is . You can use the unit circle to evaluate the sine/cosine of any given angle shown. Finally, remember that we have the following identities: I hope this will refresh something or help you get started. If this didn't help, please let me know or repost.

Money_Word_Problems/203103: Debbie plans on investing enough money so that she will have $2000 at the end of two years. How much money wll she have to invest now if her money will grow according to the followning formula: 2000=P(1.025)^4? When I work this problem I seem to come up with 0=0 2000=P(1.025)^4 1.025^4=1.103812891 I don't understand. 1 solutions Answer 153235 by jim_thompson5910(28595)   on 2009-07-14 19:56:30 (Show Source): You can put this solution on YOUR website! I'm assuming that you have the correct formula... Start with the given equation. Raise 1.025 to the 4th power to get 1.103812 Divide both sides by 1.103812 to isolate "P". Divide Rearrange the equation and round to the nearest hundredth (ie to the nearest penny). So she needs to invest $1,811.90

Radicals/203101: This question is from textbook {{cube root of 108x^3y^7/2y^3}} 1 solutions Answer 153233 by jim_thompson5910(28595)   on 2009-07-14 19:38:04 (Show Source): You can put this solution on YOUR website! Start with the given expression. Reduce the coefficients. Divide the common variable terms by subtracting the exponents. Subtract Factor 54 into 27*2 Factor into Break up the root. Take the cube root of 27 to get 3 Take the cube root of to get "x" Take the cube root of to get "y" Rearrange the terms and combine the roots. So where

Complex_Numbers/203100: This question is from textbook Imaginary numbers x^2-2x+12,x=1-2i 1 solutions Answer 153231 by jim_thompson5910(28595)   on 2009-07-14 19:31:26 (Show Source): You can put this solution on YOUR website! Start with the given expression. Plug in FOIL Distribute Combine like terms. Plug in Multiply Subtract So This means that when

Graphs/203102: #1 What is the slope of the line containing the points (-1,4) and (3,2)? a. -2 b. 1/2 c. -1/2 d. 2 -------------------------------------------------------------------------------- Question 2 (Multiple Choice Worth 1 points) Choose the best description for graphing the point (3,-4). a. Go to the left three and up four. b. Go to the right three and up four. c. Go to the right three and down four. d. Go to the left three and down four. -------------------------------------------------------------------------------- Question 3 (Multiple Choice Worth 1 points) Put the following equation in slope intercept form 2x - 5y = 10. a. x-y = (-2/5)x-2 b. y = (-2/5)x + 2 c. y = -2x + 10 d. y = (2/5)x -2 -------------------------------------------------------------------------------- Question 4 (Multiple Choice Worth 1 points) What is the slope of the line with equation y = (1/2)x -3? a. 1/2 b. -3 c. 2 d. -2 -------------------------------------------------------------------------------- Question 5 (Multiple Choice Worth 1 points) In the equation y=-2x +5, name the slope and the y intercept. a. slope 1/2, y intercept 5 b.slope -2, y intercept 5 c.slope 5, y intercept -2 d. slope 5, y intercept 1/2 -------------------------------------------------------------------------------- Question 6 (Multiple Choice Worth 1 points) What is the y intercept of the equation y= (2/3)x -6? a. 2/3 b. 6 c. -6 d.No y intercept -------------------------------------------------------------------------------- Question 7 (Multiple Choice Worth 1 points) Choose the best directions for graphing the equation y= (1/2)x + 1. a. Graph the point (0,1) and count up 1 and to the right 2 to find another point on the line. b. Graph the point (1,0) and count up 1 and to the right 2 to find another point on the line. c. Graph the point (0,1) and count to the right 1 and up 2 to find another point on the line. d. Graph the point (1,0) and count to the right 1 and up 2 to find another point on the line. -------------------------------------------------------------------------------- Question 8 (Multiple Choice Worth 1 points) What kind of line is created when you graph the equation x=2? a.Horizontal line b.Oblique line c.Vertical line d.No line is created. -------------------------------------------------------------------------------- Question 9 (Multiple Choice Worth 1 points) What is the slope of the line x=2? a.Zero b.Undefined c.2 d.Not enough information. -------------------------------------------------------------------------------- Question 10 (Multiple Choice Worth 1 points) What is the slope of the line y= -3? a.Zero b.Undefined c.-3 d.Not enough information. -------------------------------------------------------------------------------- Question 11 (Multiple Choice Worth 1 points) Choose the equation that is written in standard form. a.y= -3x +4 b.2y = x+8 c.x = 3y-6 d.4x -2y = 10 -------------------------------------------------------------------------------- Question 12 (Multiple Choice Worth 1 points) Find the x intercept of the graph represented by the equation 3x + 4y = 12. a.(0,3) b.(3,0) c.(4,0) d.(0,4) -------------------------------------------------------------------------------- Question 13 (Multiple Choice Worth 1 points) Find the y intercept of the graph represented by the equation 3x + 4y = 12. a.(0,3) b.(3,0) c.(4,0) d.(0,4) -------------------------------------------------------------------------------- Question 14 (Multiple Choice Worth 1 points) Choose ALL the points that are solutions to the equation y= 2x -5. a.(0,-5) b.(1,3) c.(-1,-7) d.(1/2, -4) -------------------------------------------------------------------------------- Question 15 (Multiple Choice Worth 1 points) Choose the equation of the line with a slope of 2/3 and y passing through the point (6,-2). a.y= (2/3)x -6 b.y= (2/3)x -2 c.y= (2/3) x -4 d.y = (3/2) x -6 -------------------------------------------------------------------------------- Question 16 (Multiple Choice Worth 1 points) Choose the equation of the line passing through the points (-2,3) and (-4,1). a.y = x -1 b.y = (-1/3)x + 7/3 c.y= x-4 d.y = x + 5 -------------------------------------------------------------------------------- Question 17 (Multiple Choice Worth 1 points) Choose the equation of the vertical line passing through the point (-2,1). a.x = -2 b.x = 1 c.y = -2 d.y = 1 -------------------------------------------------------------------------------- Question 18 (Multiple Choice Worth 1 points) Choose the equation of the line passing through the points (3,4) and (1,-6). a.y = 5x - 19 b.y = (1/5)x + 15/5 c.y = (1/5)x - 2 d.y = 5x - 11 -------------------------------------------------------------------------------- Question 19 (Multiple Choice Worth 1 points) Choose the ordered pair that is the solution to the following system of equations: y= 2x + 1 and y = x -5. a.(-6, -1) b.(-6,-11) c.(4,9) d.(4,-1) -------------------------------------------------------------------------------- Question 20 (Multiple Choice Worth 1 points) Choose the orderred pair that is the solution to the following system of equations: -3x + y = 4 and -7x + y = 12. a.(-2,-2) b.(-2,2) c.(2,-2) d.(2,2) -------------------------------------------------------------------------------- Question 21 (Multiple Choice Worth 1 points) Choose the ordered pair that is the solution to the following system of equations: 5x - 2y = 3 and 3x + 4y = 7. a.(1,-1) b.(-1,1) c.(-1,-1) d.(1,1) -------------------------------------------------------------------------------- Question 22 (Essay Worth 1 points) Give an ordered pair that is the solution to the equation y = 5x + 2. -------------------------------------------------------------------------------- Question 23 (Multiple Choice Worth 1 points) Choose the equation of the line that is perpendicular to the line shown in the graph and passes through the point shown in the graph. a.y = x - ll b.y= 3x – 11 c.y = -3x - 11 d.y = -x + 11 -------------------------------------------------------------------------------- Question 24 (Multiple Choice Worth 1 points) Choose the equation of the line that is perpendicular to the line shown in the graph and passes through the point shown in the graph. a.y = 2x + 7 b.y = -2x + 10 c.y = 2x - 10 -------------------------------------------------------------------------------- Question 25 (Multiple Choice Worth 1 points) The figure shown below looks like a trapezoid. By definition, a trapezoid is a four-sided figure with at least two parallel sides. Choose the best description below of how to determine whether the figure shown is or is not a trapezoid. a.Find the slope of the line between point (-9,5) and point (-3,8). Then find the slope of the line between point (4,-1) and point (3,5). If the slopes are opposite reciprocals, then the figure is a trapezoid. b.Find the slope of the line between point (-3,8) and point (3,5). Then find the slope of the line between point (-9,5) and point (4,-1). If the slopes are the same, then the figure is a trapezoid. c.Find the slope of the line between point (-9,5) and point (3,5). Then find the slope of the line between point (-3,8) and point (4,-1). If the slopes are the same, then the figure is a trapezoid. d.Find the slope of the line between point (-3,8) and point (3,5). Then find the slope of the line between point (-9,5) and point (4,-1). If the slopes are opposite reciprocals, then the figure is a trapezoid. 1 solutions Answer 153229 by jim_thompson5910(28595)   on 2009-07-14 19:23:05 (Show Source): You can put this solution on YOUR website! We're not here to do your homework. If that's the case, you might as well give us your degree, your job, and any money that you make from it. Not much fun is it? If you are struggling with the problems, please break them up and show that you've attempted to solve them. Also, post the problems that are giving you the most trouble.

Travel_Word_Problems/203094: Can you please help me write out and solve the equation for this word problem? A car leaves a town traveling at 60 miles per hour. How long will it take a second car traveling at 75 miles per hour to catch up to the first car if it leaves 2 hours later? 1 solutions Answer 153227 by jim_thompson5910(28595)   on 2009-07-14 18:17:27 (Show Source): You can put this solution on YOUR website! Let's make t = time (in hours) that the first car travels. First, let's set up the equation for the first car to leave. Recall that the distance "d" equals the speed "r" multiplied by the time "t". So this means that Plug in (the speed of the first car) ------------- Now let's set up the second equation (for the second car). Since the first car travels "t" hours, and it has a 2 hour head start, this means that the second car is going to travel "t-2" hours (eg, the first car travels t=6 hours and the second travels t-2=6-2=4 hours). So this means that we start with and plug in and replace "t" with "t-2" to get Note: the "d"s of both equations are the same "d" since the cars will have gone the distance when they meet. So we have the two equations and Start with the first equation. Plug in Distribute. Add to both sides. Subtract from both sides. Combine like terms on the left side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is which means that the first car traveled for 10 hours Subtract 2 from this answer to get . So this tells us that the second car traveled 8 hours. So it takes 8 hours for the second car to catch up and meet the first car.

Linear-systems/203093: 1.Use the Substitution method to solve the system of equations. x + y = 10 y = x + 8 2.Use the Substitution method to solve the system of equations. 3x + y = 5 4x - 7y = -10 3.Use the Substitution method to solve the system of equations. y - 2x = -5 3y - x = 5 can you please help me solve each step by step so i can under stand 1 solutions Answer 153222 by jim_thompson5910(28595)   on 2009-07-14 17:31:26 (Show Source): You can put this solution on YOUR website! # 1 Start with the given system of equations: Start with the second equation. ------------------------------------------- Move onto the first equation. Plug in . Combine like terms on the left side. Subtract from both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ------------------------------------------- Since we know that , we can use this to find . Go back to the second equation. Plug in . Add So the solutions are and . This means that the system is consistent and independent. Notice when we graph the equations, we see that they intersect at . So this visually verifies our answer. Graph of (red) and (green) # 2 Start with the given system of equations: Start with the first equation. Subtract from both sides. Rearrange the terms and simplify. ------------------------------------------- Move onto the second equation. Now plug in . Distribute. Combine like terms on the left side. Add to both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ------------------------------------------- Since we know that , we can use this to find . Go back to the first equation. Plug in . Multiply. Subtract from both sides. Combine like terms on the right side. So the solutions are and . This means that the system is consistent and independent. Notice when we graph the equations, we see that they intersect at . So this visually verifies our answer. Graph of (red) and (green) # 3 Start with the given system of equations: Start with the first equation. Add to both sides. Rearrange the terms and simplify. ------------------------------------------- Move onto the second equation. Now plug in . Distribute. Combine like terms on the left side. Add to both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ------------------------------------------- Since we know that , we can use this to find . Go back to the first equation. Plug in . Multiply. Add to both sides. Combine like terms on the right side. So the solutions are and . This means that the system is consistent and independent. Notice when we graph the equations, we see that they intersect at . So this visually verifies our answer. Graph of (red) and (green)

Exponential-and-logarithmic-functions/203087: Find the absolute maximum and absolute minimum values of the function below. If an absolute maximum or minimum does not exist, enter NONE. f(x) = x^2 + 250/x on the open interval (0,infinity ) I know that the absolute max is the answer NONE but I can not figure out the absolute min can someone help please thanks 1 solutions Answer 153219 by jim_thompson5910(28595)   on 2009-07-14 16:28:18 (Show Source): You can put this solution on YOUR website! Here are the basics to finding the extrema (ie the max/min) Step 1) Derive the function f(x) to get f'(x) Since this means that or simply Step 2) Set the derivative function f'(x) equal to zero Multiply both sides by and rearrange the equation. Add 250 to both sides. Take the square root of both sides Simplify the square root Evaluate the square root Now because the domain is , this means that we must reject (as it isn't in the domain) So the max/min occurs at the approximate value (or at the endpoints) Now if you graph the function (or evaluate a list of x values), you'll find that there is no max value (so you are correct). So this means that the min occurs when or the min occurs at the endpoints of the interval. Because x=0 is not defined (and infinity is not a number), this means that the min must occur at Step 3) Find the minimum value of the function f(x) From here, simply plug in the x value that generates the min value, which in this case is to get So the absolute min is approximately f(x)=31.6228 and occurs at x=15.8114 Note: if you want to keep things exact, then you can plug in to get the min value: . This would then mean that the absolute min of occurs at

Pythagorean-theorem/203082: the sum of two numbers is 12. their product is 32. find the numbers. not good at word problem so i am have trouble setting it up. once i get it set up i can work its just braking it down to set it up. please help. 1 solutions Answer 153199 by jim_thompson5910(28595)   on 2009-07-14 14:29:05 (Show Source): You can put this solution on YOUR website! Let x = first number y = second number "the sum of two numbers is 12" translates to and "their product is 32" means that So you basically have the system I'll let you take it from there. Let me know if you need more help.

Linear-systems/203073: -3x+y=3 and 2x-y=-1 what is the solution 1 solutions Answer 153190 by jim_thompson5910(28595)   on 2009-07-14 13:27:04 (Show Source): You can put this solution on YOUR website! Start with the given system of equations: Start with the first equation. Add to both sides. Rearrange the terms and simplify. ------------------------------------------- Move onto the second equation. Now plug in . Distribute. Combine like terms on the left side. Add to both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ------------------------------------------- Since we know that , we can use this to find . Go back to the first equation. Plug in . Multiply. Subtract from both sides. Combine like terms on the right side. So the solutions are and . This means that the system is consistent and independent. Notice when we graph the equations, we see that they intersect at . So this visually verifies our answer. Graph of (red) and (green)

Polynomials-and-rational-expressions/203062: I have to perform the operation of: k^2+3k-10 3 _________ x ________ 36 k^2-4 I tried to look for simplification by factoring out the first numerator to (k+5)(k-2). But I don't see how I can factor out a (k-2) from the second denominator because there's no k value in the expression. However, the choices for answers indicate that it is cancelled out somehow. Where do I go from here? 1 solutions Answer 153184 by jim_thompson5910(28595)   on 2009-07-14 13:01:45 (Show Source): You can put this solution on YOUR website! Start with the given expression. Factor the first numerator Factor the first denominator Factor the second denominator (hint: use the difference of squares formula) Combine the fractions. Highlight the common terms. Cancel out the common terms. Simplify Distribute So simplifies to In other words,

Polynomials-and-rational-expressions/203065: (n^2+5n+6/n^2+7n+12)(n^2+9n+20/n^2+11n+30)= ? 1 solutions Answer 153181 by jim_thompson5910(28595)   on 2009-07-14 12:37:08 (Show Source): You can put this solution on YOUR website! Start with the given expression. Factor to get . Factor to get . Factor to get . Factor to get . Combine the fractions. Highlight the common terms. Cancel out the common terms. Simplify. So simplifies to . In other words,

Polynomials-and-rational-expressions/203014: (3x^2 + 4x – 3) + (2x^2 – 3x -1) – (x^2 + x + 7) I just do not get this. I do not understand any algebra at all. It might as well be Chineese. I am in my 5th week of 10 and am failing. I need to pass this class. I need ti understand. The light just won't go on? Please help, Angela 1 solutions Answer 153155 by jim_thompson5910(28595)   on 2009-07-13 23:13:04 (Show Source): You can put this solution on YOUR website! When we want to add two variable expressions like "x" and "2x", what we're really doing is adding the coefficients. So x+2x=3x. Remember "2x" is the same as "x+x". So x+2x=x+x+x=3x Start with the given expression. Distribute Group like terms. Combine like terms. So

Equations/202989: 3-(x/3)=2x/5 This is actually a rational equation but i couldn't find it in the topics. 1 solutions Answer 153142 by jim_thompson5910(28595)   on 2009-07-13 19:49:10 (Show Source): You can put this solution on YOUR website! Start with the given equation. Multiply EVERY term by the LCD to clear any fractions. Multiply and simplify. Subtract from both sides. Subtract from both sides. Combine like terms on the left side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is which approximates to .

Graphs/202967: What is an equation of the points on the graph(0,5) and (1,1) ? 1 solutions Answer 153106 by jim_thompson5910(28595)   on 2009-07-13 15:27:21 (Show Source): You can put this solution on YOUR website! First let's find the slope of the line through the points and Note: is the first point . So this means that and . Also, is the second point . So this means that and . Start with the slope formula. Plug in , , , and Subtract from to get Subtract from to get Reduce So the slope of the line that goes through the points and is Now let's use the point slope formula: Start with the point slope formula Plug in , , and Distribute Multiply Add 5 to both sides. Combine like terms. Simplify So the equation that goes through the points and is Notice how the graph of goes through the points and . So this visually verifies our answer. Graph of through the points and

Linear-systems/202943: Explain how to apply elimination in solving a system of equations; apply substitution in solving a system of equations. Demonstrate each technique in solving the system 2x + 7y = 12 5x - 4y = -13 1 solutions Answer 153104 by jim_thompson5910(28595)   on 2009-07-13 14:38:43 (Show Source): You can put this solution on YOUR website! Start with the given system of equations: Start with the first equation. Subtract from both sides. Divide both sides by to isolate . Rearrange the terms and simplify. ------------------------------------------- Move onto the second equation. Now plug in . Distribute. Multiply both sides by the LCD to clear any fractions. Distribute and multiply. Combine like terms on the left side. Add to both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ------------------------------------------- Since we know that , we can use this to find . Go back to the first equation. Plug in . Multiply. Add to both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. So the solutions are and . This means that the system is consistent and independent. Notice when we graph the equations, we see that they intersect at . So this visually verifies our answer. Graph of (red) and (green)

Polynomials-and-rational-expressions/202948: Factor: 64x^3+27y^3 1 solutions Answer 153103 by jim_thompson5910(28595)   on 2009-07-13 14:36:27 (Show Source): You can put this solution on YOUR website! Start with the given expression. Rewrite as . Rewrite as . Now factor by using the sum of cubes formula. Remember the sum of cubes formula is Multiply ----------------------------------- Answer: So factors to . In other words,

Graphs/202960: Could you please help me out? Find an equation of the line having the given slope and containing the given point. m=7/8, (5,-2) The equation of the line is y = Thanks for your help!! 1 solutions Answer 153102 by jim_thompson5910(28595)   on 2009-07-13 14:34:02 (Show Source): You can put this solution on YOUR website! If you want to find the equation of line with a given a slope of which goes through the point (5, -2), you can simply use the point-slope formula to find the equation: ---Point-Slope Formula--- where is the slope, and is the given point So lets use the Point-Slope Formula to find the equation of the line Plug in , , and (these values are given) Rewrite as Distribute Multiply and to get Subtract 2 from both sides to isolate y Combine like terms and to get (note: if you need help with combining fractions, check out this solver) ------------------------------------------------------------------------------------------------------------ Answer: So the equation of the line with a slope of which goes through the point (5, -2) is: which is now in form where the slope is and the y-intercept is

Graphs/202961: I need help with this algebra problem... Find an equation of the line containing the given pair of points. (-7,-5) & (-3,-2) in slope intercept form y= Thank you for your help!! 1 solutions Answer 153101 by jim_thompson5910(28595)   on 2009-07-13 14:32:00 (Show Source): You can put this solution on YOUR website! First let's find the slope of the line through the points and Note: is the first point . So this means that and . Also, is the second point . So this means that and . Start with the slope formula. Plug in , , , and Subtract from to get Subtract from to get So the slope of the line that goes through the points and is Now let's use the point slope formula: Start with the point slope formula Plug in , , and Rewrite as Rewrite as Distribute Multiply Subtract 5 from both sides. Combine like terms. note: If you need help with fractions, check out this solver. So the equation that goes through the points and is

Expressions-with-variables/202957: Please help me solve this equation. find f(1) for f(x)=2x^3-3x^2+x-21 1 solutions Answer 153100 by jim_thompson5910(28595)   on 2009-07-13 14:28:45 (Show Source): You can put this solution on YOUR website! Note: we're not solving the equation, we're simply plugging in a value for "x" and evaluating. Start with the given equation. Plug in . Cube to get . Square to get . Multiply and to get . Multiply and to get . Combine like terms. So the answer is

Expressions-with-variables/202875: Hi I need help with an area of a trapezoid problem please. The area of a trapezoid is given by A=1/2h(a+b) Solve for a. Not even sure where to start on this one. 1 solutions Answer 153033 by jim_thompson5910(28595)   on 2009-07-12 20:54:29 (Show Source): You can put this solution on YOUR website! Start with the given equation. Multiply both sides by 2. Distribute Subtract hb from both sides. Divide both sides by "h" to isolate "a". So the solution is

Radicals/202871: This question is from textbook college algebra Can you please help me find the solution: Me: (-2+i radical 4)2 Me: (-2)2+2(-2)(i radical 4)+(i radical 4)2 Me: 4-4i radical 4 + 4i(squared) I know this is wrong... 1 solutions Answer 153032 by jim_thompson5910(28595)   on 2009-07-12 20:52:10 (Show Source): You can put this solution on YOUR website! I'm assuming that the expression is right???? Start with the given expression. Factor Break up the square root. Replace with "i". Take the square root of 4 to get 2. Rearrange the terms. FOIL. Multiply. Replace with . Note: . Multiply. Combine like terms. So .

Radicals/202863: This question is from textbook college algebra How do you solve: 1 solutions Answer 153028 by jim_thompson5910(28595)   on 2009-07-12 20:07:03 (Show Source): You can put this solution on YOUR website! Start with the given expression. Factor Break up the roots Replace with "i" Take the square root of 16 to get 4. Take the square root of 81 to get 9. Rearrange the terms. Multiply Combine like terms. So

Linear_Algebra/202859: i was wondering how would you do this problem and find its products : (x+3)^3 and thank you in advance for helping me 1 solutions Answer 153015 by jim_thompson5910(28595)   on 2009-07-12 18:52:45 (Show Source): You can put this solution on YOUR website! Method # 1: The somewhat long way Start with the given expression. Factor FOIL Expand Distribute Combine like terms. So expands and simplifies to . In other words, If you didn't like method # 1, then... Method # 2: The shorter way (if you're familiar with this method) Start with the given expression To expand this, we're going to use binomial expansion. So let's look at Pascal's triangle: 1    1   1    1   2   1    1   3   3   1    Looking at the row that starts with 1,3, etc, we can see that this row has the numbers: 1, 3, 3, and 1 These numbers will be the coefficients of our expansion. So to expand , simply follow this procedure: Write the first coefficient. Multiply that coefficient with the first binomial term and then the second binomial term . Repeat this until all of the coefficients have been written. Once that has been done, add up the terms like this: Notice how the coefficients are in front of each term. However, we're not done yet. Looking at the first term , raise to the 3rd power and raise to the 0th power. Looking at the second term raise to the 2nd power and raise to the 1st power. Continue this until you reach the final term. Notice how the exponents of are stepping down and the exponents of are stepping up. So the fully expanded expression should now look like this: Distribute the exponents Multiply Multiply the terms with their coefficients So expands and simplifies to . In other words, (which is what we got before)

Exponents/202846: Solve. e^(4x^2+12x+9)= 1 1 solutions Answer 153009 by jim_thompson5910(28595)   on 2009-07-12 18:14:43 (Show Source): You can put this solution on YOUR website! Start with the given equation. Take the natural log of both sides Evaluate the natural log of to get Evaluate the natural log of 1 to get 0 Notice that the quadratic is in the form of where , , and Let's use the quadratic formula to solve for "x": Start with the quadratic formula Plug in , , and Square to get . Multiply to get Subtract from to get Multiply and to get . Take the square root of to get . or Break up the expression. or Combine like terms. or Simplify. So the only solution is

logarithm/202843: Given Log(a)=0.0441 and log(b)=0.7964; find Log[sqrt a/b] (find log square root of (a) over (b). 1 solutions Answer 153007 by jim_thompson5910(28595)   on 2009-07-12 18:00:00 (Show Source): You can put this solution on YOUR website! Start with the given expression. Break up the log using the identity Convert to exponential notation. Rewrite the first log using the identity Plug in and Multiply Combine like terms. So when and

Exponents/202841: Solve 3^2x+1=1/27 1 solutions Answer 153006 by jim_thompson5910(28595)   on 2009-07-12 17:52:17 (Show Source): You can put this solution on YOUR website! Start with the given equation. Rewrite 27 as Rewrite as Since the bases are equal, this means that the exponents are equal. Subtract from both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is

Functions/202840: Decreasing cube. Each of the three dimensions of a cube with sides of length s centimeters is decreased by a whole number of centimeters. The new volume in cubic centimeters is given by V(s)  s3  13s2  54s  72. a) Find V(10). b) If the new width is s  6 centimeters, then what are the new length and height? c) Find the volume when s  10 by multiplying the length, width, and height. Have tried the following: For A.) v(10)=10^3-13(10)^2+54(10)-72 For B. and C. I am completely lost at where to even begin. 1 solutions Answer 153005 by jim_thompson5910(28595)   on 2009-07-12 17:49:30 (Show Source): You can put this solution on YOUR website! Hi there, Something must have not been copied correctly because there are a lot of box symbols all over the page. So I'm going to make a lot assumptions of what the problem really is. a) Start with the given function. Plug in (ie replace each "s" with 10). Cube to get . Square to get . Multiply and to get . Multiply and to get . Combine like terms. b) Start with the volume equation. Plug in the given expressions Divide both sides by . Factor the numerator Cancel out the common terms. Simplify Rearrange the equation Factor So the new length and height are and c) I'll let you attempt this one on your own. Repost if you still need help. Note: the answer is going to be the same as the answer in part a) (so you'll have something to check)

Graphs/202837: how to compute the slope of the line segment determined by the pair of points(8,3) and (-6,9) 1 solutions Answer 153004 by jim_thompson5910(28595)   on 2009-07-12 17:38:27 (Show Source): You can put this solution on YOUR website! Note: is the first point . So this means that and . Also, is the second point . So this means that and . Start with the slope formula. Plug in , , , and Subtract from to get Subtract from to get Reduce So the slope of the line that goes through the points and is