Tutor:

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

---

Recent problems solved by 'jim_thompson5910'

jim_thompson5910 answered: 28582 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

Sequences-and-series/199680: Find the x values so that the infinite series become finite. (infinite) (Sigma) 3(x-2)^(k-1) (k=1) Could you help me with this question? Thank you. 1 solutions Answer 150060 by jim_thompson5910(28594)   on 2009-06-10 18:58:21 (Show Source): You can put this solution on YOUR website! The series will converge if and only if So this means that you simply need to solve to find the set of x values (since in this case, )

Probability-and-statistics/199673: This is my last question please please help I'm not good in math with word problems. A couple plan to have exactly four children. (a) Construct a tree diagram and list the sample space. (b) Find the probability that the family has at least three boys. 1 solutions Answer 150059 by jim_thompson5910(28594)   on 2009-06-10 18:43:06 (Show Source): You can put this solution on YOUR website! a) Let's build the tree: Using the tree, we get the sample space: {b,b,b,b} {b,b,b,g} {b,b,g,b} {b,b,g,g} {b,g,b,b} {b,g,b,g} {b,g,g,b} {b,g,g,g} {g,b,b,b} {g,b,b,g} {g,b,g,b} {g,b,g,g} {g,g,b,b} {g,g,b,g} {g,g,g,b} {g,g,g,g} So for instance {b,g,g,b} means that the couple had a boy, girl, girl, and then a boy (in that order). Note: recall, the sample space is the set of ALL possible outcomes. ------------------------------- b) Since we want to know the chances of the couple having "at least three boys", this means they want to know the chances of having 3 boys OR 4 boys (since at least means that figure or more). Looking back at the list of all possible outcomes (ie the sample space) from part a), we see that we have the combinations for 3 boys: {b,b,b,g}, {b,b,g,b}, {b,g,b,b}, and {g,b,b,b} So there are 4 cases where the couple would have 3 boys. So ----------------------------- Also, since there is only ONE way to have 4 boys (of a total of 4 children), this means that ------------------------------------------------------- Now simply add the two probabilities to find the chances of either one occurring: So the probability of the couple having AT LEAST 3 boys is which is 0.3125 in decimal form which gives a 31.25% chance.

Quadratic_Equations/199645: how do I solve these problems 1. 6x^2+7x=5 2.12+5x-2x^2=0 3.5x^2+16x-12=0 1 solutions Answer 150058 by jim_thompson5910(28594)   on 2009-06-10 18:26:19 (Show Source): You can put this solution on YOUR website! # 1 Start with the given equation. Subtract 5 from both sides. Notice we have a quadratic 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 Rewrite as Add to 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 solutions are or # 2 Start with the given equation. Rearrange the terms. Notice we have a quadratic 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 Rewrite as Add to 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 solutions are or # 3 Start with the given equation. Notice we have a quadratic 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 Rewrite as Add to to get Multiply and to get . Simplify the square root (note: If you need help with simplifying square roots, check out this solver) or Break up the expression. or Reduce. So the solutions are or which approximate to or

Exponents/199666: Please help me answer this: 3^(4x) = 27^(x-2) Thanks. 1 solutions Answer 150057 by jim_thompson5910(28594)   on 2009-06-10 18:22:32 (Show Source): You can put this solution on YOUR website! Start with the given equation. Rewrite 27 as Multiply the exponents. Since the bases are equal, the exponents are equal. Distribute. Subtract from both sides. Combine like terms on the left side. ---------------------------------------------------------------------- Answer: So the solution is

Equations/199678: Solve the value of x in the equation below: 2(3x+1)=10x-6 1 solutions Answer 150056 by jim_thompson5910(28594)   on 2009-06-10 18:19:55 (Show Source): You can put this solution on YOUR website! Start with the given equation. Distribute. Subtract from both sides. Subtract from both sides. Combine like terms on the left side. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is

Quadratic_Equations/199679: please help me solve and graph this equation: 5x^2-20x-60=0 1 solutions Answer 150054 by jim_thompson5910(28594)   on 2009-06-10 18:19:15 (Show Source): You can put this solution on YOUR website! Start with the given equation. Notice we have a quadratic in the form of where , , and Let's use the quadratic formula to solve for "x": Start with the quadratic formula Plug in , , and Negate to get . Square to get . Multiply to get Rewrite as Add to 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 solutions are or

logarithm/199628: what values of x cannot possibly be solutions of the following equation? LOGa(3x+1)=2 1 solutions Answer 150020 by jim_thompson5910(28594)   on 2009-06-10 13:44:35 (Show Source): You can put this solution on YOUR website! Recall that you CANNOT evaluate the log of a negative value or 0. So this means that Start with the given inequality. Subtract from both sides. Combine like terms on the right side. Divide both sides by to isolate . This means that any values of x that are either equal to or are less than are NOT allowed.

Linear-systems/199621: The difference between two numbers is 18. If the larger number is 5 more than twice the smaller number, what is their sum? 1 solutions Answer 150018 by jim_thompson5910(28594)   on 2009-06-10 13:35:30 (Show Source): You can put this solution on YOUR website! "The difference between two numbers is 18" translates to "If the larger number is 5 more than twice the smaller number" translates to Start with the first equation. Plug in Combine like terms on the left side. Subtract from both sides. Combine like terms on the right side. Go back to the second equation. Plug in Multiply Add Now simply add up x and y to get: So

logarithm/199620: Solve 3^x-1= 9^2x 1 solutions Answer 150017 by jim_thompson5910(28594)   on 2009-06-10 13:31:18 (Show Source): You can put this solution on YOUR website! Start with the given equation. Rewrite 9 as Multiply the exponents Multiply Since the bases are equal, the exponents are equal. 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 approximates to .

absolute-value/199616: name two different numbers that have an absolute value of 7 1 solutions Answer 150016 by jim_thompson5910(28594)   on 2009-06-10 12:52:45 (Show Source): You can put this solution on YOUR website! Let x=unknown number What you want to solve is Break up the absolute value to get or Note: if , then or ) So the solutions are or which means that the absolute value of -7 or 7 is 7.

Quadratic_Equations/199600: I'm not to sure if this is the right section for my question but here goes: I'm working on factorizing quadratic trinomials and here is a question I need help with factorizing this expression. 1) 20x^2-44x+21 1 solutions Answer 150015 by jim_thompson5910(28594)   on 2009-06-10 12:32:40 (Show Source): You can put this solution on YOUR website! Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is . Now multiply the first coefficient by the last term to get . Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ? To find these two numbers, we need to list all of the factors of (the previous product). Factors of : 1,2,3,4,5,6,7,10,12,14,15,20,21,28,30,35,42,60,70,84,105,140,210,420 -1,-2,-3,-4,-5,-6,-7,-10,-12,-14,-15,-20,-21,-28,-30,-35,-42,-60,-70,-84,-105,-140,-210,-420 Note: list the negative of each factor. This will allow us to find all possible combinations. These factors pair up and multiply to . 1*420 2*210 3*140 4*105 5*84 6*70 7*60 10*42 12*35 14*30 15*28 20*21 (-1)*(-420) (-2)*(-210) (-3)*(-140) (-4)*(-105) (-5)*(-84) (-6)*(-70) (-7)*(-60) (-10)*(-42) (-12)*(-35) (-14)*(-30) (-15)*(-28) (-20)*(-21) Now let's add up each pair of factors to see if one pair adds to the middle coefficient : First Number Second Number Sum 1 420 1+420=421 2 210 2+210=212 3 140 3+140=143 4 105 4+105=109 5 84 5+84=89 6 70 6+70=76 7 60 7+60=67 10 42 10+42=52 12 35 12+35=47 14 30 14+30=44 15 28 15+28=43 20 21 20+21=41 -1 -420 -1+(-420)=-421 -2 -210 -2+(-210)=-212 -3 -140 -3+(-140)=-143 -4 -105 -4+(-105)=-109 -5 -84 -5+(-84)=-89 -6 -70 -6+(-70)=-76 -7 -60 -7+(-60)=-67 -10 -42 -10+(-42)=-52 -12 -35 -12+(-35)=-47 -14 -30 -14+(-30)=-44 -15 -28 -15+(-28)=-43 -20 -21 -20+(-21)=-41 From the table, we can see that the two numbers and add to (the middle coefficient). So the two numbers and both multiply to and add to Now replace the middle term with . Remember, and add to . So this shows us that . Replace the second term with . Group the terms into two pairs. Factor out the GCF from the first group. Factor out from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis. Combine like terms. Or factor out the common term --------------------------------------------- Answer: So factors to . Note: you can check the answer by FOILing to get or by graphing the original expression and the answer (the two graphs should be identical).

Rational-functions/199584: b^2-9/b^3-27 i know i have to use the cube root but i stilll dont get how this problem is sloved.. can you please explain this (: 1 solutions Answer 150010 by jim_thompson5910(28594)   on 2009-06-10 12:09:37 (Show Source): You can put this solution on YOUR website! Start with the given expression. Factor to get (use the difference of squares formula) Factor to get (use the difference of cubes formula) Highlight the common terms. Cancel out the common terms. Simplify. So simplifies to . In other words, where

Equations/199592: Hi all, im struggling to finish off some work at the moment. (There are 3 seperate equations all of the same type). Find x such that 3 - (3/2)x = 8 - 2x is satisfied. Find x such that 6 - (5/2)x = 12 + 10x is satisfied. Find x such that 6 + 3x = 18 + 15x is satisfied. Detailed explantaions of how to solve the previous would be greatly appreciated. Thnaks, Nick. 1 solutions Answer 150008 by jim_thompson5910(28594)   on 2009-06-10 12:06:37 (Show Source): You can put this solution on YOUR website! # 1 Start with the given equation. Multiply EVERY term by the LCD to clear any fractions. Multiply. Subtract from both sides. Add to both sides. Combine like terms on the left side. Combine like terms on the right side. ---------------------------------------------------------------------- Answer: So the solution is # 2 Start with the given equation. Multiply EVERY term by the LCD to clear any fractions. Multiply. Subtract from both sides. Subtract from both sides. Combine like terms on the left side. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is which in decimal form is . # 3 Start with the given equation. Subtract from both sides. Subtract from both sides. Combine like terms on the left side. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is

Equations/199604: 3M+20=2M+80 1 solutions Answer 150006 by jim_thompson5910(28594)   on 2009-06-10 12:01:55 (Show Source): You can put this solution on YOUR website! Start with the given equation. Subtract from both sides. Subtract from both sides. Combine like terms on the left side. Combine like terms on the right side. ---------------------------------------------------------------------- Answer: So the solution is

Matrices-and-determiminant/199607: Hello, I'm trying to find the inverse of the following matrix; please help. I have to show all my work for homework, but I'm getting confused. Thank you! (18 9) (-8 -4) 1 solutions Answer 150004 by jim_thompson5910(28594)   on 2009-06-10 12:00:56 (Show Source): You can put this solution on YOUR website! Solved by pluggable solver: Finding the Inverse of a 2x2 Matrix To find the inverse of the matrix , we can follow these steps: Step 1) Find the determinant The determinant of is . So this means that Since the determinant is equal to zero, this means that no inverse exists. Remember, the inverse of is . If , then a division by zero occurs, which is NOT allowed. So we can stop here. ================================================================= Answer: So the inverse of does NOT exist since (ie the determinant is 0).

Linear-equations/199610: This question is from textbook intermediate algebra 3x-y=-3 2x+y=-7 i am trying to figure out how to find the x,y,xy corrdinates so i can graph the equation to find out this problem has one solution, no solution, or infininate solutions 1 solutions Answer 150002 by jim_thompson5910(28594)   on 2009-06-10 11:59:10 (Show Source): You can put this solution on YOUR website! Start with the given system of equations: In order to graph these equations, we must solve for y first. Let's graph the first equation: Start with the first equation. Subtract from both sides. Divide both sides by to isolate . Rearrange the terms and simplify. Looking at we can see that the equation is in slope-intercept form where the slope is and the y-intercept is Since this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis So we have one point Now since the slope is comprised of the "rise" over the "run" this means Also, because the slope is , this means: which shows us that the rise is 3 and the run is 1. This means that to go from point to point, we can go up 3 and over 1 So starting at , go up 3 units and to the right 1 unit to get to the next point Now draw a line through these points to graph So this is the graph of through the points and ---------------------------------------------------------------------------------------- Now let's graph the second equation: Start with the second equation. Subtract from both sides. Rearrange the terms and simplify. Looking at we can see that the equation is in slope-intercept form where the slope is and the y-intercept is Since this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis So we have one point Now since the slope is comprised of the "rise" over the "run" this means Also, because the slope is , this means: which shows us that the rise is -2 and the run is 1. This means that to go from point to point, we can go down 2 and over 1 So starting at , go down 2 units and to the right 1 unit to get to the next point Now draw a line through these points to graph So this is the graph of through the points and ---------------------------------------------------------------------------------------- Now let's graph the two equations together: Graph of (red). Graph of (green) From the graph, we can see that the two lines intersect at the point . So the solution to the system of equations is . This means that the system has one unique solution. This also tells us that the system of equations is consistent and independent.

Polynomials-and-rational-expressions/199608: This question is from textbook Beginning and Intermediate Algebra Find any variables for which the rational expression is undefined. 5m + 2 _________ m(squared) + m - 6 I tried to set the denominator equal to 0, then factor, but I am stuck on the factoring. Any help is appreciated. 1 solutions Answer 150001 by jim_thompson5910(28594)   on 2009-06-10 11:55:24 (Show Source): You can put this solution on YOUR website! Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is . Now multiply the first coefficient by the last term to get . Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ? To find these two numbers, we need to list all of the factors of (the previous product). Factors of : 1,2,3,6 -1,-2,-3,-6 Note: list the negative of each factor. This will allow us to find all possible combinations. These factors pair up and multiply to . 1*(-6) 2*(-3) (-1)*(6) (-2)*(3) Now let's add up each pair of factors to see if one pair adds to the middle coefficient : First Number Second Number Sum 1 -6 1+(-6)=-5 2 -3 2+(-3)=-1 -1 6 -1+6=5 -2 3 -2+3=1 From the table, we can see that the two numbers and add to (the middle coefficient). So the two numbers and both multiply to and add to Now replace the middle term with . Remember, and add to . So this shows us that . Replace the second term with . Group the terms into two pairs. Factor out the GCF from the first group. Factor out from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis. Combine like terms. Or factor out the common term So factors to . From here, you can use the factored form to solve for "m"

Rational-functions/199580: b^2-9/b^3-27 i know i have to use the cube root but i stilll dont get how this problem is sloved.. can you please explain this (: 1 solutions Answer 149988 by jim_thompson5910(28594)   on 2009-06-10 01:14:42 (Show Source): You can put this solution on YOUR website! Start with the given expression. Factor to get (use the difference of squares formula) Factor to get (use the difference of cubes formula) Highlight the common terms. Cancel out the common terms. Simplify. So simplifies to . In other words, where

Graphs/199370: What is the equation of the line that contains (7, -9) and (7, 3)? 1 solutions Answer 149987 by jim_thompson5910(28594)   on 2009-06-10 00:41:45 (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 and is the second point . Start with the slope formula. Plug in , , , and Subtract from to get Subtract from to get Remember, you cannot divide by zero. So this means that the slope is undefined. Since the slope is undefined, this means that the equation of the line through the points and is (which is a vertical line).

Subset/199545: I review the one that was done with vertice. This one is alittle different please help. Create a graph with five vertices, no bridge and a loop. Thank you so much Terri 1 solutions Answer 149985 by jim_thompson5910(28594)   on 2009-06-10 00:38:32 (Show Source): You can put this solution on YOUR website! Start with five randomly drawn points. These are the vertices Now simply connect every vertex with a line. Make sure that EVERY point has two lines connecting to it (so a bridge doesn't form). In addition to connecting every vertex, draw a line from any one vertex to itself to form a loop (in green):

Geometric_formulas/199477: Create a graph with five vertices, no bridge, and a loop. 1 solutions Answer 149984 by jim_thompson5910(28594)   on 2009-06-10 00:37:02 (Show Source): You can put this solution on YOUR website! Start with five randomly drawn points. These are the vertices Now simply connect every vertex with a line. Make sure that EVERY point has two lines connecting to it (so a bridge doesn't form). In addition to connecting every vertex, draw a line from any one vertex to itself to form a loop (in green):

Trigonometry-basics/199577: prove this identity (5 marks): sin(x) + sin(x)cot(x)^2 = csc(x) thanks in advance! 1 solutions Answer 149980 by jim_thompson5910(28594)   on 2009-06-10 00:19:38 (Show Source): You can put this solution on YOUR website! ... Start with the given equation. ... Factor out the GCF ... Replace the terms in the parenthesis with . Note: ... Use the identity ... Multiply ... Factor ... Cancel out the common terms. ... Use the identity So we've proven the identity.

Linear-systems/199578: Solve the sytem of equations by graphing. Then classify the system. x+y=15 x-y=3 1 solutions Answer 149979 by jim_thompson5910(28594)   on 2009-06-10 00:13:42 (Show Source): You can put this solution on YOUR website! Start with the given system of equations: In order to graph these equations, we must solve for y first. Let's graph the first equation: Start with the first equation. Subtract from both sides. Rearrange the terms and simplify. Looking at we can see that the equation is in slope-intercept form where the slope is and the y-intercept is Since this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis So we have one point Now since the slope is comprised of the "rise" over the "run" this means Also, because the slope is , this means: which shows us that the rise is -1 and the run is 1. This means that to go from point to point, we can go down 1 and over 1 So starting at , go down 1 unit and to the right 1 unit to get to the next point Now draw a line through these points to graph So this is the graph of through the points and ------------------------------------------------------------------------------ Now let's graph the second equation: Start with the second equation. Subtract from both sides. Divide both sides by to isolate . Rearrange the terms and simplify. Looking at we can see that the equation is in slope-intercept form where the slope is and the y-intercept is Since this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis So we have one point Now since the slope is comprised of the "rise" over the "run" this means Also, because the slope is , this means: which shows us that the rise is 1 and the run is 1. This means that to go from point to point, we can go up 1 and over 1 So starting at , go up 1 unit and to the right 1 unit to get to the next point Now draw a line through these points to graph So this is the graph of through the points and ------------------------------------------------------------------------------ Now let's graph the two equations together: Graph of (red). Graph of (green) From the graph, we can see that the two lines intersect at the point . So the solution to the system of equations is . This tells us that the system of equations is consistent and independent.

Money_Word_Problems/199561: 2. A business invests $8,000 in a savings account for two years. At the beginning of the second year, an additional $2,500 is invested. At the end of the second year, the account balance is $11,445. What was the annual interest rate? Im not good at interest 1 solutions Answer 149978 by jim_thompson5910(28594)   on 2009-06-10 00:12:23 (Show Source): You can put this solution on YOUR website! Start with the interest formula. Note: since we're only concerned with year intervals, this means that (so we don't have to worry about exponents) Plug in (since $8,000 is the original amount in the account) Distribute So after one full year, you now have dollars in the account. Now because "an additional $2,500 is invested", this pushes up the amount to (just add 2500 to the last expression). Combine like terms to get So the new principal is Since after two years you have $11,445, this means that Go back to the original formula Plug in , and FOIL Subtract 11445 from both sides. Combine like terms. Notice we have a quadratic in the form of where , , and Let's use the quadratic formula to solve for r Start with the quadratic formula Plug in , , and Square to get . Multiply to get Rewrite as Add to to get Multiply and to get . Take the square root of to get . or Break up the expression. or Combine like terms. or Divide So the possible answers are or However, since you CANNOT have a negative interest rate, this means that the only possible answer is Now multiply by 100 to get . ============================================= Answer: So the annual interest rate is 5%

Proofs/199553: If you could help me with the problem I would really appreciate it. I finished it in 19 lines but i think I did a few steps wrong. 1. B=W 2. L>B 3. ~Bv~W /~(WvL) 1 solutions Answer 149975 by jim_thompson5910(28594)   on 2009-06-10 00:06:19 (Show Source): You can put this solution on YOUR website! Note: I'm going to use "<->" for "if and only if" and "*" for "and" 1. B <-> W 2. L -> B 3. ~B v ~W :. ~(W v L) ----------------------------------- 4. (B -> W) * (W -> B) 1 Material Equivalence 5. (W -> B) * (B -> W) 4 Commutation 6. B -> W 4 Simplification 7. W -> B 5 Simplification 8. B -> ~W 3 Material Implication 9. W -> ~W 7,8 Hypothetical Syllogism 10. ~W v ~W 9 Material Implication 11. ~W 10 Tautology 12. ~B 6,11 Modus Tollens 13. ~L 2,12 Modus Tollens 14. ~W * ~L 11,13 Conjunction 15. ~(W v L) 14 De Morgan's Law

Quadratic_Equations/199563: 3. Steve traveled 150 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 2 hours less. Find the speed of his vehicle. i think it is 30mph but I cannot figure out the equation 1 solutions Answer 149972 by jim_thompson5910(28594)   on 2009-06-09 23:54:16 (Show Source): You can put this solution on YOUR website! Remember, the distance-rate-time formula is Since he "traveled 150 miles at a certain speed", this means that the first equation is (simply plug in d=150) The statement "Had gone 20mph faster, the trip would have taken 2 hour less", tells us that the next equation is (the distance is the same, but the speed is now 20 mph faster and the time is 2 hours shorter) The goal is to use this system to solve for "r" (and if we want, "t" also) Start with the first equation. Divide both sides by r. Rearrange the equation Move onto the second equation. Plug in FOIL Multiply EVERY term by the LCD to clear out the fraction. Subtract from both sides. Combine and rearrange the terms. Notice we have a quadratic in the form of where , , and Let's use the quadratic formula to solve for "r": Start with the quadratic formula Plug in , , and Negate to get . Square to get . Multiply to get Rewrite as Add to 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 possible solutions are or Since a negative speed doesn't make any sense, this means that we must ignore ==================================================================== Answer: So the only solution is which means that he was traveling 30 mph. So you are correct, good job.

Rational-functions/199564: 4. The Hudson River flows at a rate of 5 miles per hour. A patrol boat travels 40 miles upriver, and returns in a total time of 6 hours. What is the speed of the boat in still water? 1 solutions Answer 149970 by jim_thompson5910(28594)   on 2009-06-09 23:43:39 (Show Source): You can put this solution on YOUR website! Let x = speed of boat in still water Start with the distance-rate-time formula Plug in and . This equation represents the upstream journey. Divide both sides by to isolate "t"; So the expression for the time it takes to go upstream can be represented by the expression ----------------------------------- Go back to the distance-rate-time formula Plug in and . This equation represents the downstream journey Divide both sides by to isolate "t"; So the expression for the time it takes to go downstream can be represented by the expression Now simply add the two time expressions to get: Now set that expression equal to the total time of 6 hours Multiply every term by the LCD to clear the denominators. FOIL Distribute Subtract from both sides. Add to both sides. Combine like terms Notice we have a quadratic 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 Rewrite as Add to 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 possible solutions are or However, you can't have a negative speed. So the only answer is ================================================================= Answer: So the speed of the boat in still water is 15 mph.

Equations/199568: 4 times an integer is 30 less than 5 times the next consecutive even integer. Find the two integers. 1 solutions Answer 149968 by jim_thompson5910(28594)   on 2009-06-09 23:42:07 (Show Source): You can put this solution on YOUR website! "4 times an integer is 30 less than 5 times the next consecutive even integer" translates to Start with the given equation. Distribute. Combine like terms on the right side. 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 two numbers are x=20 and x+2=20+2=22

Quadratic_Equations/199571: Can anyone help me with this one? It requires me to determine the number of solutions and classify the type of solutions for each of the following equations? Justify the answer please? x^2-8x+16=0 1 solutions Answer 149967 by jim_thompson5910(28594)   on 2009-06-09 23:39:44 (Show Source): You can put this solution on YOUR website! From we can see that , , and Start with the discriminant formula. Plug in , , and Square to get Multiply to get Subtract from to get Since the discriminant is equal to zero, this means that there is one real solution.

Quadratic_Equations/199572: Can anyone help me with this one? It requires me to determine the number of solutions and classify the type of solutions for each of the following equations? Justify the answer please? 2x^2-3x+7=0 1 solutions Answer 149966 by jim_thompson5910(28594)   on 2009-06-09 23:39:16 (Show Source): You can put this solution on YOUR website! From we can see that , , and Start with the discriminant formula. Plug in , , and Square to get Multiply to get Subtract from to get Since the discriminant is less than zero, this means that there are two complex solutions. In other words, there are no real solutions.

Quadratic_Equations/199573: Can anyone help me with this one? It requires me to determine the number of solutions and classify the type of solutions for each of the following equations? Justify the answer please? x^2-4x-77=0 1 solutions Answer 149965 by jim_thompson5910(28594)   on 2009-06-09 23:38:35 (Show Source): You can put this solution on YOUR website! From we can see that , , and Start with the discriminant formula. Plug in , , and Square to get Multiply to get Rewrite as Add to to get Since the discriminant is greater than zero, this means that there are two real solutions.