New!
Get regular updates about newly solved problems
via algebra.com's RSS system.
Recent problems solved by 'jim_thompson5910'
jim_thompson5910 answered: 28684 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 , 28590..28619 , 28620..28649 , 28650..28679 , 28680..28709, >>NextQuadratic-relations-and-conic-sections/452786: Can you help me write this in standard form?
y=x^2+2x+2 1 solutions
Answer 311180 by jim_thompson5910(28696) on 2011-05-22 19:31:15 (Show Source):
You can put this solution on YOUR website!To write  in standard form, we must complete the square for
 Start with the given expression.
Take half of the  coefficient  to get  . In other words,  .
Now square  to get  . In other words,
 Now add and subtract  . Make sure to place this after the "x" term. Notice how  . So the expression is not changed.
 Group the first three terms.
 Factor  to get  .
 Combine like terms.
So after completing the square,  transforms to  . So  .
So  is equivalent to  .
which is now in standard form.
|
expressions/452716: Factor x^2+6x+9 1 solutions
Answer 311179 by jim_thompson5910(28696) on 2011-05-22 19:29:04 (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  :
1,3,9
-1,-3,-9
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to  .
1*9 = 9
3*3 = 9
(-1)*(-9) = 9
(-3)*(-3) = 9
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 | 9 | 1+9=10 | | 3 | 3 | 3+3=6 | | -1 | -9 | -1+(-9)=-10 | | -3 | -3 | -3+(-3)=-6 |
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
 Condense the terms.
===============================================================
Answer:
So  factors to  .
In other words,  .
Note: you can check the answer by expanding  to get  or by graphing the original expression and the answer (the two graphs should be identical).
|
Triangles/452746: For a triangle with a perimeter of 12 and an area of 7 what are the lengths of the triangles legs?
Is there an equation that relates area and perimeter to give you leg lengths? Or something similar?
Some things I’ve tried:
1: Finding “a” “b” and “c” directly by converting a set of equations into a matrix in row echelon form. The issue is I can’t find a set of linear equations that relate a, b, and c to the triangle.
2: Listing possible solutuions for b and h from the area formula that are also factors of 14. 14 because . This is impossible because there are an infinite number of possibilities. So I went for positive integers [(b:1,2,7,14)(h:1,2,7,14)]. I didn't continue with this train of thought because something else occured to me.
3: When a triangle is split into 2 triangles and the line that splits them is perpedicular to one of the lines it touches then both of the triangles are right triangles. The pythagorian therum works for right triangles. Giving me another equation to work with. I couldn't figure out how to use it though.
4:So I square 14 giving me 196. Factoring 196 into primes gave me (2,2,7,7). I made a chart of the ways these numbers could be combined so when they were multiplied i got 14. This gave me , , and I came to the same problem again as i had with the pythagorean therum; i didn't know how to use the new information.
5: So i went back to the visually drawn triangles and labeled the sides of the original triangle a, b, and c. Then I labeled one of the split triangles sides (the numbers to the right of the letters are subscripts) a1, b1, and c1. The other right triangle was labeled a2, b2, and c2. With the sides labeled i tried to substitute various variables in for each other to see what i could come up with. For example the 2 new triangles both have a side that is equal to (1/2)c from the original triangle so i replace the corresponding variable on the smaller triangle with (1/2)c. I tried many more substitutions like that but none of them yeilded anything I thought I could use.
I am now taking a break from it. And it occurred to me that if there were an equation relating area with perimeter that would be very usefull. Also if there are any other equations that you think might be useful please send those to me as well.
One last thing, please put the things you think might help at the top of the email, and if you know how to solve this kind of problem and include the answer or the "how to" please write something along the lines of "spoiler alert" and skip some lines so it won't be visible. Sorry if i come across as a bit pushy but i really want to figure this one out and then see if i was right. 1 solutions
Answer 311169 by jim_thompson5910(28696) on 2011-05-22 19:00:12 (Show Source):
You can put this solution on YOUR website!Hint:
The perimeter of a triangle with sides 'a', 'b', and 'c' is  (ie add up the sides to get the perimeter). But you probably already knew that.
What you probably don't know, or maybe have seen and forgot, is that the area of a triangle with sides 'a', 'b', and 'c' is
 where  (this is the semiperimeter, which is half the perimeter)
Let me know if this is enough to help get you started.
Spoiler Alert:
I don't think there's a solution because solving for a,b, and c gives equations which can be graphed. The graphs can then show where the positive solutions lie. It turns out that not all the solutions are positive, which means that a triangle can't be constructed with these given properties.
|
Volume/452707: what is the volume of a circle with a diameter of 10 and a height of 16
1 solutions
Answer 311143 by jim_thompson5910(28696) on 2011-05-22 17:13:38 (Show Source):
You can put this solution on YOUR website!This sounds like a cylinder. Since the diameter is 10, the radius is
 Start with the volume of a cylinder formula.
 Plug in  and  .
 Replace  with  (note: Use more digits of  to get better accuracy).
 Square  to get  .
 Multiply  ,  and  to get  (this value is approximate).
 Round to the nearest thousandth.
|
Quadratic_Equations/451301: Hi! I was trying to solve this problem for quadratic equation: but in order to simplify it I get stuck on the part during the equation that looks like this (the + sign in the equation represents the + or - one in the actual equation: . I know that the square root of a negative number is (square root of the number as if it were positive) x i(the square root of - 1). I also know that there will be 2 solutions to this equation. But what do I do after if I need to simplify it as much as possible? Thanks for any help you may give me! 1 solutions
Answer 310352 by jim_thompson5910(28696) on 2011-05-18 17:07:04 (Show Source):
You can put this solution on YOUR website!
 Start with the given equation.
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
 Negate  to get  .
 Square  to get  .
 Multiply  to get
 Subtract  from  to get  (this is where you made your mistake).
 Multiply  and  to get  .
 Take the square root of  to get  .
 or  Break up the expression.
 or  Break up the fraction for each case.
 or  Reduce.
So the solutions are  or
For more help with the quadratic formula, see this solver.
|
Numbers_Word_Problems/451291: I am a 3-digit number less than 300. I am divisible by 2 and 5, but not 3. The sum of my digits is 7. What number am I? 1 solutions
Answer 310351 by jim_thompson5910(28696) on 2011-05-18 17:00:52 (Show Source):
You can put this solution on YOUR website!Since the number is "divisible by 2 and 5", this means that the number is divisible by 10 (since 2*5 = 10). So the number ends with a 0.
So the possible numbers are:
100, 110, 120, 130, 140, 150, 160, 170, 180, 190,
200, 210, 220, 230, 240, 250, 260, 270, 280, 290
Now add up the digits of each possible number. The digits that add to 7 will give us the number. It turns out that 160 and 250 fit this description since 1+6+0=7 and 2+5+0=7
So the numbers are 160 and 250.
|
Linear-equations/451296: Write an equation of the line the passes through (-5,8) and (-3,-8)
1 solutions
Answer 310350 by jim_thompson5910(28696) on 2011-05-18 16:56:16 (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
 Rewrite  as
 Distribute
 Multiply
 Add 8 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-equations/451297: Write an equation of the line the passes through (-5,8) and (-3,-8) 1 solutions
Answer 310349 by jim_thompson5910(28696) on 2011-05-18 16:56:04 (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
 Rewrite  as
 Distribute
 Multiply
 Add 8 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
|
Surface-area/451294: what is the surface area of a sphere with a diameter of 18 centimeters? express the answer in square meters. Round to the nearest hundreth. 1 solutions
Answer 310348 by jim_thompson5910(28696) on 2011-05-18 16:55:16 (Show Source):
You can put this solution on YOUR website!If the diameter is 18 cm, then the radius is  cm.
 Start with the surface area of a sphere formula.
 Plug in  .
 Replace  with  (note: Use more digits of  to get better accuracy).
 Square  to get  .
 Multiply  and  to get  .
 Multiply  and  to get  .
 Round to the nearest hundredth.
So the surface area of a sphere with a diameter of  cm is approximately  square cm.
I'll let you do the conversion to square meters.
|
test/451264: The graph of y=ax^2+bx+c intersects the x-axis at two points. Which statement is true about the seros of the function? 1 solutions
Answer 310344 by jim_thompson5910(28696) on 2011-05-18 16:07:34 (Show Source):
You can put this solution on YOUR website!If the graph of y=ax^2+bx+c intersects the x-axis at two points, then there are two distinct real zeros. This is because the x-intercepts correspond to the zeros.
|
Volume/451259: a sphere has a diameter that is 7.36 inches long. finf the volume to the nearest tenth 1 solutions
Answer 310343 by jim_thompson5910(28696) on 2011-05-18 15:55:47 (Show Source):
You can put this solution on YOUR website!Since the diameter is 7.36 inches, the radius is  inches.
 Start with the volume of a sphere formula.
 Plug in  .
 Replace  with  (note: Use more digits of  to get better accuracy).
 Cube  to get  .
 Multiply  and  to get  (this figure is approximate).
 Multiply  and  to get  (again, this figure is approximate).
 Round to the nearest tenth.
So the volume is approximately 208.8 cubic inches.
|
Volume/451242: If I know the volume of a cylinder and the volume of each ball in the cylinder can I find out how many balls the cylinder will hold? 1 solutions
Answer 310326 by jim_thompson5910(28696) on 2011-05-18 15:20:48 (Show Source):
You can put this solution on YOUR website!Divide the volume of the cylinder by the volume of a single ball to find the approximate number of balls you can fit in there. Remember to round down to the nearest integer.
For example, if a cylinder has a volume of 30 cubic inches and one ball has a volume of 7 cubic inches. Then  which rounds down to 4. So the cylinder can hold at most 4 balls. Note: this is of course assuming that the shapes match (ie the radius and height of the cylinder >= the radius of a single ball)
|
Complex_Numbers/451157: Divide.
(4+2i)/(5-6i)
Write your answer as a complex number in standard form. 1 solutions
Answer 310323 by jim_thompson5910(28696) on 2011-05-18 15:17:10 (Show Source):
You can put this solution on YOUR website! Start with the given expression.
 Multiply the fraction by  .
 Combine the fractions.
 FOIL the numerator.
 FOIL the denominator.
 Multiply.
 Replace  with -1.
 Multiply.
 Combine like terms.
 Break up the fraction.
 Reduce.
So  .
So the expression is now in standard form  where  and
|
sets-and-operations/451236: I am having some problems understanding this .. this is all new could i get someone to help....thank you
Given A = {1, 2, 3, 4}, B = {4, 5, 6,}, and C = {2, 6, 7}. Evaluate each set
a) A ∩ B
b) A U C
c) B U C
d) (A U B) ∩ C
e) A U (B U C)
f) (A ∩ B) ∩ C
g) (A ∩ B) U C 1 solutions
Answer 310318 by jim_thompson5910(28696) on 2011-05-18 15:12:32 (Show Source):
You can put this solution on YOUR website!I'll do the first three to get you started
a)
To form the intersection of the two sets A={1,2,3,4} and B={4,5,6}, simply write down all of the elements that are in BOTH sets at the same time. So let's place the first set right above the second set (for an easy visual comparison): {1,2,3,4} {4,5,6} Now highlight the elements that are in BOTH sets: {1,2,3, 4} { 4,5,6} Since the highlighted terms are 4, this means that the intersection of the two sets is {4} So if  and  , then the intersection of the two sets is:
=========================================================================
b)
To form the union of the two sets A={ 1,2,3,4} and C={ 2,6,7}, simply make a new set that contains EVERY element in both sets like so { 1,2,3,4, 2,6,7}. Take note how the red elements are from the first set while the green elements are from the second set. Now remove any duplicate elements to get the set {1,2,3,4,6,7} So if  and  , then the union of the two sets is:
=========================================================================
c)
To form the union of the two sets B={ 4,5,6} and C={ 2,6,7}, simply make a new set that contains EVERY element in both sets like so { 4,5,6, 2,6,7}. Take note how the red elements are from the first set while the green elements are from the second set. Now remove any duplicate elements to get the set {4,5,6,2,7}. Now if you want, you can sort the elements from least to greatest to get {2,4,5,6,7} So if  and  , then the union of the two sets is:
For more help with set union and intersection, check out this solver
|
Probability-and-statistics/451226: Given U = {22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32}, A = {22, 24, 26, 28, 30}, and B = {23, 24, 25, 28, 29}.
Find A′U B′. finding A’, then B’ then find their union.
1 solutions
Answer 310312 by jim_thompson5910(28696) on 2011-05-18 14:57:46 (Show Source):
You can put this solution on YOUR website!A = {22, 24, 26, 28, 30}
A' = {23, 25, 27, 29, 31, 32} .... this is the set of U but excluding everything in A (ie it's everything but set A). So start with U, then delete every element found in A.
B = {23, 24, 25, 28, 29}
B' = {22, 26, 27, 30, 31, 32}
Now union A' and B' to get
A' U B' = {22, 23, 25, 26, 27, 29, 30, 31, 32}
This is just the combination of the two sets (ie dump all the elements from both sets A' and B' into A' U B' and remove any duplicates).
|
Subset/450991: Let A and B be subsets of U with n(A)=12, n(B)=25, n(A')=67, and n(A intersect B)= 11 Find n(A U B').
I have created the Venn diagram and with two circles, where A is 12, B is 25 and the intersection of A and B is 11, but i'm not sure where to go from there. A' is given as 67, but I don't see how that provides me what I need to solve.
Thanks 1 solutions
Answer 310210 by jim_thompson5910(28696) on 2011-05-18 04:29:09 (Show Source):
You can put this solution on YOUR website!Since  and  = n(A) + n(B)) , this means that
So
which means that
So there are 79 elements in the universe U. This is a very important piece of information.
Now comes the task of creating the Venn diagram. This isn't required, but it really helps to see what's going on.
So start with drawing a rectangle. Now draw two overlapping circles within this rectangle. Label the circles as A and B and label the rectangle as U like so:
Now because  = 11) , this means that the number 11 goes in the region between the two circles (since this is where the two "intersect"). So fill in this region:
Now, we know that n(A) = 12 and we know that  = 11) . We want to know how many is in A alone and not in B. So just subtract off 11 from 12 to get 12-11 = 1. So there is 1 element in set A. So fill in the circle A with "1" like so:
Do the same for set B to get 25 - 11 = 14. So there are 14 elements that are only in B (and not in A). Fill this in to get
Now add up the numbers in the separate regions to get: 1+11+14 = 26
So there are 26 elements that belong in either set A, set B, or both. Since there are 79 elements total, this must mean that there are 79 - 26 = 53 elements that are in neither set. This number gets written outside the circles but inside the rectangle like so:
Now our Venn diagram is done. Our task is to now appropriately shade the region that pertains to  .
We do this by first shading all of set A (the first set listed) like so
then we shade  , which is everything but B, to get
Combine these shadings to get the final product of
What we get is that practically everything is shaded except the region that has a 14 in it. Now just add up the numbers that lie in the shaded regions to get: 1+11+53 = 65
So there are 65 things in the set
So
|
Graphs/450462: I have a question they want me to use the multiplication pronciple to solve this equation -6x<-18 I was thinking that you would divide both sides by -6 and the answer would be 3 not sure if it would be negative or positive or maybe I am going about it all wrong. 1 solutions
Answer 309826 by jim_thompson5910(28696) on 2011-05-16 22:41:37 (Show Source):
|
Volume/450167: What is the approximate volume of the cone? Use 3.14
8cm by 12 cm
1 solutions
Answer 309698 by jim_thompson5910(28696) on 2011-05-16 15:54:36 (Show Source):
You can put this solution on YOUR website!I'm assuming the diameter is 8 cm and the height is 12 cm. So if the diameter is 8 cm, then the radius is 4 cm.
 Start with the volume of a cone formula.
 Plug in  and
 Replace  with  (note: Use more digits of  to get better accuracy).
 Square  to get
 Multiply  and  to get  (this figure is approximate).
 Multiply  and  to get  (again, this is approximate).
 Multiply  and  to get
So the volume of the cone with radius  and height  is approximately  cubic centimeters.
|
Graphs/450164: I'm learning about graphing equations, I would like to know step by step on how to solve this problem: x-2y=1. Please & thank you. 1 solutions
Answer 309694 by jim_thompson5910(28696) on 2011-05-16 15:37:40 (Show Source):
You can put this solution on YOUR website! Start with the given equation.
 Subtract  from both sides.
 Rearrange the terms.
 Divide both sides by  to isolate y.
 Break up the fraction.
 Reduce.
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 2. This means that to go from point to point, we can go up 1 and over 2
So starting at ) , go up 1 unit
and to the right 2 units 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
|
Square-cubic-other-roots/450076: I'm not sure this belongs here but...
For the functions f(x)= sqrt 12x and g(x)= sqrt 3x. Evaluate each expression
1. (f+g)(3) 2. (f-g)(x) 3. (fg)(x) 4. (f/g)(x)
Thanks, I'm completely in the dark on this one. 1 solutions
Answer 309691 by jim_thompson5910(28696) on 2011-05-16 15:27:10 (Show Source):
|
sets-and-operations/449748: Given U = {15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}, A = {16, 18, 20, 22}, and B = {17, 19, 20, 23, 24}.
Find A′U B′. Do not skip to the answer. Show your work, finding A’, then B’ then find their union 1 solutions
Answer 309349 by jim_thompson5910(28696) on 2011-05-15 19:05:03 (Show Source):
You can put this solution on YOUR website!A = {16, 18, 20, 22}
A' = {15, 17, 19, 21, 23, 24, 25} .... this is the set of U but excluding everything in A (ie it's everything but set A). So start with U, then delete every element found in A.
B = {17, 19, 20, 23, 24}
B' = {15, 16, 18, 21, 22, 25}
Now union A' and B' to get
A' U B' = {15, 16, 17, 18, 19, 21, 22, 23, 24, 25}
This is just the combination of the two sets (ie dump all the elements from both sets A' and B' into A' U B' and remove any duplicates).
|
Equations/449749: How do I slove and check this equation 4x-8=5x 1 solutions
Answer 309346 by jim_thompson5910(28696) on 2011-05-15 19:00:39 (Show Source):
You can put this solution on YOUR website! Start with the given equation.
 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
|
|