Need Help on this math problem.....for those college grads...need ur help!!!!!?
471793
Well here's some extra info you need to solve the problem:\n\n1 cubic inch = 0.00432900431 US gallons\n\nSo one can has a volume of:\n\nV=(pi)(5)(3/2)^2=35.34291 cubic inches\n\nThey filled 86400 cans which has a volume of:\n\nVtotal = V*86400 = 35.34291*86400 = 3053627.424 cubic inches\n\nThen you need to convert this to gallons so:\n\nVgallons = 0.00432900431 * 3053627.424 = 13219.1662 Gallons
957682
half it then times it by 2! that will give u the answer
909535
1) \n76>40+4c\n36>4c\nc<9\n\n2) 1500>100+17.5p\n1400>17.5p\np<80
774792
Let the length of side "b" be x ft.\nTherefore side a=2x and side c=2x+3\nAccording to the problem,\nx+2x+2x+3=93\n=>5x+3=93\n=>5x=93-3=90\n=>x=90/5=18\nTherefore.side a=2X18=36 and side c=36+3=39\nAnswer a=36ft. b=18 ft and c=39 ft
61458
Your answer is correct, and may be fine for your professor, but for readers who are not already familiar with the problem, both the question and the solution are confusingly worded.  I suggest that in the future, you copy the original question here exactly.\n\n\nI would precede your solution with the following:\n\n\nYou can find the floor in a maximum of 14 attempts.\n\nDrop a bulb from the floors in this order:\n14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100\n\nIf the bulb breaks, stop and test starting from just above the last floor that it didn't break on, and go up one floor at a time.  If the bulb breaks on the first attempt, resume the test from the first floor.\n\nExamples:\n14 (breaks), 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\n\n14, 27, 39 (breaks), 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38\n\n14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99 (breaks), 96, 97, 98\n\n14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100 (breaks)
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
800717
(1) First one is copied wrongly by you                                            (2) Where is the equal to sign?                                                        (3)This equn. also cannot be solved                                              (4)-p/16+5=9  Multiplying all the terms by 16 ,we get -p+80=144   =>-p=144-80=66 Therefore p=66 YOU SHOULD BE MORE CAREFUL BEFORE SENDING YOUR PROBLEMS TO YAHOO ANSWERS.HOWEVER,YOU MAY SEND THE CORRECTED VERSION OF THE PROBLEMS TO ME IMMEDIATELY THROUGH YAHOO ANSWERS.
971574
This is a proportioal problem\n\nMultiply the means\n\nmultiply the extremes\n\n- - - - - - -\n\nx = the amount of men\n\n4 / 7 = x / 28\n\nthe means = 7x\n\nThe extremes = 4 x 28\n\n- - - - - - - - - - - - - - -\n\n7x = 4 x 28\n\n7x = 112\n\n7x / 7 = 112 / 7\n\nx = 16\n\nThe answer there are 16 men\n\n- - - - - - s-
419226
Let a,b and c denote the number of days that A,B and C would respectively take to finish the work, working alone. \n\nNow A and B together do it in 8 days. \nHence in one day they finish 1/8 th of the work. \n1/a + 1/b = 1/8 ...........(I)\n\nB and C do it in 12 days. \nHence in one day they finish 1/12 th of the work. \n1/b + 1/c = 1/12 ..........(II)\n\nA alone does it in 16 days. \nHence in one day he finishes 1/16 th of the work. \n1/a= 1/16 => a = 16 .........(III)\n\nfrom (I) and (III), \n1/16 + 1/b = 1/8\n=> b = 16 .......(IV)\n\nFrom (II) and (IV), \n1/16 + 1/c = 1/12\n=> c = 48...... (V)\n\nNow, A and C together working in one day \nwould finish \n1/a + 1/c of the work\ni.e 1/16 + 1/48 of the work\ni.e 1/12th of the work. \n\nHence they would finish the work in 12 days.
459843
The answer should be 64350/184756 or 34.8%\nIs that what you have?\n\nTo solve this problem you will need a function called "choose", often written as C.\n\n20C10 means "given 20 items, how many possible combinations of 10 are there?".  If there are 20 questions and the teacher chooses 10 of them, then there are 20C10 possible combinations of questions on the test.\n\n20C10 is solved by 20!/[(10!)(20-10)!]\n\nThe solution to the problem is (15C8)*(5C2)/(20C10).\n\nThere are (15C8)*(5C2) possible combinations of 8 (out of the 15) questions that you do know and 2 (out of the 5) questions that you don't know.\n\n"Choose" functions are very long calculations so you should not be expected to do them by hand.  On a TI graphing calculator, use the nCr option under the MATH/PRB menu.
517123
It's hard to show it clearly here but the solution is actually:\n\n[-b±√(b²-4ac)]/2a\n\nthere will usually be two solutions!\nThe site below has it laid out more clearly...\n\n("invited" to leave school early.. 1974)
748382
Lets convert the problem to cents.\n\nIn equation form, with c for chicken feed in tons, p for pig feed in tons, and b for cattle feed in tons, we have\n47c+123p+144b=10000\nc+p+b = 100\n\nEliminating c, we arrive at \n76p+97b = 5300.\n\nSo p = (5300-97b)/76 = 69-b+(56-21b)/76\nLet z = (56-21b)/76 then\n76z + 21b = 56\nSo b = (56-76z)/21 = 2-3z+(14-13z)/21\nLet x = (14-13z)/21\nThen 21x+13z=14\nSo z=(14-21x)/13 = 1-x+(1-8x)/13\nLet y = (1-8x)/13 then\n13y+8x=1, x=(1-13y)/8 = -y+(1-5y)/8\nBecause this can go on for quite some time, let y=5, then x = -8, z = 14, b=-48, and p=131.  Shifting answers by the original values (76 and 97), b=-48+76, p=131-97, so b=28, p=34; therefore c=38.  This is a long winded solution, but it illustrates the general rule for solving linear diophantine equations.\n\nSteve
780311
You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
912853
1.) Ms. Jordan 275 / 2500 = 0.11 \n     0.11  x  100 = 11%\n\n2.) 900 students x 1.07 = 963 students .\n\n3.) Virus Scan  300 sec / 0.40 = 750 sec for 100%\n\n4.) Gasoline... 8882 / 0.85 = 10,449 BBLs per day.\n\n5.) Jerry's new salary is\n     18,500 x 1.105 = $ 20,442.50\n\n6.) Chen's Investment wound up being \n     6400 x 0.918 = $ 5875.20\n\nHave a nice day,\nZahbudar
604123
I figured it out... the answer is 1/(13*(2-3x))\nIf you need an explanation tell me
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
775488
Ok, Victoria, if you're going to ask for help you have to write the questions correctly.  Problem 1 cannot be solved using algebra the way you wrote it, I doubt you can use complex integration or any other advanced tool to solve that one.  Unfortunately, I just spent the last half hour manipulating it before I remembered that little fact.\n\nPut parentheses in.  Had you put the parentheses in the problem would have been trivial.  Kudos to the helpers who were smarter than I in figuring out the constants were in the exponent.  It's not that hard Victoria, especially when you're asking for help.\n\nWrite the question correctly.\n\n\nPS. Irma, you forgot to multiply the 2 from 4=2^2 through the whole parenthetical statement.  The answer is x=4.\n.