Croissants are produced in the desert by workers with a (free) solar oven. Debbie owns a shop where croissants are produced according to Equation 2:\n\n(Equation 2) minutes of labor= [C x C] x 3.5.\n\nHow do I Calculate and graph the number of worker minutes needed for each additional croissant.
213326
The above answer is basically correct.  Plug in a few more values, such as (21,1543.5), (22,1694), and so on for a few more values of the graph.
93094
Actually I had something to help you but x=0 which makes it pretty difficult.\n\nI guess you could try to talk about tomatoes or something.  Say Tom (X) has 1/2 the amount of tomatoes that Jill has, plus four.  Or in other words, if you take Jill's amount of tomatoes, multiply it by two, and take Tom's amount and multiply it by three, there total amount of tomatoes will equal 12.  How many tomatoes does Jill have?  How many does Tom have?
407823
I worked this up, but Jess beat me to it.\n\nFollow her very clear explanation...excellent presentation.\nBut correct her math...\n10 mols x 0.928 = 9.28 mols SO3 produced\nMW of SO3 is 80.06\nSo, 9.28 x 80.06 = 743 g SO3 produced\n\nDon't be harsh...it's easy to transpose digits.  :-))
671688
A machine is simply an object that does work. A machine transforms electrical energy into mechanical energy (simple machines excluded).\n\nWhen you think of "work" being done you usually think of some sort of movement occurring.\n\nAn oven is not a machine in that no "work" is being done. An oven just takes electrical energy & transforms it into heat energy.
340767
Assuming exactly 52 weeks in a year (although it is more like 52 1/7)...\n\n3.75 x 10^7 = 37,500,000 (37.5 million sheets a year).\n\nIn a year you have 15 hours x 5 days x 52 weeks = 3,900 hours (no vacations, no holidays, etc.)\n\n37,500,000 / 3,900 ≈ 9,615 sheets per hour.\n\nNow if your factory isn't open for the standard 10 holidays, that would be 3,750 hours per year, then your equation is:\n(3.75 x 10^7) / (3.75 x 10^3).\n\nThis becomes an even 10^4 or 10,000 sheets per hour.
318490
Manager is taking advantage.\nThis is a violation of labor laws.\nI don't remember what the exact numbers are, but its like 10 or 15 minutes every few hours, and you must get a meal break (like min of 30 minutes) if you work more than 6 or 8 hours.\nManager is shooting himself in the foot anyway. He just has a shop full of tired, unmotivated people, instead of the busy productive employees he thinks he is creating.
32898
First, we make two equations. \nLet's say p=the number of peanut butter cookie boxes\nand m=the number of mint cookie boxes.\nso p+m=25 because there was a total of 25 boxes.\nalso 3p+4m=88 because each peanut butter cookie box was $3 and each mint cookie box was $4 and the total earned was $88.\nso now we have :\np+m=25\n3p+4m=88\nwe multiply the first equation by 4 to get:\n4p+4m=100.\nthen we subtract the second equation from it get:\n{4p+4m=100}-{3p+4m=88}={p=12}\nso there were 12 boxes of peanut butter cookies. and $36 were earned from that because 12*$3=$36.\nand the rest of the money was earned from mint cookies. \n$88-$36=$52\nso $52 were earned from mint cookie boxes. and since they were sold for $4 each:\n$52/$4=13\nso 13 mint cookie boxes were sold\nand 12 peanut butter cookie boxes were sold\nI hope you understood my explanation. by the way, i used a method called elimination to solve the equations. If you had a graphing calculator, this probably would have been much easier, but i wasn't sure if you had one, so i just explained it the long way.
439200
Tom baked 4 batches of fairy cakes. Each batch consisted of x cakes. He put 3 cherries on top of each cake in the first batch, but couldn't resist taking off and eating 5 of them. He did the same again with the second batch, then the third, then the fourth! Meanwhile, Sarah baked 5 batches of fairy cakes, again with x cakes in each batch. She put 2 cherries on each cake and an extra two on the final one. She repeated this procedure with each of the batches. She managed to resist eating any at all! Tom and Sarah discovered that Tom's cakes had exactly the same number of cherries in total as Sarah's. Set up an equation for this.\n\nThen: \n\nNumber of cherries on Tom's cakes = 4(3x-5)\nNumber of cherries on Sarah's cakes=5(2x+4)\n\nBut number of cherries is equal so:\n\n4(3x-5)=5(2x+4)\n\nPhew!! I think that works!! Lol!
58267
Ok - question number 1:\n\nB takes 2 hours to do 500 dishes, T takes 3 hours to do 450 dishes. In 2 hours, they will have done 800 dishes altogther (because T does 300 dishes in 2 hours and B does 500 dishes). Now B is working harder than T: we now need 200 more dishes and they need to finish at the same time. Now use algebra:\n\nB works at the rate: 250 per hour (call this b)\nT works at the rate: 150 per hour (call this t)\n\n(b + t)*time = 200 dishes, substitute in above figures and get:\n\nt=1/2 hour. So altogether the time takes 2 1/2 hours.\n\n\n\nQuestion 2:\n\nSet up the folowing equations. The total amount of work to be done is W:\n\n(b+t)*(8 hours)=W, b*time1=W, t*time2=W\n\nI used these because in equation 1 - b and t take 8 hours to do the work, equation 2 - b takes time1 to do the work and T takes time2. Furthermore, time2=time1-12 (we are told this). Now solve this simultaneously: (ie put b=W/time1 and t=W/(time1-12) into the 1st equation) and you get time1 = 4 or 24 hours. Well since 4 is not sensible (4-12 = - 8hours!) then you choose 24 hours. So b takes 24 hours and t takes (24-12)=12 hours.\n\n\n\nQuestion 3: (follows same format as previous questions)\n\nM-100mins, S-120 mins\nEquations are \nM*time(m)=W,  S*time(s)=W, L*time(l)=W\n\nThe equation describing what happens is:\n\n30*(M+S) + 20*(M+S+L)=W\n\nSince we know time(m) and time(s) we substitute in and get:\n\n50*(M+S) + 20L = W\n50*(W/100 + W/120) + 20W/time(l) = W\n\ncancel W's and get time(l)=240mins or 4 hours\n\n\n\nQuestion4: 1st man is A and new man is B\n\n6*A=W, 8*B=W\n\n2*(A+B) + B*time(b) = W\n2*(W/6 +W/8) +W*time(b)/8 = W\n1/3 + 1/4 + time(b)/8 = 1\n\ntime(b) = 3 1/3 hours = 3 hours, 20mins.
486790
Given facts:  salad = 3* cookies (s=3c)\nDrink = 2* cookies (d=2c)\ntotal cost was $5.40\n\nassume the cookie was the base of the meal.  \n1 cookie = c\n1 salad =  3c\n1 drink =    2c\nc+3c+2c=$5.40\n6c=$5.40\nc= $.90\ncookie = $.90\nsalad = 3* $.90 = $2.70\ndrink = 2* $.90 = $1.80\nto check the answers, add the totals:  $2.70+$1.80+$.90 =$5.40
480105
The easiest way is to figure the number of minutes (to get them the same), then figure the number of hours (to get them the same).\n\nFor example, you need to move from 10:30 to 11:15...\n\nMentally move the hand forward from 10:30 to 10:35, then 10:40, then 10:45, etc.  You'll see that to get to 11:15 you have to move *9* five minute increments.  (9 x 5 minutes = 45 minutes).  So now you have made the minutes the same :15 and :15.\n\nNext you need to go from 11:15 to 1:15.  You can ignore the minutes now... you should count two spots from 11 to 12 to 1.  That's 2 hours.\n\nSo your final answer is 2 hours and 45 minutes.
230016
1800 is minimum number of apples in the harvest.\n\n(There are larger solutions but I think this is what you had in mind). Give 10 points to whomever you wish; I don't care. \n\nThe key is to use the information given concerning the number of apples that were exhanged between the farmers to set up a system of eqautions which can be solved.\n\nLet a = the no of apples produced by each tree in the 14th\ntree orchard; b = # by the 13th orchard; c the 11th; d the 9th; and e the 7th. \n\nThen, from the problem conditions we have the equation sequence:\n\n11c-1 = 7e+1 = 14a-6 = 13b+3 = 9d+3\n\nwhich implies 13b = 9d so b=9k, d=13k for k>=1\n\nif k=1, then d=13, b=9, and 13x9+3=14a-6 so a=9. \n\nIf a=9 then 14x9-6= 7e+1 and so e=17\n\nIf e=17 then 7x17+1=11c-1 and c= 11\n\nNow, this would imply that the rest of the orchards produced a total of 120 apples each but if this is a solution then the remaining 10 orchards: 1,2,3,4,5,6,8,10,12 and 15 must each produce 120 apples. But since the number of apples produced must be a multiple of the # of the orchard, these 10 numbers must divide into 120 with no remainder and this is the case.\n\nTotal no. of apples produced:\n\n10 orchards produce 120 each for a total of 1200\nThe other 5 must also average 120 so that's 600 more, for a total of 1800 (minimum).\nNote, from the above equation sequence orchards 11,7,14,13 and 9 produce respectively: 121,119,126,117,117 = 5x120 = 600.
142119
I think you solve it this way:\n\nx= number of burritos\ny= number of tacos\nz= profit\n\nthen you build three equations in order to get the three numbers you need.\n\nprofit equation\n\n40y+65x=z\n\n40 cents per number of tacos plus 65 cents per number of burritos\n\nground beef equation\n\n2y+x=22\n\n2oz per tacos plus 1oz per burritos = 22 oz\n\nchoppet tomatoes equation\n\ny+3x=36\n\n1oz per tacos plus 3oz per burritos = 36oz\n\nthen solving the three equations simultanously you get these answers\n\nx=10\ny=6\nz=890\n\nyou should make 10 burritos and 6 tacos to get the max profit of 890.\n\nI hope i helped you, but i slightly suspect that you should derive in this problem, i just can't figure it out yet, but if it helps, i think you can do it by multiplying the beef equation and the tomato equation, then you sove the equation for one of the variables. After doing that you substitute it on the profit equation. Then you derive that equation. After deriving the equation, you equal it to cero and sove for the variable, with that variable solved you get the other variable to finaly substitue on the original profit equation and then you get the maximum profit. I tried to do it but for some reason i didn't get a reasonable answer but maybe you can get it :$.
499889
Rate = 2 houses / (4 workers * 3 days)\nRate = 1/6 house per worker per day.\n\n3 workers * n days * 1/6 house/worker = 5 houses\nn = 5 * 6 / 3\nn = 30 / 3\nn = 10 days
186618
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.
108175
Let H be the nunber of hens. Assume rate of food consumption per hen is equal for all hens. Then, if X days feed left at decision point and r is feed consumption per hen per day:\n\nSell: (H-75)*r*(X+20) is feed left\n\nBuy:  (H+100)*r*(X-15) is feed left\n\nSo (H-75)(X+20)= (H+100)(X-15)\n\nHe has to know how much feed he has to say what he says. So if we know X we can find H
458485
1)  Notice that 6^2=6*6=36; 6^3=36*6=216; everytime you multiple a power of 6 by 6, the units digit remains 6.  Now what about powers of 5?\n\n2)  If r is total minutes and the first 3 minutes is $1.00, and the rest of r costs $0.20 per minute, then the total cost would be $1.00 + $0.20(r-3) = \n1 +0.2r + 0.6 =\n1.6 + 0.2r\n\n3)  3*10^4 = 3*10,000 = 30,000, and 4*103 = 412.\nNow divide 30,000 by 412 to get their ratio, and then multply that number by 100 to convert to percent difference.\n\n4)  Not enough info - you need the height too.  If the same height, obviously the can with the larger radius\n\n5.1)  65*20% =65*0.2 = 13\n\n5.2)  Again, not enough info. From the info given, all you can say about one-half the number of people who preferred green was that it was between 0 and (65-13)/2, aka 26