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HELPP!! been working on it for a week! due tomorrow! i feel so dumb :(

i have this math problem, "you have 128 ounces of chocolate to make chocolate chip cookies and chocolate chip muffins. one dozen chocolate chip cookies require 6 ounces of chocolate and one dozen chocolate chip muffins require 4 ounces of chocolate. A. write and graph an equation for the possible number of dozens of choco. cookies and choco. muffins you can make. B. describe the meaning of the slope and intercepts in the context of the problem." will someone please tell me or help with the answer. im going crazy!

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HELPP!! been working on it for a week! due tomorrow! i feel so dumb :(

start: 1. y hav 128 oz choklut 2. 12 kookees yuze 6 oz choklut 3. 12 muffins yuze 4 oz choklut: If yuze all choklut for kookees, get 128/6=21.33 duzen if yuze all choklut for muffins, get 128/4=32 duzen muffins In tween yu get strate lines av number av eech let x= duzen kookees & y=duzen muffins then 4y=128-6*x or duzens av muffins=32- 1.5*duzens av kookees
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how to solve a vinndiagram

vinn diagrams show two things unions and intersections. the data is usually in sets for example: set A {a,b,c,d,e} set B {d,e,f,g,h} to put it in the diagram you do the following. any common values are put in the middle (Common area) Now the intersection is the set {d,e} values in both sets A,B. the union is the set {a,b,c,d,e,f,g,h}  all the values in the diagram.
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3 weeks at 28 hours a week. week 1 for 3 days. week 2 for 4 days. week 3 was for 5 days atn 5, 6, 7 and 8 hour shifts, what are the possible hours she worked to add up to 28 hours for each week?

Week 1: She cannot make up 28 hours because she would need to work an average of 9.3 hours a day, and the maximum shift is 8 hours. The only way to do this is to spread a shift overnight, so that there are four periods of work rather than 3. The question changes to 4 shifts over 3 days. The average length of shift is 7 hours. 3 24-hour days is 72 hours, divided by 4 is 18 hours. Allowing 8 hours a day for sleep, we have 10 hours remaining. 4 periods of 10 hours is 40 hours. Part of that time is eating and travelling. 3 1-hour time slots for eating per period reduces the time by 12 hours to 28 hours for working and travelling. So the most reasonable solution is 4 shifts of 7 hours each in Week 1. Day 1: 10am to 5pm; Day 2: 4am to 11am; 10pm to 5am the next day; Day 3: 4pm to 11pm. This pattern (as an example) allows 11 hours minimum between shifts. Week 2: (5,7,8,8), (6,6,8,8), (7,7,7,7), (6,7,7,8), i.e., any mixture of 5, 7, 8, 8 hour shifts on the 4 days, etc. These working hours can be extended to, and adopted for the 4 working periods suggested for Week 1. Week 3: (5,5,5,5,8), (5,5,5,6,7), (5,5,6,6,6) and any mixture.  
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139.04 multiplied by .005%

I understand the 401k reference to be the U.S. pension contribution scheme. I understand that $139.04 is your withheld tax and 401k contribution for about two weeks' work. Your work allowance of 1040 hours per year is 1000*10.25=$10250 plus 40*10.25=$10660 gross, as you said. By part-time I guess that means that over a year you can only work 1040 hours in total = 26 40-hour weeks, and when you are working your regular work hours are 40 hours a week. You haven't stated your normal weekly hours. You could, for example, work 20 hours a week for a year, because 20*52=1040 hours. Your biweekly take-home pay of $820 implies that the gross amount is 820+139.04=$959.04, assuming your employer deducts tax withholding and 401k contribution before paying you. $959.04 is $479.52 a week or $11.99 an hour in a 40-hour week. Tax withholding and 401k contribution is 139.04/959.04=0.145 or about 14.5% of your gross pay. Therefore, if you know your gross biweekly pay, you can calculate the tax withholding and 401k contribution by multiplying the gross amount by 14.5%. Multiplying the deductions by 0.005%, which is a very small amount, seems pointless, because you want the tax withholding and 401k contribution, which you already have! Dividing by 0.005% seems more appropriate, although it could only give you something related to your gross pay. Since 0.005% is equivalent 0.00005 you would end up with a figure 20000 times bigger, giving you an amount 20000*139.04 or around 278000 which is way in excess of your annual earnings! 14.5% of your annual earnings is 10660*0.145=$1545.70, or $59.45 biweekly, but your biweekly deductions are known to be $139.04, more than twice as much. $820 is equivalent to 80 hours of work (over two weeks) at $10.25 an hour, so your work rates may have been quoted as what you take home after deductions have been made (net pay), rather than the gross amount you indicated. There should be enough information in your payslips and your annual pay statement to work out how and what deductions are made before you receive your earnings, and exactly what your are paid gross by the hour. Perhaps your mother-in-law could also help to justify her calculations, particularly since I'm unfamiliar with the American tax system. Can I offer another possible explanation and calculation that may fit your information better? Let's say your gross pay is $11.99 per hour, then for 1040 hours per year, you would receive $12469.60 gross pay. Divide this by 26 to get the biweekly wage: $479.60. Take out the deductions of 14.5% and that leaves 85.5% of $480=$410.06 and the deductions are $69.94. The deductions for $820.12 (2*$410.06) would therefore be 2*$69.94=$139.88 and the net hourly rate for a 40-hour week paying $410.06 would be $10.25. I'm not saying this is the true case, but if we change gross to net in your question, the figures seem to fit the facts better, don't you think?
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Algebra - work problems

If Michael takes M weeks to build a house by himself and Thomas takes T weeks, then we can say that if they build a house together in 6 weeks, in one week they can build 1/6 of a house. By himself in a week, Michael can build 1/M of a house and by himself Thomas can build 1/T of a house, so working together: 1/M+1/T=1/6; multiplying through by 6MT: 6T+6M=MT; MT-6T=6M. Also M=T-5, because Michael takes 5 fewer weeks than Thomas to build a house. Substituting for M we have: T^2-5T-6T=6T-30; T^2-17T+30=0; (T-15)(T-2)=0, T=15 or 2. M=T-5, so M=10 or -3. We can reject -3 as being meaningless, so M=10 weeks. Check: 1/M+1/T=1/10+1/15=3/30+2/30=5/30=1/6. They can build a house together in 6 weeks.
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An employee works three weeks on, one off

There are 52 weeks in a year. Working 3 weeks and then taking a week off means working 3 weeks out of 4, which is 3/4. 3/4*52=39 weeks per year.
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A ball is fired at 30 m/s from a height of 10 m/s. How far from it's origin will it land.

Wot direkshun du yu fire that dang thang??????????????? horizontal???? vertikal??? at angel tween em ???????
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What is the value of n Im (17•25)•4=n•(25•4)

I guess that n=17 because it's a demonstration of the associative property of numbers and multiplication. A*(B*C)=(A*B)*C. In words, it means no matter what order you decide to do multiplication (or addition) the result is always the same. The operation in parentheses is done first then the result is multiplied by, or is used to multiply, what's outside the parentheses, but the result is the same no matter which you do first. I think that's what the teacher was trying to tell you. The large dot to me means multiplication. Similarly, A+(B+C)=(A+B)+C. This doesn't work for division and subtraction and you can't mix addition and multiplication. For example, A*(B+C) is not the same as (A*B)+C. By the way, perhaps for extra credit, you should know that A*B=B*A, which is the commutative property of numbers and multiplication. When you combine the two properties you can see that A*(B*C)=(A*C)*B or A+(B+C)=(A+C)+B and other combinations, as long as you don't mix addition and multiplication in the same sum. These rules apply in algebra, too, where variables (unknowns) represent numbers. I hope this important concept is a bit clearer now, and that you continue to improve your grades.
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math triangle sums equals 20 on all sides using the numbers 1 - 9 only once

Note that if we take the numbers 1 to 19 in 9 pairs, the sums of the paired numbers come to 20: (1,19), (2,18), (3,17), (4,16), (5,15), (6,14), (7,13), (8,12), (9,11). Using the numbers 1 to 9 only, we can break down 4 of these pairs by splitting the larger number of each pair: (3,8,9), (4,7,9), (5,6,9), (5,7,8). These are the only unique groups of three, but they do not include 1 and 2. But if we include 1 and 2 we know the remaining 2 numbers to make up a group of 3 must add up to 19, which can't be achieved using a pair of numbers 1 to 9. Therefore, in some cases we need 4 numbers to make up the sum. We can only use  the numbers 1 to 9 and we need to use them all and we only have 3 sides of the triangle, so if we have 4 numbers on one side, one of the other sides will have only 2 numbers, because we're only left with 5 when we've taken out 4 for one side. What to do? We have to use the vertices so that the numbers at each vertex are used twice. Let's assume that all three vertices contain such numbers, then we will have 4 numbers a side. Instead of looking at groups of three numbers, we have to look for groups of 4: (1,2,8,9), (1,3,7,9), (1,4,6,9), (2,3,6,9), (1,4,7,8), (2,3,7,8), (1,5,6,8), (2,4,5,9), (2,4,6,8), (3,4,5,8), (3,4,6,7). Out of these we need to find where different groups contain a common number. These common numbers will form the vertices. Let's try putting the numbers 1, 2 and 3 at the vertices. This gives us a choice of 5 groups for each of the numbers 1, 2 and 3 (we can arrange the order so that the numbers occupy the first element in each set). Here are the sets: (1,2,8,9), (1,3,7,9), (1,4,6,9), (1,4,7,8), (1,5,6,8) (2,1,8,9), (2,3,6,9), (2,3,7,8), (2,4,5,9), (2,4,6,8) (3,1,7,9), (3,2,6,9), (3,2,7,8), (3,4,5,8), (3,4,6,7) Now we have to rearrange the numbers so that there are two vertices within each set, so that means we pick groups containing 1 and 2, 2 and 3, and 3 and 1. Yes, there are some, but unfortunately some contain duplicates of the two numbers that are not at the vertices. Look at the following, for example: (1,8,9,2), (2,6,9,3), (3,7,9,1) 9 is duplicated and we have no 4 or 5. In fact, we can"t have 1 as a vertex with 2 and 3 as the other vertices, because there's always a duplicate to spoil the arrangement. The same thing happens with 2, 3 and 4, and 3, 4 and 5. Symmetry is a common occurrence in mathematics and there's nothing more pleasing to a mathematician than an "elegant" solution. So I'm going to assume that symmetry applies to this problem and go for vertices in the middle of the range 1 to 9. The middle is occupied by numbers 4, 5 and 6. Let's see what happens if we use them as vertices: (4,2,9,5), (5,1,8,6), (6,3,7,4) This time it works! There are no duplications in the central pair of numbers so all the numbers from 1 to 9 are there. The other combination of 4 and 5 in the same group, (4,3,8,5), leads to a duplicated 8 in (5,1,8,6). So one answer is:  The vertices are A, B and C=4, 5 and 6 respectively, with 2 and 9 along AB, 1 and 8 along BC, and 3 and 7 along AC.  
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find all reconstuctions of the sum: ABC+DEF+GHI=2014 if all letters are single digits

There is no solution if A to I each uniquely represent a digit 1 to 9, because the 9's remainder of ABC+DEF+GHI (0) cannot equal the 9's remainder of 2014 (=7). Let me explain. The 9's remainder, or digital root (DR), is obtained by adding the digits of a number, adding the digits of the result, and so on till a single digit results. If the result is 9, the DR is zero and it's the result of dividing the number by 9 and noting the remainder only. E.g., 2014 has a DR of 7. When an arithmetic operation is performed, the DR is preserved in the result. So the DRs of individual numbers in a sum give a result whose DR matches. If we add the numbers 1 to 9 we get 45 with a DR of zero, but 2014 has a DR of 7, so no arrangement can add up to 2014. If 2 is replaced by 0 in the set of available digits, the DR becomes 7 (sum of digits drops to 43, which has a DR of 7). 410+735+869=2014 is just one of many results of applying the following method. Look at the number 2014 and consider its construction. The last digit is the result of adding C, F and I. The result of addition can produce 4, 14 or 24, so a carryover may apply when we add the digits in the tens column, B, E and H. When these are added together, we may have a carryover into the hundreds of 0, 1 or 2. These alternative outcomes can be shown as a tree. The tree: 04 >> 11, >> 19: ............21 >> 18; 14 >> 10, >> 19: ............20 >> 18; 24 >> 09, >> 19: ............19 >> 18. The chevrons separate the units (left), tens (middle) and hundreds (right). The carryover digit is the first digit of a pair. For example, 20 means that 2 is the carryover to the next column. Each pair of digits in the units column is C+F+I; B+E+H in the tens; A+D+G in the hundreds. Accompanying the tree is a table of possible digit summations appearing in the tree. Here's the table: {04 (CFI): 013} {09 (BEH): 018 036 045 135} {10 (BEH): 019 037 046 136 145} {11 (BEH): 038 047 137 056 146} {14 (CFI): 059 149 068 158 167 347 086 176 356} {18 (ADG): 189 369 459 378 468 567} {19 (BEH/ADG): 379 469 478 568} {20 (BEH): 389 479 569 578} {21 (BEH): 489 579 678} {24 (CFI): 789} METHOD: We use trial and error to find suitable digits. Start with units and sum of C, F and I, which can add up to 4, 14 or 24. The table says we can only use 0, 1 and 3 to make 4 with no carryover. The tree says if we go for 04, we must follow with a sum of 11 or 21 in the tens. The table gives all the combinations of digits that sum to 11 or 21. If we go for 11 the tree says we need 19 next so that we get 20 with the carryover to give us the first two digits of 2014. See how it works? Now the fun bit. After picking 013 to start, scan 11 in the table for a trio that doesn't contain 0, 1 or 3. There isn't one, so try 21. We can pick any, because they're all suitable, so try 489. The tree says go for 18 next. Bingo! 567 is there and so we have all the digits: 013489576. We have a result for CFIBEHADG=013489576, so ABCDEFGHI=540781693. There are 27 arrangements of these because we can rotate the units, tens and hundreds independently like the wheels of an arcade jackpot machine. For example: 541+783+690=2014. Every solution leads to 27 arrangements. See how many you can find using the tree and table!
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