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write the first quantity as fraction of 2nd quantity a)3 days;the month of january

Math solve to find 2nd quantity

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Problem of the Month: Fractured Numbers - Inside Mathematics


Problem of the Month Fractured Numbers Page 3 ... and 3 packages of meat. How many days will the dog be fed ... larger fraction, the next natural number ...
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Fraction Concepts - Conceptual Mathematics


Each day when the date is written have a student write out what fraction of the month, ... first: the fraction of the counted number of coins, second: ...
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Calendar Math Worksheets - Super Teacher Worksheets


Answer questions about the January calendar. ... Write an ordinal number for each month given. ... and "Write the date that is fifteen days after May 25."
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Months - Math is Fun


Month Number. Month. In 3 letters ... See the days in each month ... the second King of Rome, added January (for the god Janus) ...
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Math Forum: Ask Dr. Math FAQ: Calendar and Days of the Week


How do I find the day of the week for ... but a normal calendar year is only 365 days. The extra fraction of a ... k is the day of the month. Let's use January ...
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Write the decimal as a fraction or mixed number in simplest ...


Write the decimal as a fraction or mixed number in ... The graph below shows the experimental probability of a runner finishing in first place (1), in second ...
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Writing Classes and Javadoc - Carnegie Mellon School of ...


... //Scan every earlier month, summing the # of days in that month... for (int m=JANUARY; ... number (fraction) ... 3 on the first die and a 5 on the second) ...
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Fraction (mathematics) - Wikipedia


A mixed number can be converted to an improper fraction as follows: Write the mixed number ... The second fraction, ... as decimal fractions were first used five ...
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Suggested Questions And Answer :


write the first quantity as fraction of 2nd quantity a)3 days;the month of january

????? yu want 3/31 ???=0.096774193
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i = 0.072/4 how do you simplify the fraction?

Let's consider the monthly contributions. 10 years ago she started paying $625 to the retirement fund. After another 5 years that initial payment would increase to $625*(1+0.0384/12)^(15*12). The second month accumulates $625*(1+0.0384/12)^(15*12-1), one month less than the initial payment. The third month's period of growth will be measured over 15*12-2=178 months. So it continues up to the last payment which will only accumulate a month's interest. The monthly rate is 0.0384/12=0.0032, so the growth factor is 1.0032^p where p is the number of months. Now we can write a series, S, that takes into account the accumulated no interest. Let's use a general expression, where A is the monthly contribution amount, r the monthly rate as a fraction, and p=12n where n is the number of years: S=A(1+r)^p+A(1+r)^(p-1)+A(1+r)^(p-2)+...+A(1+r)^2+A(1+r) S=A(1+r)((1+r)^(p-1)+...+1). This equation contains a geometric progression which can be summed. (1+r)S=A(1+r)((1+r)^p+(1+r)^(p-1)+...+(1+r)^2+(1+r)). So (1+r)S-S=A(1+r)((1+r)^p-1); rS=A(1+r)((1+r)^p-1). This formula can be applied to both parts of this question. a. Same rate applies for the whole 15 years. 0.0032S=625*1.0032(1.0032^180-1)=625*1.0032*0.7773=487.3504, so S=$152,297. b. 0.0032 applies for ten years: 0.0032S=625*1.0032*(1.0032^120-1)=292.963, so S=$91,550.94. Now start a new series with A=1000 and r=0.0772/12=0.0064333... or 0.0193/3 and p=60. 0.006433S=1000*1.006433(1.006433^60-1)=472.286, so S=$73,412.33. However, we're not through yet, because $91,550.94 continues to earn interest over 5 years at the new rate. This comes to 91550.94*1.006433^60=$134,512.78. The combined amount is this plus 73,412.33=$207,925.10 approx. (Approximate figures throughout.)
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HOW TO SOLVE LINEAR EQUQTIONS WITH FIVE UNKNOWN

It's clear that the percentages exceed 100%, so I conclude that the dry matter content includes the crude ingredients, which are supplemented by another ingredient that is not specified. I also conclude that the dry matter content indicates that water is present, because none of the percentages for dry matter are 100%. First, remove water content and recalculate percentages: to do this we divide each percentage of the components by the percentage of dry matter. The result is expressed as a percentage. This enables us to mix the feed on a dry basis.  The adjusted percentages are: CORN: CP: 9.4909%, CF: 2.3812%, CFB: 1.1339% RICE: CP: 13.4427%, CF: 13.7776%, CFB: 8.3150% SBM: CP: 48.0682%, CF: 6.3636%, CFB: 5.1136% WHEAT: CP: 18.5698%, CF: 5.1903%, CFB: 8.5352% COPRA: CP: 20.7792%, CF: 13.2035%, CFB: 12.4459% We also have to adjust the percentages on the required feed: CP: 18.1818%, CF: 4.5455%, CFB: 13.6364% The last letter of each ingredient will be used to signify the fraction of that ingredient needed to make up the required feed (N, E, M, T, A). So we can write: CP REQUIREMENT: 9.4909N+13.4427E+48.0682M+18.5698T+20.7792A=18.1818 CF REQUIREMENT: 2.3812N+13.7776E+6.3636M+5.1903T+13.2035A=4.5455 CFB REQUIREMENT: 1.1339N+8.3150E+5.1136M+8.5352T+12.4459A=13.6364 N+E+M+T+A=1 (The ingredient fractions must add up to 1.) [To show these equations are valid, let's move away from percentages and consider actual amounts in the dry mix. For example, n grams of corn contains 9.4909n/100 grams (0.094909n grams) of crude protein. Similarly for the other ingredients: e grams of rice contains 13.4427e/100 grams (0.134427e grams) of crude protein. When we add together the crude protein contributions for all the ingredients we get the required amount, 18.1818x/100 (0.1818x grams), where x grams is the weight of the dry mix=n+e+m+t+a. So we can multiply through by 100 to leave the percentage numbers; if we divide through by x we get n/x, e/x, etc., and we can replace these by N, E, etc., where these are fractions of the total feed: N=n/x, E=e/x, ... X=x/x=1.] There are five variables but only four equations. The amount of copra is A=1-(N+E+M+T), so the equations can be reduced to omit A: CP REQUIREMENT: 11.2883N+7.3366E-27.2890M+2.2094T=2.5974 CF REQUIREMENT: 10.8222N-0.5741E+6.8398M+8.0132T=8.6580 CFB REQUIREMENT: 11.3120N+4.1308E+7.3323M+3.9107T=-1.1905 This last equation is suspicious because all the values on the left are positive but the value on the right is negative. Therefore at least one assumption in the logic is false. To resolve this difficulty we have to change the requirements to include inequalities. For example, the mix must contain at least a certain amount of an essential ingredient, or no more than a certain amount, as well as exactly a certain amount. We have no guide in the question to make any assumptions. Nevertheless, we'll continue with the solution and see where it leads us. The plan now is to treat T as an independent variable or constant, and to find each of N, E and M in terms of T. T is therefore a parameter from which the other fractions can be derived. We can eliminate M from CP and CF: 6.8398(11.2883N+7.3365E-27.2890M+2.2094T) + 27.2890(10.8223N-0.5714E+6.8398M+8.0132T)=6.8398*2.5974+27.2890*8.6580 77.2104N+50.1808E+15.1122T+295.3277N-15.6668E+218.6706T=17.7658+236.2681 372.5381N+34.5140E+233.7828T=254.0338 And we can similarly eliminate M from CF and CFB: 7.3323(10.8223N-0.5714E+6.8399M+8.0132T) - 6.8398(11.3120N+4.1308E+7.3323M+3.9107T)=7.3323*8.6580+6.8398*1.1905 79.3514N-4.2095E+58.7544T-77.3719N -28.2542E-26.7486T=63.4827+8.1427 1.9795N-32.4637E+32.0059T=71.6253. We now have two equations involving N, E and T, and we can eliminate E between them and so find N in terms of T: 32.4637(372.5381N+34.5140E+233.7828T) + 34.5140(1.9795N-32.4637E+32.0059T)=32.4637*254.0338+34.5140*71.6253, 12162.2997N+8694.1124T=10718.9626, 12162.2997N=10718.9626-8694.1124T, N=0.8813-0.7148T. If we substitute this value of N in 1.9795N-32.4637E+32.0059T=71.6253, we can find E in terms of T: 1.9795(0.8813-0.7148T)-32.4637E+32.0059T=71.6253; 1.7445-1.4149T-32.4637E+32.0059T=71.6253 -32.4637E+30.5910T=69.8808; E=(30.5910T-69.8808)/32.4637, E=0.9423T-2.1526. Substituting for E and N we can find M then A: using the CP requirement equation: 11.2883N+7.3366E-27.2890M+2.2094T=2.5974 11.2883(0.8813-0.7148T)+7.3366(0.9423T-2.1526)-27.2890M+2.2094T=2.5974, 9.9484-8.0689T+6.9133T-15.7928-27.2890M+2.2094T=2.5974, 1.0538T-27.2890M=8.4418, M=(1.0538T-8.4418)/27.2890=0.0386T-0.3093. We now have N, E and M in terms of T. We can find A from A=1-(N+E+M+T): A=1-(0.8813-0.7148T+0.9423T-2.1526+0.0386T-0.3093+T)=2.5806-1.2661T. All the ingredients have now been found in terms of T (wheat). These values are based on equality in the ingredient equations.  Summary N=0.8813-0.7148T, E=0.9423T-2.1526, M=0.0386T-0.3093, A=2.5806-1.2661T. These are all supposed to be positive fractions<1, and clearly they are not! This implies that the exact amounts required in the mix cannot be achieved. I've checked the arithmetic and logic fully and I can find no errors, so it does not appear to be possible to mix the desired quantities of the essential ingredients. It may be possible to get the right quantity of one particular ingredient at the expense of the others, or with an excess of at least one of the other ingredients. An excess could be as undesirable as a deficiency.
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An amount of 10 000 was invested in a special savings account

15% per annum is 3.75% per quarter. 15 May is included in the 2nd quarter of the year, an even period. At the end of the quarter 10000 would accumulate a whole quarter's interest: 375, making the total savings 10375. This is the starting amount for July, so by October, if the simple interest is calculated on the original 10000 a further 375 would accumulate making the savings 10750. October is the start of the last quarter and the 7-month period ends in December. 10750 would accumulate by compound interest, becoming 11153.125 at the end of the quarter. So the interest is 1153.13. If compound interest is applied for the whole 7 months, then taking this to be 3 quarter periods, the interest is 1167.71. If simple interest is applied for 7 months the interest is 875, based on 7/12*0.15*10000.  
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of tgere is 4 mondays and 4 fridays what day is january 1

In a month of 31 days there will be 4 dates where the same day of the week falls on that date if d+28>31, so d>3 where d is the earliest date in the month. So, for example, if d=4, then the dates are 4, 11, 18, 25. Also, d<8, because if d=8, then there would be an earlier date, d=1. Therefore, 3 Read More: ...

Describe two methods for converting a mixed number to a decimal

Represent the mixed number as N a/b where N is the whole number part and a over b is the fraction. Method 1 Write down N and follow it by a decimal point. Now you need to divide b into a, but because a is smaller than b you need to write a, a decimal point, and as many zeroes as you wish for accuracy. Some decimals terminate (divide exactly) and some recur (a pattern repeats indefinitely). Example: a=3 and b=4 so we have 3/4. Divide b into 3.0000... Treat 3.0 as 30 and divide by 4, putting the answer immediately after the decimal point. So we have .7 and 2 over making 20 with the next zero. Now divide 4 into 20 and write the result after the 7: .75. This is an exact division so we don't need to go any further. Let's say N was 5 so the mixed number is 5 3/4. We've already written N as 5. so now we continue after the decimal point with what we just calculated giving us 5.75. Another example: 7 5/6. We write 7. first. Divide 6 into 5.0000... and we get .8333. This is a recurring decimal. We keep getting the same carryover. Now we attach the whole number part to get 7.83333... Another example: 2 1/16. We write 2. first. But be careful in the next part: 16 divided into 1.0000... 16 doesn't go into 10 so we write .0 as our first number. Then we divide 16 into 100. This is 6 remainder 4. So we have .06. 16 into 40 goes 2 remainder 8. That gives us .062. Finally 16 into 80 is exactly 5 so we have .0625. Attach this to the whole number: 2.0625. Method 2 For N a/b we make the improper fraction b*N+a over b, then divide by b. Example: 2 1/16: 16*2+1=33. So we divide 33.000.... by 16. We end up with 2.0625. Example: 7 5/6: divide 6*7+5=47 by 6. We end up with 7.83333...
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Susan is writing a linear equation for the cost of her cell phone plan.

We assume a linear relationship between the cost, y, and the time, x, in the form: y=ax+b, where a is the charge rate in $ per minute and b is a standing fixed charge. So we have two equations: 19.41=52a+b and 45.65=380a+b. Subtract the first eqn from the second: 328a=26.24, so a=26.24/328=$0.08 or 8 cents per minute (answer for part A). We can find b by substituting a=0.08 in the first equation: 19.41=0.08*52+b, so b=19.41-4.16=$15.25. Susan's plan (answer to part B) is based on y=0.08x+15.25. Check that the second eqn fits: 45.65=0.08*380+15.25.
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10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
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what is 9/13 6/8 3/5 7/10 5/7 from least to greatest

One way to do this is to use the lowest common denominator of the factions. First, though, 6/8 can be reduced to 3/4. Since the LCD of 4, 5 and 10 is 20 and there are no other common factors in the other denominators, the LCD=1820 (=20*13*7). We then multiply each fraction by 1820, and write down the answers: 1260, 1365, 1092, 1274, 1300. These are easy to put in order: 1092, 1260, 1274, 1300, 1365. We now arrange the associated fractions in the same order: 3/5, 9/13, 7/10, 5/7, 6/8 (3/4). [When we multiply each fraction by 1820, we divide the denominator into 1820 and multiply the result by the numerator.] Another way is to convert each fraction into decimal and compare decimals. We only need a couple of places of decimals to make the comparisons: 0.69, 0.75, 0.60, 0.70, 0.71. This is the same as comparing: 69, 75, 60, 70, 71. The order is 60, 69, 70, 71, 75. An easy way to organise the fractions is to write the fractions on separate pieces of paper. On the back of each piece of paper write the number you are going to use in place of the fraction. Now you can easily arrange the papers in order according to the order of the replacement numbers. All you have to do then is to turn the papers over to see what the fraction order is.
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why in februry there 28 or 29 days and other month 30 or 31?

Please dont put these kinds of questions here anymore but here is the answer i copied it off of an amazon ask.com web sit which you could have just googled for./ ********************************************************************************** To meticulous persons such as ourselves, having the calendar run out in December and not pick up again until March probably seems like a pretty casual approach to timekeeping. However, we must realize that 3,000 years ago, not a helluva lot happened between December and March. The Romans at the time were an agricultural people, and the main purpose of the calendar was to govern the cycle of planting and harvesting. February has always had 28 days, going back to the 8th century BC, when a Roman king by the name of Numa Pompilius established the basic Roman calendar. Before Numa was on the job the calendar covered only ten months, March through December. December, as you may know, roughly translates from Latin as "tenth." July was originally called Quintilis, "fifth," Sextilis was sixth, September was seventh, and so on. Numa, however, was a real go-getter-type guy, and when he got to be in charge of things, he decided it was going to look pretty stupid if the Romans gave the world a calendar that somehow overlooked one-sixth of the year. So he decided that a year would have 355 days--still a bit off the mark, admittedly, but definitely a step in the right direction. Three hundred fifty five days was the approximate length of 12 lunar cycles, with lots of leap days thrown in to keep the calendar lined up with the seasons. Numa also added two new months, January and February, to the end of the year. Since the Romans thought even numbers were unlucky, he made seven of the months 29 days long, and four months 31 days long. ********************************************************************************** But Numa needed one short, even-numbered month to make the number of days work out to 355. February got elected. It was the last month of the year (January didn't become the first month until centuries later), it was in the middle of winter, and presumably, if there had to be an unlucky month, better to make it a short one. ********************************************************************************** Many years later, Julius Caesar reorganized the calendar yet again, giving it 365 days. Some say he made February 29 days long, 30 in leap year, and that Augustus Caesar later pilfered a day; others say Julius just kept it at 28. None of this changes the underlying truth: February is so short mainly because it was the month nobody liked much.
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