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Subtract. Simplify your answer and write it as a proper fraction or as a whole or mixed number.

11 1/2 - 9 4/5=

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Subtract. Simplify your answer and write it as a proper ...


Resources / Answers / Subtract. Simplify your a... ... Subtract. Simplify your answer and write it as a proper fraction or as a whole or mixed number. 12 1/12 ...
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Fractions Calculators - Online Calculator Resource


Fractions calculators to add, subtract, ... to mixed numbers. Simplify proper and improper fractions, showing the work and the answer as a fraction or mixed number.
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How to Subtract Fractions from Whole Numbers: 10 Steps


How to Subtract Fractions from Whole ... the whole number into a fraction, or subtract 1 from that whole ... a mixed number (optional). If your answer is an ...
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Subtracting Fractions - Math Is Fun


Subtracting Fractions . ... Simplify the fraction (if needed). ... Subtract the top numbers and put the answer over the same denominator:
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Converting Fractions to Mixed Numbers


Let's look at some more examples of converting fractions to mixed numbers ... a mixed number because it is a proper fraction. ... write the mixed number four ...
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Adding and Subtracting Mixed Fractions - Maths Resources


... A Mixed Fraction is a whole number and a fraction combined, ... Adding Mixed Fractions. ... Now Subtract: 189 12 − 106 12 = ...
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Subtracting mixed numbers (video) | Khan Academy


Subtracting mixed numbers ... We're asked to subtract, simplify the answer and write as a ... or you can subtract the whole number part and then the fraction ...
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Simplifying Fractions - Southwest Virginia Governor's School


The student will simplify a proper fraction to its ... is the whole number in the mixed fraction. ... not change the fraction. Step 6: Write your answer.
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How to Subtract Fractions: 6 Steps (with Pictures) - wikiHow


How to Subtract Fractions. ... Simplify your answer. ... Turn the whole number into a fraction with the same denominator as the fraction that is being subtracted.
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Suggested Questions And Answer :


how do i add fractions

To add or subtract fractions, obtain a least common denominator. Subtract the numerators in the correct order and retain the same least common denominator for your answer. Simplify. To multiply fractions, multiple the numerators. The product will be the numerator of your answer. Repeat with denominators. Simplify. To divide fractions, take the reciprocal of what you are dividing by. Multiply the reciprocal with the initial number (see above for multiplication process). Simplify. Evaluate means to solve. You can solve fraction problems using the above processes. You can only simplify if both the numerator and denominator are divisible by the same number. If the denominator is odd, you can only simplify it if the numerator also is divisible by a same number. Ex. 88/33. Although the denominator is odd, both the numerator and denominator are divisible by 11 resulting in 8/3 as the simplified answer. To pace yourself during a test do the following. Find out how long you have for the test. Divide this by the total number of problems on the test. Example. 1 hour for 20 problems on your test. This means you have 3 minutes per problem. If you spend more than 3 minutes on a problem, skip it. Continue until you attempt all the problems. Go back with the remainder of the time to retry these problems you skipped. Most likely they are the most difficult, hence why you spent alot of time on them. This method of pacing allows you to skip the hard problems at first, attempt all problems, and finish the easier problems for sure.
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Multiply. Write the answer as a mixed number, whole number or proper fraction: 4/5 x 3 3/4

4/5 x 3 3/4 = 4/5 x 15/4 = 60/20 = 3
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how can I write this as a whole or mixed number 1,300 cellular phones per 1,000 person

a corporation reported profits of approximately 1,500 million with approximately 47,000 million in revenues. compare the profit to revenue by writing as a fraction in lowest terms. the ratio of profits to revenue is.. type an integer,or a simplified fraction.
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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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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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Subtract. Simplify your answer and write it as a proper fraction or as a whole or mixed number.

start: 11 1/2 - 9 4/5 =(23/2)- 49/5 =(115/10) -(98/10) =17/10 or 1 & 7/10 or 1.7
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With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
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what is the difference between 2 1/8 x 3 3/8 and 2 3/16 x 3 11/16?

Convert the mixed numbers to improper fractions: 17/8*27/8-35/16*59/16=459/64-2065/256. 1836/256-2065/256=-229/256. The difference is 229/256 (the sign depends on which way you do the subtraction). [To convert a mixed number to an improper fraction, let's write a mixed number as A b/c and convert it: (cA+b)/c. In words, multiply the whole part of the mixed number by the fraction denominator and add on its numerator; then write the result over the denominator. To convert 459/64 to 1836/256 I just multiplied top and bottom by 4. The reason for this conversion was to put both fractions over the same denominator to do the arithmetic: 1836/256-2065/256 is (1836-2065)/256=-229/256.]
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howto write the improper fraction as a mixed number or whole number 14/3

???????? "improper frakshun" ?????????? aent no such anamal 14/3=4.666666... or =4 & 2/3 or even 46.7% & all av that is PROPER
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howto write the improper fraction as a mixed number or whole number 14/3

????????? "improper frakshun" ??????????? aent no such anamal 14/3=4.66666... or 467% or 4 & 2/3 & all is PROPER
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