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what will 6 divided by 10 reduce to

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Reduce 6/10 to the simplest form - CoolConversion.com


Reduce 6/10 to the simplest form. ... First divide both (numerator/denominator) by 2 to get 6/30. Divide both of those by 2 to get 3/15, then,
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What is 6 divided by 10 - Answers.com


What is 6 divided by 10? SAVE CANCEL. already exists ... 7/10 divided by 1 1/6 Since 1 and 1 over 6 is a mixed fraction we have to convert it into a improper ...
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6. Divide 7/24 by 35/48 and reduce the quotient to the lowest ...


6. Divide 7/24 by 35/48 and reduce the quotient to the lowest fraction. ... A. 245/1152 B. 42/48 C. 2/5 D. 4/10 Weegy: (7/24) / (35/48) = 2/5 Mayang30|Points 3378|
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Fractions: Comparing and Reducing Fractions - Page 3


Comparing and Reducing Fractions. ... 10 divided by 2 is 5. Since there's no number that 4 and 5 can be divided by, we can't reduce 4/5 any further.
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What is 5/6 divided by 10/3? - Welcome to Research Maniacs


It can sometimes be difficult to divide fractions, such as 5/6 divided by 10/3. But it's no problem! We have displayed the answer below:
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Math Forum - Ask Dr. Math


Improper Fractions and Mixed Numbers ... Divide the 6 into the 7, ... you'll see that I could reduce 1 and 2/6 to 1 and 1/3, ...
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Fractions - Math League


The following numbers are all fractions 1/2, 3/7, 6/10, ... Add or subtract the fractions. Reduce if necessary. Example: ... 1/3 is 1 divided by 3.
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Fractions and Mixed Numbers - Grade 6 Math Questions and ...


Fractions and Mixed Numbers Grade 6 Math Questions and Problems With Solutions and Explanations. ... 1 and 2/10 = 1 2/10 = 1 1/5 , reduce fraction 2/10 = 1/5
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Simplify 7/10 to the simplest form - Conversion Calculators


Click here to simplify or reduce 7/10 to lowest terms or form ... Simplify 7/10 to the simplest form. ... by 2 to get 6/30. Divide both of those by 2 to get ...
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UNIT 9: FRACTIONS - REDUCING TO LOWEST TERMS


... FRACTIONS - REDUCING TO LOWEST TERMS. ... Reduce these fractions to their lowest terms: ... 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24 ...
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Suggested Questions And Answer :


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 are greatest common factors?

All non-prime integers (composite numbers or integers) can be broken down into factors. Fractions are formed when two integers are arranged into a fraction: a/b, where a and b are different integers. If a and b are composite integers, they can each be replaced by the product of their factors. If they have a common factor, that factor can be used to reduce the fraction to a simpler form. The greatest common factor is the largest of the common factors, which is the product of all the common prime factors (prime means that the factor cannot be reduced further). The GCF is also useful in ratios as a means of simplifying the ratio. So GCF only applies if there are two or more integers. Example: 420 and 378. 420=2*2*3*5*7 and 378=2*3*3*3*7. These numbers have three factors in common: 2, 3 and 7. If we multiply them together we get 2*3*7=42. This is the GCF. If we saw the fraction: 378/420 we can reduce it by dividing top and bottom by 42 to get 9/10. Example: 24:84:144 can be reduced to 2:7:12 by dividing each number by the GCF 12.
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Dierdre's work to solve a math problem is shown below. Problem: How many 1 1/2- foot pieces can be cut from a 12-foot length of a ribbon?

Dierdre's work to solve a math problem is shown below. Problem: How many 1 1/2- foot pieces can be cut from a 12-foot length of a ribbon? Step 1: 1 1/2 divide by 12 = n Step 2: 3/2 divide by 12 = n Step 3: 3/2 X 1/12 = n Step 4: 3/24 = n Answer: 1/8 = n What was Dierdre's first error? A. She switched the divisor and the dividend in step 1 when creating an equation to model the problem. B. She used an improper fraction in step 2 that is not equivalent to 1 1/2. C. She replaced the divisor by its reciprocal in step 3 instead of re[;acing the dividend by its reciprocal. D. She reduced the fraction in step 4 incorrectly. Her first error was A. She switched the divisor and the dividend in step 1 when creating an equation to model the problem.
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How do you reduce 32x³y /24xy⁴to the simplest form?

32x^3y --------- 24xy^4 Look at this in parts: the numbers first: 32/24 divide by 2: 16/12 divide by 2: 8/6     divide by 2: 4/3 Now look at the x's: x^3/x  subtract the exponents: x^2 Last the y's: y/y^4  subtract the exponents 1/y^3 The answer is: 4x^2 ------- 3y^3
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How can $3375 equal 9/16,0.562,and 56.2%

How can $3375 equal 9/16,0.562,and 56.2%.  To write it as a fraction count the number of places and put a one in front of zeros.  4 places, this is the bottom. 3375/10000  Reduce the fraction divide by 5/5 675/2000   reduce by 5/5 again 135/400   reduce by 5/5 27/80  now divide it and get .3375
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divide fraction 7 over 12 by 1 over 5?

or can 1/5 divide by 7/12 and reduced to lowest terms?
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7/8 (9/10 divided by 5/8) =???? leave as as improper fraction. Need help

1) 7/8 (9/10 divided by 5/8) =???? leave as as improper fraction. Need help start in the parenthesis to divide you switch the top and bottom of the second fraction and multiply 9/10 * 8/5 = 72/50  reduce this: 36/25 now multiply by 7/8 7/8 * 36/25 = 7/4 * 18/25 = 126/100 = 63/50 2). 14a^2/10b^2 divided by 21a^2/15b^2 again flip the second fraction 14a^2/10b^2 * 15b^2/21a^2 rewrite with common together 14a^2/21a^2 * 15b^2/10b^2 = 2/3 * 3/2 = 1 3). 7x^3/3 divided by 14x^2/6 7x^3/3 * 6/14x^3 rewrite 7x^3/14x^3 * 6/3 = 1/2 * 2/1 = 1 4). a^2-2a+1/6 divided by (a^2-1) (a - 1)(a - 1)/6 * 1/(a - 1)(a + 1) = (a - 1) /6(a + 1)
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5x+10y=35 and 6x+12y=48 substitution method

First eqn can be reduced to x+2y=7 by dividing through by 5. Second eqn can be similarly reduced by dividing through by 6: x+2y=8. But x+2y cannot be both 7 and 8 so the eqns are inconsistent and there can be no solution.
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What is 5/40?

How do I reduce 16/24? divide both the numerator and the denominater by the greatest common number 16/ 24 divide both by 8 and we have 2/3  
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How do I solve: 9a/7a-21 + 3a/21a -63

   9a              3a ---------    =  -------- 7a - 21        21a - 63  ============================ reduce a 3 out of the right side   9a                   a ----------  =  ---------- 7a - 21          7a - 21 ==================================      9                   1              divide both sides by a ---------  =   ------------ 7a - 21         7a - 21 ================================= cross multiply 9(7a - 21) = 7a - 21   distribute the 9 63a - 189 = 7a - 21    subtract 7a from both sides 63a - 7a - 189 = -21 56a - 189 = -21          add 189 to both sides 56a = -21 + 189       56a = 168                   divide both sides by 56 a = 168/56                  Reduce the answer a = 84/28 = 42/19 or 2 4/19
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