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what is 659͵782 nearest one thousand

Nearest one thousand

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Rounding to Hundred Thousands


An interactive math lesson about rounding numbers to the nearest 100,000. Rounding to the Nearest ... To round numbers to the nearest hundred thousand, ...
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Rounding to the Nearest Hundred Thousand – Grade 4 Common ...


Rounding to the Nearest Hundred Thousand – Grade 4 Common Core Standards. 4th Grade Common Core Standards Navigation. Choose: Drills View | Activities View ...
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Decimals - Thousandths - HelpingWithMath.com


Decimals - Thousandths; Rounding; To Nearest 10; To Nearest 100; ... Home > By Subject > Place Value > With Decimals - Thousandths ... One (1) ...
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Rounding Numbers to the Nearest Tens Place Packet


603 _____ 782 _____ 911 ... 516 _____ 813 _____ 659 _____ 980 _____ 754 ... Rounding Numbers to the Nearest Tens Place Worksheets G ...
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Rounding to the Nearest 10th - Softschools.com


Rounding to the Nearest 10 th. ... When this happens the nine becomes a zero and the place to the left is one bigger. Here is a look at the 9 becoming a 10.
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Rounding to Hundred Thousand - Math A Tube


Rounding numbers to the nearest hundred thousand. Look at the ten thousand place digit of the number. If.. ... one, and the rest of the digits go to zero.
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Rounding whole numbers - Basic mathematics


This lesson explains rounding whole numbers to the nearest ten, hundred, or thousand. Homepage; Math blog; ... Rounding whole numbers to the nearest thousand
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Rounding to nearest 100 (video) | Khan Academy


If we're rounding to the nearest 100, ... If we're trying to round to the nearest 100, we would look to one place to the right. We'll look at the tens place.
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Suggested Questions And Answer :


round 49.285371 to the nearest thousand

49.285371 = 0049.285371 there's a 0 in the thousands place. one digit to the right, in the hundreds place, is also 0. the digit in the hundreds place is less than 5, so we leave the digit in the thousands place alone. everything to the right of the thousands place is replaced with 0s 0000.000000 = 0   if you meant to the nearest thousandth, then: 49.285371 there's a 5 in the thousandths place one digit over, in the ten thousandths place, is a 3. 3 is less than 5, so we leave the thousandths place alone and replace everything to the right with 0s. 49.285000 or just 49.285
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what is 0.3 repeating rounded to the nearest thousand

0.3333333333 (3's forever) rounded to the nearest thousand: 0.3333333333 = 0000.33333333333 The left-most 0 is in the thousands place.  The next digit over is the 0 in the hundreds place.  Because that digit (0) is less than 5, the 0 in the thousands place does not round up to 1.  We have a 1 in the thousands place and everything to the right of that just becomes 0s. Result:  0000.333333333 becomes 0000.00000000 or just 0. Answer:  0 . If you meant "round to the nearest thousandth," then: 0.33333333 (3's forever) Counting from the left, the third 3 is in the thousandths place.  One digit to the right, in the ten-thousandths place, is another 3.  Because that digit (3) is less than 5, the 3 in the thousandths place does not round up to 4. Answer:  0.333
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complete the following statement and round to the nearest thousand 2x^2+5 square root x-2

2x^2+5sqrt(x-2)-8=0 completes the equation so that at x=2 the equation is satisfied. Rounding to the nearest thousand or thousandth isn't necessary for this solution, since it's exact. If the constant 8 is increased, there may be more than one solution for x. For example, 2x^2+5sqrt(x-2)=23 or 2x^2+5sqrt(x-2)-23=0 has two solutions, one of which is 3. The other is irrational, 3.8614, and so would be rounded off to 3.861. (There are two solutions if sqrt delivers a positive or negative result, for example, sqrt(1) is -1 or +1.)
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how do you round 49,052 to the nearest thousand?

The 9 is in the thousands place. Look one digit to the right, in the hundreds place. If that digit is 5 or greater, then add 1 to the 9. If that digit is 4 or less, then leave the 9 alone. Replace everything to the right of the 9 with 0s. 0 in the hundreds place. 0 is 4 or less, so leave the 9 alone. Answer:  49,000   If the problem had been to round 49,552 to the nearest thousand, then the 5 in the hundreds place would have made the 9 go up by 1 and the answer would have been 50,000.
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Round off 9704.8372 to the: nearest thousand,nearest whole number , two decimal places , three significant figures

Nearest thousand is 10000, which is nearer than 9000. 9704.8372 is 704.8372 away from 9000; but only 295.1628 away from 10000. The guide is usually the digit in the hundred position: if it's 5 or more then we go up to the next thousand (9 to 10 in this case), otherwise we just take the thousand digit as it is (9 in this case). 7, the hundred digit, is bigger than 5, so we take 10 thousand. Nearest whole number is 9705, which is nearer than 9704, because the next digit 8 is bigger than 5 and the 4 in the ones position goes up to 5. 2 dec places is 9704.84, which is nearer than 9704.83. 3 sig figures is 9700. The 3 figures are 9, 7 and 0, because the zero is occupying a place separating it from the next digit 4, so it is significant. The 4 would make it 4 sig figures. The size of the number has to be preserved because, for example, 970 would not suggest the right number of thousands. Leading zeroes in whole numbers or mixed numbers are always discounted because they do not have any effect on the magnitude of the number.
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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 14071.75 to the nearest 1000

Well since 0<5 so you wouldn't add one to the thousands place so the answer is 14,000
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What number that is less than 50 000

ABCDE is the number. E=C+4 (ones digit is 4 more than hundreds digit). A=E-5. B=D=0. So A=C-1. A B C D E 1 0 2 0 6 2 0 3 0 7 3 0 4 0 8 4 0 5 0 9 E is half of what? Is it 12, 14, 16 or 18? This would give one of the numbers 10206, 20307, 30408, 40509. If C is half of E the answer is 30408. If E is half the combined digits AC (ten thousands and hundreds digits) then the answer is 10206.
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What is 50 rounded to the nearest thousandths

50 rounded to the nearest thousand. 50 is the same thing as 0050. The leftmost 0 is in the thousands place.  The next digit over, a 0, is less than 5, so the 0 in the thousands place doesn't round up. Answer:  0000 or just 0.
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rounding a four digit number to nearest ten thousand

The 0's at the front can be there, but we almost never write them or use them.  4 is the same thing as 0000000000004.  1,996 is the same thing as 01,996. We're trying to round 01,996 to the nearest ten thousand. The 0 is in the ten thousands place.  The next digit over is 1, which is less than 5, so the 0 doesn't round up. Answer:  00,000 or just 0.
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