2.34 grams
2.34 grams
It depends on the number of digits accuracy required. Round off to: 3 digits: 12.426 2 digits: 12.43 1 digit: 12.4 Round number: 12
After determining whether to round up or down, the digits, to the right of the place, are discarded.
It is 708.
if the last two digits are lesser than 50 (e.g. 8634), you round down by making the last two digits 0 (in this case, 8600).If the last two digits are greater than 50 (e.g. 8662), you round up by adding 1 to the hundreds and making the last two digits 0 (in this case, 8700).If the last two digits are exactly 50, you can round up or down, your choice.
2.34 grams
2.45
Approximately 9,980, but that should probably round to 10,000 if you use 2 significant digits.
=ROUND(Number, Number of Digits) Number is the number you are trying to round. Number of Digits is the amount of digits you want to round it to. So for example: =ROUND(41.98662,3) That will give you 41.987 as the answer.
It depends on the number of digits accuracy required. Round off to: 3 digits: 12.426 2 digits: 12.43 1 digit: 12.4 Round number: 12
If necessary.
After determining whether to round up or down, the digits, to the right of the place, are discarded.
0.40
It is 708.
5.2g When you add or subtract using significant figures, you round the answer to the fewest number of decimal places as the measurement with the fewest decimal places.
if the last two digits are lesser than 50 (e.g. 8634), you round down by making the last two digits 0 (in this case, 8600).If the last two digits are greater than 50 (e.g. 8662), you round up by adding 1 to the hundreds and making the last two digits 0 (in this case, 8700).If the last two digits are exactly 50, you can round up or down, your choice.
i dont think ANYONE knows why there are so many digits to pi! There are so many digits to pi because it is the ratio to the circumference of the diameter of a circle. It is the measurement of any circle and TRUST ME it is hard to come up with a number for how round a circle is. Maybe you should ask a math professor or teacher that can assist you with this question.