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Rounded to nearest whole number it is 3.

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โˆ™ 2008-10-24 21:22:43
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Algebra

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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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3.74
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Q: 2.68 rounded to the nearest whole number?
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Related questions

What is 268 rounded to the nearest hundreds?

300


What is 268 rounded to the nearest 100?

300


How do you round 268 to the nearest hundreds?

As the digit in the tens position is a 6, it is rounded up to give 300 as the answer.


Which number is around to the nearest ten 268 274 255 269?

200


What is 268 divisible by?

The positive whole-number factors of 268 are:1, 2682, 1344, 67


What are 25 rounding to ten problems?

Twenty-five rounding to ten's place are: What is 12 rounded to the nearest tens? What is 17 rounded to the nearest tens? What is 231 rounded to the nearest tens? Is 268 rounded to the nearest tens 260? Is 1074 rounded to the nearest tens 1100? Why is 648 rounded to the nearest tens not 640? How do you know that 7386 rounded to the nearest tens is not 7380? What is 36988374647 rounded to the nearest tens? How do you decide what 89473 should be rounded to? Round the following numbers to the nearest tens place" 13 783987 173676 98731 392 69837 76734 25555 876355 76754 777 54 769 47659 38


How do you round 268 to the nearest hundred?

300 - 268 = 32 268 - 200 = 68 so as 32 is smaller than 68 then the nearest hundred is 300


What is 268 to the nearest hundred?

300


What is 268 is closer to?

To the nearest ten it is 270.To the nearest hundred it is 300.


How do you round 268 to the nearest tenth?

268.0


Is 268 divisible to any of these numbers 2 3 4 5 6 9 10?

268 is divisible by 2 and 4 the rest of the numbers 268 is not divisible by with a whole number being the dividend.


What is 270 rounded to nearest ten?

It's already there.

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