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3 years ago

What is 101,949 rounded to the nearest hundred thousand?

Mathematics

2 answers:

Maslowich3 years ago

6 0

Answer:

The nearest hundred thousand of the provided number 101,949 is 100,000.

Step-by-step explanation:

Consider the provided number 101,949.

Rule of rounding a number is:

If 0, 1, 2, 3, or 4 follow the number, then no need to change the rounding digit.

If 5, 6, 7, 8, or 9 follow the number, then rounding digit rounds up by one number.

Here, the number at the ten thousand place is 0, so no need to change the rounding digit.

Thus, the number can be rounded to the nearest hundred thousand is shown as:

100,000

The nearest hundred thousand of the provided number 101,949 is 100,000.

kobusy [5.1K]3 years ago

5 0

Well the answer is 103,000 because when you round it the 1 becomes a 2

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The correct answer is: " √x − 2√b " .
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The "conjugate" of " √x + 2√b " is: " √x − 2√b " .
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Explanation:
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In an expression with 2 (TWO) terms; that is, in a "binomial expression",
the "conjugate" of that expression refers to that very expression — with the "sign" in between those two terms—"reverse" (e.g. "minus" becomes "plus" ; or, "plus" becomes "minus" .) .
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→ So: We are given: " √x + 2√b " .

 → Note that this is a "binomial expression" ;

→ that is, there are 2 (TWO) terms: " √x " ; and: " 2√b " .
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To find the "conjugate" of the given binomial expression:

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→ We simply change the "+" {plus sign} to a "−" {minus sign} ; and rewrite:
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→ " √x − 2√b " ;

→ which is the "conjugate" ; and is the correct answer:
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→ " √x − 2√b " ; is the "conjugate" of the expression: " √x + 2√b " .
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→ {that is: " √x − 2√b " ; is the conjugate.}.
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3 years ago

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