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4 years ago

Round 656.692169912 to 1 decimal place. I am so confused.

2 answers:

8 0

You would round it to 656.70 because the 9 behind the 6 is greater than 5 so it rounds up and makes 6 and 7 and all the numbers behind 6 zero.

Airida [17]4 years ago

6 0

Answer:

Hence, the required number is 656.7

Step-by-step explanation:

Consider the provided number.

656.692169912

We need to round the number to the 1 decimal place.

There are 9 digits after the decimal place.

First understand the means of rounding to the 1 decimal place.

The meaning of rounding to the 1 decimal place is Round the number value after the integer number, restricted to one decimal place.

For example: Consider the number 8.14 is rounded to 8.1

The rule of rounding a number is:

If 0, 1, 2, 3, or 4 follow the number, round it down.

If 5, 6, 7, 8, or 9 follow the number, round it up.

Now consider the number 656.692169912

The number at 2 decimal place was 9, so to round the 1 decimal place round the number up.

Thus round the number 656.692169912 to the 1 decimal place is:

656.7

Hence, the required number is 656.7

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3 years ago

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dlinn [17]

Answer with Step-by-step explanation:

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Suppose that P(n) is true for n=k >1

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Now, we shall prove that p(n) is true for n=k+1

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LHS

$P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2$

Substitute the value of P(k)

$P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)$

$P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})$

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$P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS=RHS

Hence, P(n) is true for all n >1.

Hence, proved

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3 years ago

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