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2 years ago

Round 1627187 to the nearest ten?

Mathematics

2 answers:

Eduardwww [97]2 years ago

7 0

Answer:

1627190

Step-by-step explanation:

(see attached for reference)

Given the number 1627187, we can see that the number in the tens place is the number 8.

How we round this depends on the number immediately to the right of this number. (i.e the digit in the ones place)

Case 1: If the digit in the ones place is less less than 5, then the number in the tens place remains the same and replace all the digits to its right with zeros

Case 2: If the digit in the ones places is 5 or greater, then we increase the digit in the tens place and replace all the digits to its right with zeros.

In our case, the digit in the ones places is 7, this greater than 5, hence according to Case 2 above, we increase the digit in the tens place by one (from 8 to 9) and replace all the digits to its right by zeros giving us:

1627190

n200080 [17]2 years ago

4 0

Answer:

1627190 is your answer. Have a nice day!

Step-by-step explanation:

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Step-by-step explanation:

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So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

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$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

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$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

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$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
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