All categories

3 years ago

1726 rounded to the nearest 100

Mathematics

2 answers:

ipn [44]3 years ago

6 0

Answer:

1726 to the nearest 100 = 1700

daser333 [38]3 years ago

4 0

Answer:

1700

700 is the nearest 100

2 lower than 5>0

if 6>+1

You might be interested in

What’s 1,234,948 rounded to the nearest 1,000,000

garik1379 [7]

Your answer would be 1,001,000

4 0

3 years ago

Calculate the following limit:

aleksklad [387]

$\lim_{x\to\infty}\dfrac{\sqrt x}{\sqrt{x+\sqrt{x+\sqrt x}}}=\\
\lim_{x\to\infty}\dfrac{\dfrac{\sqrt x}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{\sqrt{x^2}}}}=\\$
$\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{x+\sqrt x}{x^2}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{\sqrt x}{\sqrt{x^4}}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{x}{x^4}}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{1}{x^3}}}}}=\\
=\dfrac{1}{\sqrt{1+\sqrt{0+\sqrt{0}}}}=\\$
$=\dfrac{1}{\sqrt{1+0}}=\\
=\dfrac{1}{\sqrt{1}}=\\
=\dfrac{1}{1}=\\
1
$

8 0

3 years ago

96,240 is divisible by 2, 4, and 8. true or false?

VikaD [51]

Answer:

false

Step-by-step explanation:

96,240

a-9

2,3

8,9

4 0

3 years ago

Name the following segment or point.

Mamont248 [21]

Answer:

Step-by-step explanation:

It's where the altitudes meet

5 0

3 years ago

Karla has spent $42 on food. She bought

Mandarinka [93]

Answer:

What is the question?

Step-by-step explanation:

4 0

2 years ago