Order the decimals from least to greatest 3.4897, 3.3487, 1.5467, 3.348

Mathematics

1 answer:

Anna71 [15]3 years ago

5 0

Answer:

1.5467, 3.348, 3.3487, 3.4897

Step-by-step explanation:

Remember to compare each one number at a time!

For the first number, only one number is 1 while the others are 3 making it the lowest.

Then when comparing 3.348 and 3.3487, it can be implied that there is a 0 at the end of 3.348 because it makes the numbers have the same # of decimal places without changing the value of the number (3.3480). That number is less than 7.

Finally, 3.4897 has a 4 in the tenths place instead of a 3 making it the highest number.

Hope this helps!

You might be interested in

Need help ASP TWT……….

Ksivusya [100]

Answer:

C srry if im wrong

Step-by-step explanation:

6 0

2 years ago

Marcos is organizing a fundraiser for his youth group. If he spent $38.00 on paper products, $187.00 on food, and $40.00 on deco

lys-0071 [83]

If he bought all that then he would have spent 265$

3 0

3 years ago

Read 2 more answers

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$