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Nikolay [14]
3 years ago
12

Find the largest integer which belongs to the following interval: [−∞, 31]

Mathematics
1 answer:
ELEN [110]3 years ago
8 0

Answer:

Largest integer in the interval [−∞, 31] is <em>31</em>.

Step-by-step explanation:

Given the interval: [−∞, 31]

To find: The largest integer in this interval.

Solution:

First of all, let us learn about the representation of intervals.

Two kind of brackets can be used to represent the intervals. i.e. () and [].

Round bracket means not included in the interval and square bracket means included in the interval.

Also, any combination can also be used.

Let us discuss one by one.

1. [p, q] It means the interval contains the values between p and q. Furthermore, p and q are also included in the interval.

Smallest p

Largest q

2. (p, q) It means the interval contains the values between p and q. Furthermore, p and q are not included in the interval.

Smallest value just greater than p.

Largest value just smaller than q.

3. [p, q) It means the interval contains the values between p and q. Furthermore, p is included in the interval but q is not included in the interval.

Smallest value p.

Largest value just smaller than q.

4. (p, q] It means the interval contains the values between p and q. Furthermore, p is not included in the interval but q is included in the interval.

Smallest value just greater than p.

Largest value q.

As per above explanation, we can clearly observe that:

The largest integer which belongs to the following interval: [−∞, 31] is <em>31</em>.

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Step-by-step explanation:

We can write the equation of a circle in the form

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According to Greg, perfect cherry pies have a ratio of 240 cherries to 3 pies. How many cherries does Greg need to make 9 perfec
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4 0
3 years ago
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)

Which gives;

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2022}  \right)

\displaystyle  A = 2 \times 1011 \cdot  \left(1 - \frac{1}{2022}  \right) = \frac{1032231}{511} \approx \mathbf{2020.022}

  • A ≈ <u>2020.022</u>

Learn more about the sum of a series here:

brainly.com/question/190295

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