A division equation that has solution of 1,350

Mathematics

2 answers:

d1i1m1o1n [39]2 years ago

4 0

How about
13,500 ÷ 10 = 1,350

monitta2 years ago

3 0

2,700 divided be 2 = 1,350
hope this helps!

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Set D is the set of positive two-digit even numbers less than 25 that do not contain the digit 0.

gavmur [86]

The required set of positive two-digit even numbers less than 25 that do not contain the digit 0 is 12,14,16,18,22,24.

<h3>What is integer?</h3>

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Zero is not a fraction or decimal of any number. It is neither positive nor negative.

Given:

Set D is the set of positive two-digit even numbers less than 25 that do not contain the digit 0.

According to given question we have

2-digit numbers are the numbers that have two digits and they start from the number 10 and end on the number 99.

Starting at 12 going up the two digit numbers even numbers less than 25 that do not contain the digit 0.

12,14,16,18,22,24

Therefore, the required set of positive two-digit even numbers less than 25 that do not contain the digit 0 is 12,14,16,18,22,24.

Learn more details about integer here:

brainly.com/question/15276410

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4 0

1 year ago

Define fn : [0,1] --> R by the

sasho [114]

Answer:

The sequence of functions $\{x^{n}\}_{n\in \mathbb{N}}$ converges to the function

$f(x)=\begin{cases}0&0\leq x$ .

Step-by-step explanation:

The limit $\lim_{n\to \infty }c^{n}$ exists and converges to zero whenever $\lvert c \rvert$ . But, if $c=1$ the sequence $\{c^{n}\}$ is constant and all its terms are equal to $1$ , then converges to $1$ . Using this result, consider the sequence of functions $\{f_{n}\}$ defined on the interval $[0,1]$ by $f_{n}(x)=x^{n}$ . Then, for all $0\leq x$ we have that $\lim_{n\to \infty}x^{n}=0$ . Now, if $x=1$ , then $\lim_{n\to \infty }x^{n}=1$ . Therefore, the limit function of the sequence of functions is