<img src="https://tex.z-dn.net/?f=%5Csqrt%2018-%20%5Csqrt%202" id="TexFormula1" title="\sqrt 18- \sqrt 2" alt="\sqrt 18- \sqrt 2

All categories

vovangra [49]

2 years ago

" align="absmiddle" class="latex-formula">

Mathematics

2 answers:

zimovet [89]2 years ago

4 0

The answer is 2√2

√18 - √2
3 √2 - √2 = 2√2

Alex787 [66]2 years ago

4 0

Answer:

2√2

Step-by-step explanation:

18 factors down to 3 * 3 * 2

The rule for taking the square root of a number factored into primes is if they are in pairs, take one of the pairs out of the root sign and throw the other one away. Any odd number of primes always has one prime left under the root sign.

3√2 - √2

Root 2 is a common factor on either side of the minus sign.

√2*(3 -1)

√2 * 2

You might be interested in

Is 7/9 an irrational number?

Mila [183]

Answer:

7/9 is a rational number

Step-by-step explanation:

x/y where x=7 and y=9 aand an reatinal number is something that is composed of an integer

3 0

3 years ago

Read 2 more answers

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