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3 years ago

Which list contains ONLY irrational numbers?

Mathematics

2 answers:

Black_prince [1.1K]3 years ago

8 0

The first choice is the contains only irrational number

chubhunter [2.5K]3 years ago

4 0

Answer:

The correct option is A.

Step-by-step explanation:

A real number is called a rational number if it can be represented as p/q, where p and q are two integers and q≠0. For example: 2,4/5, 0.2.

A real number is called an irrational number if it can not be represented as p/q, where p and q are two integers and q≠0. For example: π, √2, 2.333....

In option A, all numbers are irrational numbers.

Therefore option A is correct.

In option B,

$\sqrt{9}=3$

3 is a rational numbers, so √9 is a rational number.

Therefore option B is incorrect.

In option C,

$\sqrt{144}=12$

12 is a rational numbers, so √144 is a rational number.

Therefore option C is incorrect.

In option D,

$\sqrt{121}=11$

11 is a rational numbers, so √121 is a rational number.

Therefore option D is incorrect.

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Write a polynomial that represents the area of the shaded region

marusya05 [52]

<h3>Answer: x^2-3x+36</h3>

=========================================

Explanation:

The larger rectangle has area of (x+1)(x+1) = x^2+2x+1 through the use of the FOIL rule or distribution

If you use distribution, then it might help to let y = x+1 so we'd have y(x+1) lead to xy+1y which becomes x(x+1)+1(x+1). From there it might be easier to see how to get x^2+2x+1 after everything distributes again and simplifies.

The smaller rectangle has area 5x-35 which is found by distributing 5(x-7)

To get the shaded area, we subtract the two rectangle areas found above

shaded area = (larger area) - (smaller area)

shaded area = (x^2+2x+1) - (5x - 35)

shaded area = x^2+2x+1 - 5x + 35

shaded area = x^2-3x+36

6 0

3 years ago

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$