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To which subsets of real numbers does the following number belong? square root of 30

LenaWriter [7]

In order to determine which subset it belongs, we must rewrite

$\sqrt(30)$

as

$\sqrt(30)=\sqrt(2\times \times 3 \times 5)$

$\Rightarrow \sqrt(30)=\sqrt(2 \times 3\times 5)$

$\Rightarrow \sqrt(30)=\sqrt(2) \times \sqrt(3) \times \sqrt(5)$

All these three numbers are irrational numbers, hence their product is also irrational

<h2> $\Rightarrow \sqrt(30)$ </h2><h2>belongs to the set of irrational numbers</h2>

6 0

3 years ago

Read 2 more answers

PLZ ANSWER ASAP for brainliest!

Darya [45]

Answer:

I think it is C because it keeps on going and it doesnt stop at 0

hope this :)

Step-by-step explanation:

Mark as brainllest if it helps

5 0

3 years ago

Read 2 more answers

Two pounds of grapes costs $6 as shown in the table. At this rate, how many pounds of grapes can you buy with $1? Round your ans

sergiy2304 [10]

Well 1 pound should cost $3 so do the math lol

5 0

3 years ago

Find the volume of the rectangle 4 2/3cm,9 1/8cm, 2cm

amm1812

Answer:

$\large\boxed{V=85\dfrac{1}{6}\ cm^3}$

Step-by-step explanation:

The formula of a volume of a rectangular prism:

$V=lwh$

l - length

w - width

h - height

We have

$l=4\dfrac{2}{3}\ cm=\dfrac{4\cdot3+2}{3}\ cm=\dfrac{14}{3}\ cm\\\\w=9\dfrac{1}{8}\ cm=\dfrac{9\cdot8+1}{8}\ cm=\dfrac{73}{8}\ cm\\\\h=2\ cm$

Substitute:

$V=\dfrac{14}{3}\cdot\dfrac{73}{8}\cdot2=\dfrac{7}{3}\cdot\dfrac{73}{2}\cdot1=\dfrac{511}{6}=85\dfrac{1}{6}\ cm^3$

6 0

3 years ago

Can you find the limits of this

Pavel [41]

Answer:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{-3}{8}$

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Constant]: $\displaystyle \lim_{x \to c} b = b$

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

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

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

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

We are given the following limit:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16}$

Let's substitute in <em>x</em> = -2 using the limit rule:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{(-2)^3 + 8}{(-2)^4 - 16}$

Evaluating this, we arrive at an indeterminate form:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{0}{0}$

Since we have an indeterminate form, let's use L'Hopital's Rule. Differentiate both the numerator and denominator respectively:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \lim_{x \to -2} \frac{3x^2}{4x^3}$

Substitute in <em>x</em> = -2 using the limit rule:

$\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{3(-2)^2}{4(-2)^3}$

Evaluating this, we get:

$\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{-3}{8}$

And we have our answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

3 years ago

Help plssssssssssssssss