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3 years ago

What is 1/3 divided by 2/3 ?

Mathematics

2 answers:

kumpel [21]3 years ago

6 0

1/2 is your answer :) hope this helped♥

insens350 [35]3 years ago

4 0

1/3 because 1/3 kcf and then 3/2 is 1/3

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A cab company charges $0.25 per 1 mile as well as a flat fee of $2.25 just for taking a ride in the

vivado [14]

Answer:

y - intercept = 2.25

Step-by-step explanation:

y = mx + c

Where,

y = Total cost

x = additional cost

m = slope

c = y - intercept

y = 0.25x + 2.25

Relating with the above equation,

y - intercept = 2.25

Slope, m = 0.25

7 0

2 years ago

Use the product property of roots to choose the expression equivalent to 3√5x*3√25x^2?

nignag [31]

Answer: $\sqrt[3]{125x^3}$ .

Step-by-step explanation: Given radical expression

$\sqrt[3]{5x} \times \sqrt[3]{25x^2}$ .

According to the product property of roots.

$\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{a \times b}$

On applying above rule, we get

$\sqrt[3]{5x} \times \sqrt[3]{25x^2} = \sqrt[3]{5x \times 25x^2}$

5 × 25 = 125 and

$x \times x^2 = x^3$

Therefore,

$\sqrt[3]{5x \times 25x^2}= \sqrt[3]{125x^3}$

<h3>So, the correct option would be second option $\sqrt[3]{125x^3}$ .</h3>

3 0

3 years ago

Read 2 more answers

Dot got a 75% on her test. She missed 12 questions. how many questions were on the test

AleksAgata [21]

Well, if it was out of 50, it would be 76%.

So the answer might be 50 questions total, but maybe there was 49 questions total.

I would probably go with 50 questions total.

7 0

2 years ago

Consider the sets below. u = {x | x is a real number} a = {x | x is an odd integer} r = {x | x = 3, 7, 11, 27} is r ⊂ a?

-BARSIC- [3]

The correct option is (B) yes because all the elements of set R are in set A.

<h3>What is an element?</h3>

In mathematics, an element (or member) of a set is any of the distinct things that belong to that set.

Given sets:

U = {x | x is a real number}
A = {x | x is an odd integer}
R = {x | x = 3, 7, 11, 27}

So,

A = 1, 3, 5, 7, 9, 11... are the elements of set A.
R ⊂ A can be understood as R being a subset of A, i.e. all of R's elements can be found in A.
Because all of the elements of R are odd integers and can be found in A, R ⊂ A is TRUE.

Therefore, the correct option is (B) yes because all the elements of set R are in set A.

Know more about sets here:

brainly.com/question/2166579

#SPJ4

The complete question is given below:
Consider the sets below. U = {x | x is a real number} A = {x | x is an odd integer} R = {x | x = 3, 7, 11, 27} Is R ⊂ A?

(A) yes, because all the elements of set A are in set R

(B) yes, because all the elements of set R are in set A

(D) no, because each element in set R is not represented in set A

8 0

1 year ago

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

scoray [572]

Answer:

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

General Formulas and Concepts:

<u>Calculus</u>

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:

<u>Step 1: Define</u>

<em>Identify given limit</em>.

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

<u>Step 2: Find Limit</u>

Let's start out by <em>directly</em> 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:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$

∴ we have <em>evaluated</em> the given limit.

___

Learn more about limits: brainly.com/question/27807253

Learn more about Calculus: brainly.com/question/27805589

___

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

Unit: Limits

3 0

1 year ago