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3 years ago

Which set is an example of like fractions? A. 2⁄1 and 2⁄3 B. 1⁄2 and 3⁄2 C. 7⁄4 and 4⁄7 D. 10⁄10 and 5⁄5

Mathematics

2 answers:

ratelena [41]3 years ago

6 0

The answer is B. 1⁄2 and 3⁄2!

Sloan [31]3 years ago

5 0

Fractions which have the same denominator are called like fractions.
The answer is B. 1⁄2 and 3⁄2

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In the diagram, the measure of angle 8 is 124°, and the measure of angle 2 is 84°.

Margaret [11]

m∠8 + m∠7 = 180°

therefore

m∠7 = 180° - m∠8

m∠8 = 124°

substitute:

m∠7 = 180° - 124° = 56°

Answer: 56 degrees

6 0

3 years ago

Read 2 more answers

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:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$

∴ we have evaluated 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

A unit rate is a rate in which the unit in the denominater is?

Sidana [21]

Answer: a unit rate is a rate with 1

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

La potencia que se obtiene de elevar a un mismo exponente un numero racional y su opuesto es la misma verdadero o falso?

malfutka [58]

Answer:

Falso.

Step-by-step explanation:

Sea $d = \frac{a}{b}$ un número racional, donde $a, b \in \mathbb{R}$ y $b \neq 0$ , su opuesto es un número real $c = -\left(\frac{a}{b} \right)$ . En el caso de elevarse a un exponente dado, hay que comprobar cinco casos:

(a) El exponente es cero.

(b) El exponente es un negativo impar.

(c) El exponente es un negativo par.

(d) El exponente es un positivo impar.

(e) El exponente es un positivo par.

(a) El exponente es cero:

Toda potencia elevada a la cero es igual a uno. En consecuencia, $c = d = 1$ . La proposición es verdadera.

(b) El exponente es un negativo impar:

Considérese las siguientes expresiones:

$d' = d^{-n}$ y $c' = c^{-n}$

Al aplicar las definiciones anteriores y las operaciones del Álgebra de los números reales tenemos el siguiente desarrollo:

$d' = \left(\frac{a}{b} \right)^{-n}$ y $c' = \left[-\left(\frac{a}{b} \right)\right]^{-n}$

$d' = \left(\frac{a}{b} \right)^{(-1)\cdot n}$ y $c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{(-1)\cdot n}$

$d' = \left[\left(\frac{a}{b} \right)^{-1}\right]^{n}$ y $c' = \left[(-1)^{-1}\cdot \left(\frac{a}{b} \right)^{-1}\right]^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c = (-1)^{n}\cdot \left(\frac{b}{a} \right)^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left[(-1)\cdot \left(\frac{b}{a} \right)\right]^{n}$

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left[-\left(\frac{b}{a} \right)\right]^{n}$

Si $n$ es impar, entonces:

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = - \left(\frac{b}{a} \right)^{n}$

Puesto que $d' \neq c'$ , la proposición es falsa.

(c) El exponente es un negativo par.

Si $n$ es par, entonces:

$d' = \left(\frac{b}{a} \right)^{n}$ y $c' = \left(\frac{b}{a} \right)^{n}$

Puesto que $d' = c'$ , la proposición es verdadera.

(d) El exponente es un positivo impar.

Considérese las siguientes expresiones:

$d' = d^{n}$ y $c' = c^{n}$

$d' = \left(\frac{a}{b}\right)^{n}$ y $c' = \left[-\left(\frac{a}{b} \right)\right]^{n}$

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{n}$

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = (-1)^{n}\cdot \left(\frac{a}{b} \right)^{n}$

Si $n$ es impar, entonces:

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = - \left(\frac{a}{b} \right)^{n}$

(e) El exponente es un positivo par.

Considérese las siguientes expresiones:

$d' = \left(\frac{a}{b} \right)^{n}$ y $c' = \left(\frac{a}{b} \right)^{n}$

Si $n$ es par, entonces $d' = c'$ y la proposición es verdadera.

Por tanto, se concluye que es falso que toda potencia que se obtiene de elevar a un mismo exponente un número racional y su opuesto es la misma.

3 0

3 years ago

Express the following as a unit rate : 5 croissants for $10

raketka [301]

Answer:

$2/croissant

Step-by-step explanation:

$10÷ 5croissants = $2/ 1 croissant

7 0

3 years ago