All categories

3 years ago

Marvin rides his bike at an average speed of 13 miles per hour. How far can he bike in 8 hours?

Mathematics

1 answer:

Viefleur [7K]3 years ago

8 0

Answer:

104 miles

Step-by-step explanation:

13 miles * 8 hours = 104 miles

You might be interested in

Does anyone want to help me with my math work?

Lerok [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What are the possible factors of 39

Marizza181 [45]

1,3,13,39 are possible factors

8 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

59-(2c+3)=4(c+7)+c.

alisha [4.7K]

Step-by-step explanation:

${:}\leadsto$ $\sf 59-(2c+3)=4 (c+7)+c$

${:}\leadsto$ $\sf 59-2c-3=4c+28+c$

${:}\leadsto$ $\sf 56-2c=5c+28$

${:}\leadsto$ $\sf 5c+28=56-2c$

${:}\leadsto$ $\sf 5c+2c=56-28$

${:}\leadsto$ $\sf 7c=28$

${:}\leadsto$ $\sf c={\dfrac {28}{7}}$

${:}\leadsto$ $\sf c=4$

4 0

2 years ago

Read 2 more answers

Which statement is true about whether Z and B are

Dmitrij [34]

Answer:

Z and B are independent events because P(Z I B) = P(Z)

Step-by-step explanation:

If Z and B are independent, Z does not affect B and B does not affect Z.
This means that, to know something about B does not help predicting Z, and to know something about B does not help predicting Z.
Then, the probability P (Z/B) = P(Z) because B does not add any information about Z, therefore, the probability of Z only depends on itself.

6 0

3 years ago

Read 2 more answers