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3 years ago

Simplify 81^5 = 3x x=?

Mathematics

2 answers:

dangina [55]3 years ago

8 0

Um
81^5=3x
3,486,784,401=3x
x=1,162,261,467

Anon25 [30]3 years ago

5 0

its 20, got it on e dgenuity

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What is the least common multiple (LCM) of 6 and 10? OA. 2. OB. 20 O C. 30 OD. 60

Naily [24]

Answer:

Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

How do I solve this problem? -17=2x+7

Airida [17]

Answer
x=-12
Explanation

Move the terms
-17-2x=7

Add the numbers
-2x=24

Divide both sides of the equation by -2
x=-12

Hope this helps:)

4 0

3 years ago

What is the solution of the system?y=-8x-2 and y=-6x+4

Tcecarenko [31]

Hello :
y=-8x-2 and y=-6x+4
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8 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:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$

∴ 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 150 no container can hold 23.6 g of aluminum what is the density of the aluminum?

Drupady [299]

Answer:

# The density of the aluminum is 59/375 ≅ 0.157 g/cm³

# The volume of the container is 4/17 ≅ 0.235 cm³

# The mass of iron is 1225 g

Step-by-step explanation:

* Lets explain how to solve the problem

- Density is a measurement that compares the amount of matter

of an object to its volume [ Density = mass/volume]

- Density is found by dividing the mass of an object by its volume

- The unit of measuring density is gram per centimeter³ (milliliter)

* Lets solve the problems

# A container of volume 150 cm³ can hold 23.6 g of aluminum

∵ Density = mass/volume

∵ The mass = 23.6 g

∵ The volume = 150 cm³

∴ The density = 23.6/150 = 59/375 ≅ 0.157 g/cm³

* The density of the aluminum is 59/375 ≅ 0.157 g/cm³

# The density of mercury is 13.6 g/cm³ at 23 C°

- A container can hold 3.2 g of mercury at this temperature

∵ Density = mass/volume

∵ The mass = 3.2 g

∵ The density = 13.6 g/cm³

∴ 13.6 = 3.2/volume

- By using cross multiplication

∴ The volume = 3.2/13.6 = 4/17 cm³

* The volume of the container is 4/17 ≅ 0.235 cm³

# The density of the iron is 4.9 g/cm³

- The beaker has a volume 250 ml

∵ Density = mass/volume

∵ The ml = cm³

∴ The volume of the beaker is 250 cm³

∵ The density = 4.9 g/cm³

∴ 4.9 = mass/250

- Multiply both sides by 250

∴ mass = 1225 g

* The mass of iron is 1225 g

7 0

3 years ago