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elena-14-01-66 [18.8K]

2 years ago

9

the amount of rent paid by a teacher is two-fifths of the teacher's monthly income. The rent paid by the teacher is $1100. What

is the teacher's monthly income?

1 answer:

miv72 [106K]2 years ago

3 0

Answer:

ඞ

Step-by-step explanation:

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Which function represents an exponential

Leviafan [203]

I think its A!! But im not sure

6 0

3 years ago

A cheetah can run at its top speed for about 25 seconds. Complete the table to represent a cheetah running at a constant speed.

olchik [2.2K]

Answer:

Cheetahs can go from 0 to 60 miles per hour in just 3.4 seconds and reach a top speed of 70 miles per hour. While they are the fastest land animal in the world, they can only maintain their speed for only 20 to 30 seconds.

Step-by-step explanation:

7 0

2 years ago

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Line FG goes through the points (4,9) and (1,3). Which equation represents a line that is perpendicular to FG and passes through

stiks02 [169]

Answer:

$x + 2y= 2$

Step-by-step explanation:

Given

Points:

$F = (4,9)$

$G = (1,3)$

Required

Determine the equation of line that is perpendicular to the given points and that pass through $(2,0)$

First, we need to determine the slope, m of FG

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where

$F = (4,9)$ --- $(x_1,y_1)$

$G = (1,3)$ --- $(x_2,y_2)$

$m = \frac{3 - 9}{1 - 4}$

$m = \frac{- 6}{- 3}$

$m =2$

The question says the line is perpendicular to FG.

Next, we determine the slope (m2) of the perpendicular line using:

$m_2 = -\frac{1}{m}$

$m_2 = -\frac{1}{2}$

The equation of the line is then calculated as:

$y - y_1 = m_2(x - x_1)$

Where

$m_2 = -\frac{1}{2}$

$(x_1,y_1) = (2,0)$

$y - 0 = -\frac{1}{2}(x - 2)$

$y = -\frac{1}{2}(x - 2)$

$y = -\frac{1}{2}x + 1$

Multiply through by 2

$2y = -x + 2$

Add x to both sides

$x + 2y= -x +x+ 2$

$x + 2y= 2$

Hence, the line of the equation is $x + 2y= 2$

8 0

3 years ago

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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

Help please! 15 pts!

Afina-wow [57]

Answer: attached below

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers