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2 years ago

Which is the independent and which is the dependent?

2 answers:

7 0

Independent - You and a friend go to see the most recent superhero movie.

dependent- the length of the movie is 109 minutes, but of course, there will

also be previews.

sdas [7]2 years ago

3 0

Answer:

Independent: The Number of Minutes taken up by previews

Dependent: The Number of Minutes you spend watching previews and the movie

Step-by-step explanation:

=D Good Luck!

You might be interested in

(x-3)(2x+1) Can someone help me on this please

a_sh-v [17]

2x^2-5x-3

This is the answer I believe, I apologize if it’s incorrect

6 0

2 years ago

Find the value of the discriminant for <img src="https://tex.z-dn.net/?f=7x%5E%7B2%7D%20%2B5x%2B1%3D0" id="TexFormula1" title="7

kondor19780726 [428]

Answer:

No real roots

Step-by-step explanation:

Given

7x² + 5x + 1 = 0 ← in standard form

with a = 7, b = 5, c = 1

To determine the nature of the roots use the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real and distinct

• If b² - 4ac = 0 then roots are real and equal

• If b² - 4ac < 0 then the roots are not real

Here

b² - 4ac = 5² - (4 × 7 × 1) = 25 - 28 = - 3

Thus the 2 roots are not real

7 0

2 years ago

Lucy obtains a 1-year payday loan for $5000.00 at 12% interest compounded monthly. To get the loan, she

AURORKA [14]

Answer:

An origination fee is typically 0.5% to 1% of the loan amount and is charged by a lender as compensation for processing a loan application. Origination fees are sometimes negotiable, but reducing them or avoiding them usually means paying a higher interest rate over the life of the loan

Step-by-step explanation:

3 0

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