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konstantin123 [22]

3 years ago

7

Graph in slope intercept form 2x-5y=5

1 answer:

elena55 [62]3 years ago

4 0

y= $\frac{2}{5}$ x-1

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Need help ASAP NO LINKS

matrenka [14]

Answer:

9

Step-by-step explanation:

5(n-2)=35

5n-10=35

5n=35+10

5n=45

n=45/5

n=9

Please mark me as brainliest

6 0

3 years ago

Read 2 more answers

PLEASE HELP ME GUYS OR I WONT PASS <br>this calculus!!!!

KonstantinChe [14]

Answer:

b. $\displaystyle \frac{1}{2}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Algebra I</u>

Functions
Function Notation
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$ <u> </u>

<u>Calculus</u>

Derivatives

Derivative Notation

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</em>

<em /> $\displaystyle H(x) = \sqrt[3]{F(x)}$ <em />

<em />

<u>Step 2: Differentiate</u>

Rewrite function [Exponential Rule - Root Rewrite]: $\displaystyle H(x) = [F(x)]^\bigg{\frac{1}{3}}$
Chain Rule: $\displaystyle H'(x) = \frac{d}{dx} \bigg[ [F(x)]^\bigg{\frac{1}{3}} \bigg] \cdot \frac{d}{dx}[F(x)]$
Basic Power Rule: $\displaystyle H'(x) = \frac{1}{3}[F(x)]^\bigg{\frac{1}{3} - 1} \cdot F'(x)$
Simplify: $\displaystyle H'(x) = \frac{F'(x)}{3}[F(x)]^\bigg{\frac{-2}{3}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle H'(x) = \frac{F'(x)}{3[F(x)]^\bigg{\frac{2}{3}}}$

<u>Step 3: Evaluate</u>

Substitute in <em>x</em> [Derivative]: $\displaystyle H'(5) = \frac{F'(5)}{3[F(5)]^\bigg{\frac{2}{3}}}$
Substitute in function values: $\displaystyle H'(5) = \frac{6}{3(8)^\bigg{\frac{2}{3}}}$
Exponents: $\displaystyle H'(5) = \frac{6}{3(4)}$
Multiply: $\displaystyle H'(5) = \frac{6}{12}$
Simplify: $\displaystyle H'(5) = \frac{1}{2}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e

5 0

3 years ago

Which expression is equivalent to log Subscript c Baseline StartFraction x squared minus 1 Over 5 x EndFraction?

Harrizon [31]

The equivalent expression of $\log_c(\frac{x^2 - 1}{5x})$ is $\log_c(x^2 - 1) - \log_c(5x)$

<h3>How to determine the equivalent expression?</h3>

The logarithmic expression is given as:

$\log_c(\frac{x^2 - 1}{5x})$

The law of logarithm states that:

log(a) - log(b) = log(a/b)

This means that the expression can be split as:

$\log_c(\frac{x^2 - 1}{5x}) = \log_c(x^2 - 1) - \log_c(5x)$

Hence, the equivalent expression of $\log_c(\frac{x^2 - 1}{5x})$ is $\log_c(x^2 - 1) - \log_c(5x)$

Read more about equivalent expression

brainly.com/question/2972832

#SPJ1

3 0

2 years ago

Find limit as x approaches 2 from the left of the quotient of the absolute value of the quantity x minus 2, and the quantity x m

Molodets [167]

Answer:

$\lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1$ .

Step-by-step explanation:

We want to find $\lim_{x \to 2^-} \frac{|x-2|}{x-2}$ .

By definition:

$|x-2|=\left \{ {{x-2,\:if\:x\:>\:2} \atop {-(x-2),\:if\:x\:$

Since we want to find the Left Hand Limit, we use f(x)=-(x-2)

$\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} \frac{-(x-2)}{x-2}$ .

$\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} (-1)$ .

The limit of a constant is the constant.

$\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1$ .

8 0

3 years ago

Question 20 fast please

kipiarov [429]

Answer:

Square root of 4 is 2

so 1+ $\frac{[1+\sqrt{2x-1]}^2 }{4}$ =4

frac{[1+sqrt{2x-1]}^2 }{4}=3

so 1+sqrt{2x-1]}^2=12

so 1+sqrt{2x-1}=V12

sqrt{2x-1}=V12-1

2x-1=(V12-1)^2=12-2V12+1=13-2V12

2x=13+1-2V12

2x=14-2V12

x=7-V12

x=7-2V3

I put an attachment to explain better

Step-by-step explanation:

6 0

3 years ago