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3 years ago

Simplify the expression below.

Mathematics

2 answers:

mario62 [17]3 years ago

5 0

3 · -8 + 5 - 4 ÷ 2

Use PEMDAS

P=parenthesis

E=exponents

M= multiplication

A= addition

S=subtraction

-24+5- 4÷ 2

-19-2

=-21

Answer is B-21

Margaret [11]3 years ago

4 0

Answer:

B. -21

Step-by-step explanation:

You have to follow PEMDAS

3*-8=-24

So, -24+5-4÷2

-4÷2=-2

So, -24+5-2

Then add and subtract from left to right

-24+5=-19

-19-2=-21

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Answer:

Subtract

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Step-by-step explanation:

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3 years ago

Plss help for a brainlist :)))

sveticcg [70]

Answer:

The slope is 24

Step-by-step explanation:

Use any two points in the graph to find the slope of the graph
- - - - - - - - - - - - - - - - - - - -
Step 1. Use two points

Ex: (10, 320) & (5, 200)

Step 2. Find the slope using the two points

m = slope

m = change in y/change in x

m = (y2 - y1)/(x2 - x1)

m = (200 - 320)/(5 - 10)

m = -120/-5

m = 120/5

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The slope of the graph/line is 24

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2 years ago

Read 2 more answers

-3/5x - 7/20x + 1/4x= -56

romanna [79]

Answer:

x = 80

Step-by-step explanation:

Combine − 3/5x and 7/20x to get -19/20x.

Then combine -19/20x and 1/4x and you get -7/10x.

Then you multiply both sides by -10/7 which is the reciprocal of -7/10.

Make -56(-10) one fraction.

Multiply -56 and -10 and you get 560.

Then divide 560 by 7 to get 80.

8 0

3 years ago

Read 2 more answers

Which exponential function is represented by the values in the table?

svlad2 [7]

Here we use the equation

$y = a(b)^x$

Taking points (0,4), (1,2)

Substituting the point (0,4) , we will get

$4 = a(b)^0
\\
4 = a(1) \\
a =4$

Substituting (1,2) we will get

$2 = 4 (b)^1
\\
2/4 = b
\\
b = 1/2$

So we have

$a = 4, b = 1/2$

Therefore , required equation is

$y = 4(1/2)^x$

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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:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

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.

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Learn more about limits: brainly.com/question/27807253

Learn more about Calculus: brainly.com/question/27805589

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

Unit: Limits

3 0

2 years ago