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3 years ago

A rectangle bulliten board is 1 1/5 meters wide and 2 meters long

Mathematics

1 answer:

Vadim26 [7]3 years ago

7 0

Well what’s the question asking. And if that’s all then good for the board

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A simple random sample of size n=40 from a population with mu = 74 and standard deviation= 7. Does the population need to be nor

Maurinko [17]

Answer:

Step-by-step explanation:

How does lack of sleep affect risk of injury

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3 years ago

there's some number that if you subtract 15 from it first, and then multiply that total by 7, the result is 28. find the number

yulyashka [42]

(x-15)*7=28

x-15=4

x=19

Final answer: 19

6 0

3 years ago

Read 2 more answers

the coordinates of points a and b are (2,4) and (-2,-4) respectively. what is the slope of the line that connects the two points

solmaris [256]

we are given two points as

(2,4) and (-2,-4)

Let's assume

first point as (x1,y1)=(2,4)

so, x1=2 and y1=4

Second point as (x2,y2)=(-2,-4)

so, x2=-2 and y2=-4

now, we can use slope formula

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

now, we can plug values

$m=\frac{-4-4}{-2-2}$

now, we can simplify it

$m=2$

so, slope is 2 ..................Answer

4 0

3 years ago

Deﬁne the function add_mn that takes two integers m, n as arguments and returns m + 10 * n. For instance, add_mn 3 5 = 3 + 10*5

jok3333 [9.3K]

Answer:

// C++ Program to arithmetic operationf on 2 Numbers using Recursion

// Comments are used for explanatory purpose

#include <bits/stdc++.h>

using namespace std;

// add10 recursive function to perform arithmetic operations

int add10(int m, int n)

{

return (m + product(n, 10)); //Result of m + n * 10

return 0;

}

// Main Methods Starts here

int main()

{

int m, n; // 2 Variables m and n declared as integer

cin>>m; // accept input for m

cin>>n; // accept input for n

cout << "Result : "<<add10(m,n); // Print results which is calculated by m + 10 * n

return 0;

}

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.

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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

1 year ago