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3 years ago

What is the distance between A and B? Round your answer to the nearest tenth.

1 answer:

4 0

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-1}~,~\stackrel{y_1}{5})\qquad 
B(\stackrel{x_2}{4}~,~\stackrel{y_2}{1})\qquad \qquad 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AB=\sqrt{[4-(-1)]^2+[1-5]^2}\implies AB=\sqrt{(4+1)^2+(1-5)^2}
\\\\\\
AB=\sqrt{5^2+(-4)^2}\implies AB=\sqrt{41}$

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Try the numbers 22, 23, 24, 25 in the equation 4/3 = 32/d to test whether any of them is a solution. a. 23 is a solution. c. 24

S_A_V [24]

Answer:

c. 24 is a solution

Step-by-step explanation:

Given equation:

$\frac{4}{3}=\frac{32}{d}$

To test the numbers 22, 23, 24, 25 for the solution.

Solution:

In order to test the given number for the solution, we will plugin each number in the unknown variable $d$ and see if it satisfies the equation.

1) $d=22$

$\frac{4}{3}=\frac{32}{22}$

Reducing fraction to simplest form by dividing the numerator and denominator by their G.C.F.

$\frac{4}{3}=\frac{32\div 2}{22\div 2}$

$\frac{4}{3}=\frac{16}{11}$

The above statement can never be true and hence 22 is not a solution.

2) $d=23$

$\frac{4}{3}=\frac{32}{23}$

The fractions can no further be reduced.

The statement can never be true and hence 23 is not a solution.

3) $d=24$

$\frac{4}{3}=\frac{32}{24}$

Reducing fraction to simplest form by dividing the numerator and denominator by their G.C.F.

$\frac{4}{3}=\frac{32\div 8}{24\div 8}$

$\frac{4}{3}=\frac{4}{3}$

The above statement is always true and hence 24 is a solution.

4) $d=25$

$\frac{4}{3}=\frac{32}{25}$

The fractions can no further be reduced.

The statement can never be true and hence 25 is not a solution.

4 0

3 years ago

What is your estimate for the number of airline flights in a year?

Firdavs [7]

The answer is between 45,000,000-65,000,000

8 0

3 years ago

When Colton commutes to work, the amount of time it takes him to arrive is normally distributed with a mean of 41 minutes and a

N76 [4]

Answer:

(34, 48)

Step-by-step explanation:

According to the Empirical Rule, 95% of normally distributed data lie within two standard deviations of the mean. That, in turn, means 95% of the data in this problem lie within 2(3.5 min), or 7 min, of the mean:

41 - 7 < mean < 41 + 7, or

34 < mean < 48, or simply (34, 48)

6 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:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$