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3 years ago

Find the midpoint of the line segment with the given endpoints. (14, -6), (13, 14)

Mathematics

2 answers:

Cloud [144]3 years ago

3 0

Answer:

Step-by-step explanation:

Flura [38]3 years ago

3 0

Answer:

( $\frac{27}{2}$ , 4 )

Step-by-step explanation:

Use the midpoint formula

midpoint = [ $\frac{1}{2}$ (x₁ + x₂), $\frac{1}{2}$ (y₁ + y₂ ) ]

with (x₁, y₁ ) = (14, - 6) and (x₂, y₂ ) = (13, 14), hence

[ $\frac{1}{2}$ (14 + 13), $\frac{1}{2}$ (- 6 + 14) ]

= ( $\frac{27}{2}$ , 4)

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How do you find the area of a figure

Nostrana [21]

Answer:

Any figure?

Step-by-step explanation:

3 0

3 years ago

Need answer 25 and answer 27

UNO [17]

$\\ \bull\tt\dashrightarrow \dfrac{-3}{8}x-20+2x>6$

$\\ \bull\tt\dashrightarrow \dfrac{-3}{8}x+2x=6+20=26$

$\\ \bull\tt\dashrightarrow \dfrac{-3+16}{8}x=26$

$\\ \bull\tt\dashrightarrow \dfrac{13}{8}x=26$

$\\ \bull\tt\dashrightarrow x=26\times \dfrac{8}{13}$

$\\ \bull\tt\dashrightarrow x=8(2)=16$

#27

$\\ \bull\tt\dashrightarrow 0.5x-4-2x\leqslant 2$

$\\ \bull\tt\dashrightarrow 0.5x-2x\leqslant=2+4=6$

$\\ \bull\tt\dashrightarrow -1.5x\leqslant 6$

$\\ \bull\tt\dashrightarrow x\leqslant\dfrac{6}{-1.5}$

$\\ \bull\tt\dashrightarrow x\leqslant 4$

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