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3 years ago

54) |1-2/3x|<1

Mathematics

1 answer:

Naily [24]3 years ago

4 0

$|a| < b\iff a < b\ \wedge\ a > -b\\\\|a| > b\iff a > b\ \vee\ a < -b$

$\left|1-\dfrac{2}{3}x\right| < 1\iff1-\dfrac{2}{3}x < 1\ \wedge\ 1-\dfrac{2}{3}x > -1\ \ \ |\text{subtract 1 from both sides}\\\\-\dfrac{2}{3}x < 0\ \vedge\ -\dfrac{2}{3}x > -2\ \ \ \ |\text{change the signs}\\\\\dfrac{2}{3}x > 0\ \wedge\ \dfrac{2}{3}x < 2\ \ \ \ |\text{multiply both sides by 3}\\\\2x > 0\ \wedge\ 2x < 6\ \ \ \ |\text{divide both sides by 2}\\\\x > 0\ \wedge\ x < 3\to0 < x < 3\to x\in(0,\ 3)$

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How do I find the equation of the perpendicular bisector between (-2,5) and (2,-1)?

Arturiano [62]

Answer:

$y = \frac{2}{3} x + 2$

Step-by-step explanation:

Start by finding the slope of the given line.

$\boxed{ slope = \frac{y _{1} - y_2 }{x_1 - x_2} }$

Slope of given line

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

$= \frac{5 + 1}{ - 4}$

$= \frac{6}{ - 4}$

$= - \frac{3}{2}$

A perpendicular bisector cuts through the line at its midpoint perpendicularly.

The product of the slopes of two perpendicular lines is -1.

Let the slope of the perpendicular bisector be m.

$- \frac{3}{2} m = - 1$

$m = - 1 \div ( - \frac{3}{2} )$

$m = - 1 \times ( - \frac{2}{3} )$

$m = \frac{2}{3}$

$y = \frac{2}{3}x + c$ , where c is the y-intercept.

To find the value of c, we need to substitute a pair of coordinates that lies on the perpendicular bisector into the equation. Since the perpendicular bisector passes through the midpoint of the given line, we can use the midpoint formula to find the coordinates.

$\boxed{midpoint = ( \frac{x _{1} + x _2}{2} , \frac{y_1 + y_2}{2} )}$

Midpoint of given line

$= ( \frac{ - 2+ 2}{2} , \frac{5 - 1}{2} )$

$=( \frac{0}{2} , \frac{4}{2} )$

= (0, 2)

$y = \frac{2}{3} x + c$

When x= 0, y= 2,

2= ⅔(0) +c

2= 0 +c

c= 2

Thus, the equation of the perpendicular bisector is $y = \frac{2}{3} x + 2$ .

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aleksandr82 [10.1K]

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$f'(c):=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

So the value of this limit is exactly the value of the derivative of $f(x)=e^{\sin x}$ at $x=\pi$ .

You have

$f'(x)=\cos x\,e^{\sin x}\implies f'(\pi)=\cos\pi\,e^{\sin\pi}=-1$

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