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FinnZ [79.3K]
3 years ago
7

Which of the following equations represents the perpendicular bisector of WX graphed below?

Mathematics
1 answer:
kotykmax [81]3 years ago
8 0
The coordinates of the 2 given points are W(-5, 2), and X(5, -4).

First, we find the midpoint M using the midpoint formula:

\displaystyle{ M_{WX}= (\frac{x_1+x_2}{2},  \frac{y_1+y_2}{2} )=  (\frac{-5+5}{2},  \frac{2+(-4)}{2} )=(0, -1).

Nex, we find the slope of the line containing M, perpendicular to WX. We know that if m and n are the slopes of 2 parallel lines, then mn=-1.

The slope of WX is \displaystyle{ m= \frac{y_2-y_1}{x_2-x_1}= \frac{2-(-4)}{-5-5}= \frac{6}{-10}= -\frac{3}{5}.

Thus, the slope n of the perpendicular line is \displaystyle{  \frac{5}{3}.

The equation of the line with slope \displaystyle{ n= \frac{5}{3} containing the point M(0, -1) is given by:

\displaystyle{ y-(-1)=\frac{5}{3}(x-0)

\displaystyle{ y+1= \frac{5}{3}x

\displaystyle{ 3y+3=5x

\displaystyle{ 5x-3y-3=0

Answer: 5x-3y-3=0
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Se corta una manzana en 12 partes, se comen ocho partes. Qué fracción simplificada a su mínima expresión representa la cantidad
lakkis [162]

Answer:

A. 1/3

Step-by-step explanation:

Dado que:

una manzana se corta en 12 partes y se comen ocho partes.

La fracción de manzanas consumidas = 8/12

= 2/3

La fracción de manzanas que quedan = valor original de la manzana que se corta - la fracción de la manzana que se come

La fracción de manzanas que quedan  =1-  \dfrac{8}{12}

La fracción de manzanas que quedan  =\dfrac{12-8}{12}

La fracción de manzanas que quedan =\dfrac{4}{12}

A la fracción más baja; obtenemos = 1/3

6 0
3 years ago
Use the long division method to find the result when 3x^3+23x^2+23x+6 is divided by 3x+2.
horsena [70]

Answer:

x^{2} + 7x+3

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
1) Suppose f(x) = x2 and g(x) = |x|. Then the composites (fog)(x) = |x|2 = x2 and (gof)(x) = |x2| = x2 are both differentiable a
Rufina [12.5K]

Answer:

This contradict of the chain rule.

Step-by-step explanation:

The given functions are

f(x)=x^2

g(x)=|x|

It is given that,

(f\circ g)(x)=|x|^2=x^2

(g\circ f)(x)=|x^2|=x^2

According to chin rule,

(f\circ g)(c)=f(g(c))=f'(g(c)g'(c)

It means, (f\circ g)(c) is differentiable if f(g(c)) and g(c) is differentiable at x=c.

Here g(x) is not differentiable at x=0 but both compositions are differentiable, which is a contradiction of the chain rule

3 0
3 years ago
Help me this assignment is due soon.
AlekseyPX
So maybe it’s 5, it’s equivalent to the side with 6, just shorter
6 0
3 years ago
Write an exponential equation in the form y = abx whose graph passes through points (2, 18) and (5, 60.75).
yan [13]
18  = ab^2
60.75 = ab^5
from first equation:- a = 18/b^2
so
60.75 = (18/b^2) * b^5
60.75 = 18b^3
b = cube root (60.75 / 18)
b = 1.5  

so a =  18/1.5^2 = 8

so the required equation is y = 8(1.5)^x
8 0
3 years ago
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