Mr. Smith wants to figure out which fast food restaurant is the most popular

When a transversal crosses parallel lines, all of the acute angles are congruent, and all of the obtuse angles are congruent. When it crosses at right angle, all of the angles are right angles.

2) All of the angles are right angles.

11x +13 = 90

11x = 77 . . . . . . subtract 13

x = 7 . . . . . . . divide by 11

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3) The two marked angles are acute.

16x = 80 . . . . the acute angles are congruent

x = 5 . . . . . . divide by 16

5 0

2 years ago

Please help me find the slope thank u

Rzqust [24]

Y2-Y1
Over X2-X1 and then there is your slode

6 0

3 years ago

Halla la tasa de variación de cada funcion en el intervalo [-4,3] e indica si es positiva , negativa o nula A) f(x)=x2-2x+4 B) f

masya89 [10]

Answer:

$A) \hspace{3}Rate\hspace{3}of\hspace{3}change=-5\hspace{3}Negative\\\\B)\hspace{3}Rate\hspace{3}of\hspace{3}change=-21\hspace{3}Negative$

Step-by-step explanation:

Given a function $f(x)$ , we called the rate of change to the number that represents the increase or decrease that the function experiences when increasing the independent variable from one value " $x_1$ " to another " $x_2$ ".

The rate of change of $f(x)$ between $x_1$ and $x_2$ can be calculated as follows:

$Rate\hspace{3}of\hspace{3}change=f(x_2)-f(x_1)$

For:

$f(x)=x^2-2x+4$

Let's find $f(x_1)$ and $f(x_2)$ , where:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=(-4)^2-2(4)+4=16-8+4=12\\f(x_2)=f(3)=(3)^2-2(3)+4=9-6+4=7$

So:

$Rate\hspace{3}of\hspace{3}change =7-12=-5\hspace{3}Negative$

And for:

$f(x)-3x+2$

Let's find $f(x_1)$ and $f(x_2)$ , where:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7$

So:

$Rate\hspace{3}of\hspace{3}change =-7-14=-21\hspace{3}Negative$

<em>Translation:</em>

Dada una función $f(x)$ , llamábamos tasa de variación al número que representa el aumento o disminución que experimenta la función al aumentar la variable independiente de un valor " $x_1$ " a otro " $x_2$ ".

La tasa de variación de $f(x)$ entre $x_1$ y $x_2$ , puede ser calculada de la siguiente forma:

$Tasa\hspace{3}de\hspace{3}variacion=f(x_2)-f(x_1)$

Para:

$f(x)=x^2-2x+4$

Encontremos $f(x_1)$ y $f(x_2)$ , donde:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7$

Entonces:

$Tasa\hspace{3}de\hspace{3}variacion =7-12=-5\hspace{3}Negativa$

Y para:

$f(x)-3x+2$

Encontremos $f(x_1)$ y $f(x_2)$ , donde:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7$

Entonces:

$Tasa\hspace{3}de\hspace{3}variacion=-7-14=-21\hspace{3}Negativa$

8 0

3 years ago