9. Look at the map. Explain why Elm Street and Birch Street are not perpendicular.

A. The slope of Birch Street is 0. To be perpendicular to Elm, Birch would have to have a slope that is undefined.

B. The slope of Birch Street is 0. To be perpendicular to Elm, Birch would have to have a slope of -4/7.

C. The slope of Birch Street is 0. To be perpendicular to Elm, Birch would have to have a slope of 7/4.

B. The slope of Birch Street is 0. To be perpendicular to Elm, Birch would have to have a slope of -4/7.

Step-by-step explanation:

I cannot see the choices, so, I will explain this using two different concepts and you can apply the explanation on the given choices

1- Explanation using angle formed:

Imagine that Elm street and Birch street are two lines.

Now, for to lines to be perpendicular, these two lines must intersect forming a 90 degree angle.

Now, noticing the intersection between the two streets, we will find that the angle, formed is not 90 degrees

2- Explanation using slopes:

For two lines to be perpendicular, the slopes of the two lines must be negative reciprocals of each other.

This means that, if the slope of one line is m, then the slope of the other must be -1/m for the two lines to be perpendicular.

Now, Birch street is a horizontal line, this means that its slope is zero.

For Elm street to be perpendicular on it, the slope of Elm street must be equal to -1/0. This means that the slope of Elm street has to be undefined which means that Elm street must be a perfectly vertical line.

However, from the graph, we can note that Elm street is an inclined line which means that its slope is not undefined.

This means that the two streets are not perpendicular to each other.

Hope this helps :)

The slope of Birch Street is 0. To be perpendicular to Elm, Birch would have to have a slope of -4/7.

step-by-step explanation:

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

ver más en -

step-by-step explanation: