Arectangle is transformed according to the rule r0, 90º. the image of the rectangle has vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4). what is the location of q?
a. (-4,-3)
b. (-3,-4)
c. (3,4)
4. (4,3)

The location of Q is:

          Option: D

      D.    (4,3)

The transformation R0,90° means that the figure is rotated 90 degree in the anticlockwise direction about the origin.

Hence, the rule of the transformation is given by:

                (x,y) → (-y,x)

Here we are asked to find the location of Q.

Let Q be located at (x,y)

This means that the coordinates of the image of point after transformation i.e. Q' is:

(-y,x)

From the given information we have the coordinates of Q' are: (-3,4)

This means that:

(-y,x)= (-3,4)

i.e. y=3 and x=4

Hence, the coordinates of Q are:

                   (4,3)

(-4, -3) if a clockwise transformation was made

(4, 3) if a counterclockwise transformation was made

Step-by-step explanation:

The rule R0, 90º means: 90° rotation about the origin, but there are 2 options clockwise and counterclockwise.  

Clockwise transformation maps from point (x,y) to point (y, -x)

Counterclockwise transformation maps from point (x,y) to point (-y, x).

If a clockwise transformation was made, then Q coordinates are (-4, -3), so that, after transformation Q' is point (-3,4).

If a counterclockwise transformation was made, then Q coordinates are (4, 3), so that, after transformation Q' is point (-3,4)

c > -10.55

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

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step-by-step explanation: