Х | 40 yd Parking Lot In Exercises 49-52, use the following information. Your school plans to increase the area of the parking lot by 1000 square yards. The original parking lot is a rectangle, as shown. The length and width of the original parking lot will each increase by x yards. The width parking of the original parking lot is 40 yards, and the length lot of the original parking lot is 50 yards. 49. Find the area of the original parking lot. 50 yd 50. Find the total area of the parking lot with the new space. 51. Write an equation that you can use to find the value of x. 52. Solve the equation from Exercise 51. By how many yards should the length and width of the parking lot increase? need help with 52 I have the answer for 51 just a little lost on how to 52

c-4 is the final answer! : )

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:

answer: none of these answers are correct only the second half is

step-by-step explanation:

a reflection across the x axis and a 90 degree rotation left and right.