Can someone really help me please !

Step-by-step explanation:

The markings shown are typical of rulers graduated in inches and fractions. Metric rulers and engineering scales are typically graduated differently, often in units of 10ths or 100ths. Architectural scales are often graduated in ways that facilitate creating and reading inches and feet from drawings with different scale factors.

First, you notice that the markings are of different lengths. The longest marks have numbers beside them. These are whole inches.

First Ruler

The first ruler shown is measuring a length of 3 inches.

Halfway between the inch marks are marks that are slightly shorter. These are the 1/2 inch marks. The one midway between 4 and 5 represents 4 1/2 inches, for example.

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The next shorter marks are in the middle of the two spaces created by the 1/2-inch marks. These are the 1/4 inch marks. The one to the left (toward the lower number) represents 1/4; the one to the right (toward the higher number) represents 3/4. Of course, the 1/2 inch mark also corresponds to 2/4.

Slightly shorter than the 1/4-inch marks, and in the middle of the spaces created by the 1/4-inch (and 1/2-inch) marks, are the 1/8-inch marks. You can see that each set of marks divides the previous distances in half.

Second, Third Rulers

On the second and third rulers shown, the end of the green bar is at the first 1/8-inch mark after the number. That means the length is the number plus 1/8. Those two rulers show lengths of 4 1/8 and 8 1/8, respectively.

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Each successive level of marks is 1/2 the length of the previous level. So, the marks just shorter than 1/8 are 1/16. Finally, the shortest marks are 1/32 inch.

Fourth Ruler

The fourth ruler shows a length that is 1/16 inch less than 1/4 inch less than 5 inches. The length shown there is ...

  5 -1/4 -1/6

  = 5 - 4/16 -1/16

  = 5 -5/16

  = 4 11/16

You can also recognize that length as being ...

  4 1/2 + 1/8 + 1/16 = 4 + (8 +2 + 1)/16 = 4 11/16

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: