Arectangle swimming pool is 4 ft deep. one side of the pool is 2.5 times longer than the other. the amount of water needed to fill the swimming pool is 1960 cubic feet. find the dimensions of the pool. (find largest and smallest)

The dimensions of the pool are length is 35 feet and width is 14 feet.

Given that,

A rectangle swimming pool is 4 ft deep.

One side of the pool is 2.5 times longer than the other.

The amount of water needed to fill the swimming pool is 1960 cubic feet.

We have to determine,

Find the dimensions of the pool. (find largest and smallest).

According to the question,

The area of the pool by dividing the volume by the depth:

The pool is 490 square feet.

And The area is found by multiplying the length by the width.

Let, The width of the pool is X.

One side of the pool is 2.5 times longer than the other.

Length is 2.5X (2.5 times longer)

Then,

The width of the pool is 14 feet,

And The length is = 2x = 2.5 (14) = 35 feet.

Hence, The required dimensions of the pool are length is 35 feet and width is 14 feet.

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Find the are of the pool by dividing the volume by the depth:

1960 / 4 = 490

The pool is 490 square feet.

Area is found by multiplying the length by the width.

Let the width = X

We are told the length is 2.5X ( 2.5 times longer)

So we now have 2.5x * x = 490

2.5x * x = 2.5x^2

Now we have 2.5x^2 = 490

Divide both sides by 2.5:

x^2 = 490/2.5

x^2 = 196

find X by taking the square root of 196:

x = √196

x = 14

The width is 14 feet

The length is 2.5 * 14 = 35 feet

christopher is 272,195 year old

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: