I NEED ASAP! PLEASE.. god bless ❤️ The cylindrical water tank on a semitrailer has a length of 20 feet. The volume of the tank is equal to the product of π, the square of the radius of the tank, and the length of the tank.

Let V represent the volume of the tank, r represent the radius of the tank, and h represent the length of the tank.

1.) Create and equation that could be used to find the volume, V, of the cylindrical tank.

2.) Rewrite the volume formula to create an equation that can be used to calculate the radius, r, of the water tank.

Drag the terms to the correct locations in the equation. Not all terms will be used.

r = √_blank_
blank

Options: 20V, 400πh, h, 20h, V, 20π

3.) Graph the radical equation that can be used to calculate the radius, r, if the tank.

4.) Suppose the cylindrical water tank has a radius of 12 feet. Use this information and the equation modeling the radius of the tank to complete these statements

The equation modeling this situation has (one, no, two) extraneous solution/s.

The volume of the water tank is about (754, 2,880, 400, 9,048) cubic feet.

That’s all! If you answer correctly you’ll receive BRAINLIEST! :)

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  1) V = πr^2·h

  2) r = √(V/(20π))

  3) see attached

  4) 9048 ft^3; no extraneous solutions

Step-by-step explanation:

1) The problem statement tells you the formula for the volume is ...

__

2) Rewritten to find r, given that h=20, we have ...

__

3) see the attachment for a graph of r(V). Volume is on the horizontal axis; radius is on the vertical axis.

__

4) There are no extraneous solutions. The volume of the tank is about 9048 cubic feet. The graph shows it as 2880π cubic feet. The numerical value of that is about 9047.79 ft³ ≈ 9048 cubic feet.

Filling in r=12 in the equation of (2), we have ...

  12 = √(V/(20π))

  144(20π) = V . . . . . . square both sides; multiply by 20π

This is the same equation as in (1). It has no extraneous solutions.

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: