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devlian [24]
3 years ago
7

An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the

pool at the rate of 12 gallons per minute.
Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?

Mathematics
1 answer:
erastovalidia [21]3 years ago
5 0

An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the pool at the rate of 12 gallons per minute.

Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the pool at the rate of 12 gallons per minute.

Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the pool at the rate of 12 gallons per minute.

Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the pool at the rate of 12 gallons per minute.

Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?An average swimming pool contains 19200 gallons of water. The pool needs drained at the end of the summer. Water drains from the pool at the rate of 12 gallons per minute.

Which equation can be used to find the volume, V, of the water in the pool m minutes after draining begins?

61y)6y \times 2 \div  \sqrt[ \geqslant  | | \frac{ >  >  <  =  {348)48)98)8(978). >  \\  | >  | \sqrt[m \: \% log( \beta  \tan( \\  log_{ log_{ \binom{ \\  \\ {}cbc log(\pi \beta  \sec( \cos( \tan( \sin( \infty e log( ln( log( ln( > .15 - 756 { >  | >  | <  \sqrt[ < 5x6y6)9x9yyyy \div  \frac{ \\ 2697930046y - xy >  \geqslant \:  \binom{ log( \binom{ \sec( \gamma  \\ }m{cx > 03y9)y \times ) }{?} ) }{?}  \times \frac{?}{?} }{?}  \times \frac{?}{?} ]{?} | | }^{?} ) ) ) ) ) ) ) ) ) }{?} }(?) }(?) ) ) ]{?} | | }^{?} }{?} | | ]{?}

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<em><u>follow</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>my</u></em><em><u> </u></em><em><u>friend</u></em><em><u>. </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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