Ask question

Ask question

All categories

kondor19780726 [428]

4 years ago

12

A swimming pool contains 1000 gallons of water. 20 gallons is added every hour. The expression 1000 + 20h represents the number

of gallons of water in the pool after a certain number of hours?

1 answer:

nadya68 [22]4 years ago

3 0

Specify question

Please specify the question it depends on how much gallons they were adding per hour

You might be interested in

Write two equivalent fractions

Fynjy0 [20]

I love Algebra so I dont mind answering this. I will start simple: 1/1 and 2/2, 1/2 and 2/4, 1/3 and 2/6. Hope this helped

4 0

4 years ago

Read 2 more answers

3.14 times 16 divided by 3

AleksAgata [21]

Answer:

The answer is 16.746667

Step-by-step explanation:

1) Simplify 3.14 × 16 to 50.24.

$50.24 \div 3$

2) Simplify 50.24 ÷ 3 to 16.746667.

$16.746667$

<u>Therefor</u><u>. </u><u>the</u><u> </u><u>answer</u><u> </u><u>is</u><u> </u><u>16.746667.</u>

5 0

3 years ago

Find the solution to the system

Tatiana [17]

Answer:

m

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Round 2.1132 to the nearest hundredth.<br> Answer:

Schach [20]

Answer:

2.11

Step-by-step explanation:

hope this helps :)

4 0

3 years ago

Read 2 more answers

Of the 45 animals on the farm, there are only pigs and chickens. The animals have a total of 124 legs. How many pigs and how man

Bess [88]

Answer:

$on \: this \: farm \: there \: are \to \\ \underline{ \boxed{17 \: pigs }}\: \\ and \\ \underline{ \boxed{28 \:chickens}}$

Step-by-step explanation:

$\\ \underline{ \boxed{solution \: to \: question \: \boxed{ A}}} \\ \\ let \: the \: pigs \: be \to \: p \\ let \: the \: chicks \: be \to \: c \\let \: the \: total \: animals \: be \to \: 45 \\ hence \to \\ p + c = 45.......eq(1) \\ since \: chickens \: have \: two \: legs : \\ while \: pigs \: have \: four \: legs : \\ let \: animals \: with \: 2 \: legs \: be \to \: 2c \\ let \: animals \: with \: 4\: legs \: be \to \: 4p \\ the \: total \: number \: of \: animal \: legs = 124 \\ hence \to \\ 4p + 2c = 124..........eq(2) \\ therefore \: the \: system \: of \: equations \: \\ to \: describe \: the \: situation \: is \to \\ \\ p + c = 45 \\ 4p + 2c = 124 \\ \\ \\ \underline{ \boxed{solution \: to \: question \: \boxed{B }}} \\ \\ \\ we \: could \: apply \: any \: of \: the \: simultanious \\ principles \to \\ here : am \: appling \: the \: elimination \: method \to \\ p + c = 45 \\ 4p + 2c = 124 \\ to \: elimnate \: one \: varaible \: we \: multiply \: by \to \\ (4) \times p + c = 45 \\ (1) \times 4p + 2c = 124 \\ hence \to \\ - \frac{4p + 4c = 180}{4p + 2c = 124} \\next \to \\ 2c = 56 \\ c = \frac{56}{2} \\ \boxed{ c = 28} \\ lets \: work \: for \: p \to \\ p + c = 45 \\ p + 28 = 45 \\ p = 45 - 28 \\ \boxed{p = 17}$

6 0

3 years ago