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shusha [124]
3 years ago
6

Help help help please please

Mathematics
1 answer:
irinina [24]3 years ago
4 0

Answer:

5x - 2 < 8 \\ 5x < 8 + 2 \\ 5x < 10 \\ x <  \frac{10}{5}  \\   \color{red}\boxed{x  < 2}

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A pool has 2 pipes, one to fill and one to empty it. Ms.Charles wants to fill the pool, but she mistakenly turns on both pipes a
dolphi86 [110]

-- The filler pipe can fill 1/6 of the pool every hour.

-- The drainer pipe can drain 1/10 of the pool every hour.

-- When they're filling and draining at the same time, the filler pipe
will win eventually, because it finishes more of the pool in an hour
than what the drain pipe can finish in an hour.

-- When they're filling and draining at the same time, then every hour,
1/6 of the pool fills and 1/10 of it empties.  The difference is   (1/6) - (1/10).

To do that subtraction, we need a common denominator.
The smallest denominator that works is 30.

       1/6  =  5/30

     1/10  =  3/30 .

So in every hour,  5/30 of the pool fills, and  3/30 of the pool empties.
The result of both at the same time is that  2/30 = 1/15  fills each hour.

If nobody notices what's going on and closes the drain pipe, it will take
<em><u>15 hours</u></em> to fill the pool.


If the drain pipe had <em><u>not</u></em> been open, the filler pipe alone could have filled
the pool <em><u>2-1/2 times</u></em> in that same 15 hours.  With both pipes open,
1-1/2 pool's worth of water went straight down the drain during that time,
and it was wasted. 

I would say that the school should take the cost of 1-1/2 poolsworth out
of Ms. Charles' pay at the rate of $5 a week.  I would, but that would
guarantee her more job security than she deserves after pulling a stunt
like that.

I hope this did not take place in California.


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3 years ago
What is the median number of pairs of shoes owned by the children ?
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Step-by-step explanation:

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Read 2 more answers
David y Angie tienen dos cartulinas iguales. David corta la suya en 3 trozos iguales y Angie la corta en 7 trozos. Los dos usan
Bess [88]

Answer:

(i) David ha empleado \frac{2}{3} de la cartulina.

(ii) Angie ha empleado \frac{2}{7} de la cartulina.

(iii) David ha usado más cartulina.

Step-by-step explanation:

El problema indica que David y Angie emplean una cartulina del mismo tamaño cada uno. David corta la suya en 3 pedazos iguales, mientras que Angie obtiene 7 pedazos iguales. Finalmente, cada uno emplea dos de sus pedazos. A continuación, respondemos a las preguntas del enunciado:

(i) <em>¿Qué fracción de cartulina ha usado David?</em>

La fracción de cartulina empleada por David es igual a los pedazos utilizados divididos por el total de pedazos. Esto es:

x = \frac{2}{3}

David ha empleado \frac{2}{3} de la cartulina.

(ii) <em>¿Qué fracción de cartulina ha usado Angie?</em>

Aplicando el mismo procedimiento del punto anterior, tenemos que:

x = \frac{2}{7}

Angie ha empleado \frac{2}{7} de la cartulina.

(iii) <em>¿Quién ha usado más cartulina?</em>

La persona que ha usado la mayor cantidad de cartulina es aquella que tiene el menor denominador. Por tanto, David ha usado más cartulina.

7 0
3 years ago
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