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andrew11 [14]
3 years ago
13

A pond is being drained by a pump. After 3 hours, the pond is half empty. A second pump is put into operation and together the t

wo pumps finish emptying the pond in half an hour. How long would it take the second pump to drain the pond if it had to do the same job alone?
Mathematics
2 answers:
wariber [46]3 years ago
8 0

Answer:

The second pump to drain the pond if it had to do the same job alone is \frac{3}{5} hour or 0.6 hour.

Step-by-step explanation:

Given : A pond is being drained by a pump. After 3 hours, the pond is half empty. A second pump is put into operation, and together the two pumps finish emptying the pond in half an hour.

To find : How long would it take the second pump to drain the pond if it had to do the same job alone?

Solution :

According to question,

Pump A drained in 3 hours.

So, Work done by pump in 1 hour is \frac{1}{3}

Together the two pumps finish emptying the pond in half an hour.

i.e. Pump A + Pump B drained in \frac{1}{2} hour

Work done by pump in 1 hour is \frac{1}{\frac{1}{2}}=2 hour

The work done by pump B is 2-\frac{1}{3} hour.

The work done by pump B is \frac{5}{3} hour.

So, Pump B drained in \frac{3}{5} hour.

Therefore, The second pump to drain the pond if it had to do the same job alone is \frac{3}{5} hour or 0.6 hour.

aleksandr82 [10.1K]3 years ago
5 0

The first pump empties half the pond in 3 hours, so in 1/6 that time (1/2 hour), it empties (1/6)·(1/2) = 1/12 of the pond.

The second pump empties the other 5/12 of the pond in that half hour, so has a pumping rate of (1/2 h)/(5/12 pond) = (6/5 h)/pond.

The second pump could do the entire job alone in 1 hour and 12 minutes.

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What term is 1/1024 in the geometric sequence,-1,1/4,-1/6..?
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Answer:

\large\boxed{\text{sixth term is equal to}\ \dfrac{1}{1024}}

Step-by-step explanation:

The explicit formula for a geometric sequence:

a_n=a_1r^{n-1}

a_n - n-th term

a_1 - first term

r - common ratio

r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=...=\dfrac{a_n}{a_{n-1}}

We have

a_1=-1,\ a_2=\dfrac{1}{4},\ a_3=-\dfrac{1}{6},\ ...

The common ratio:

r=\dfrac{\frac{1}{4}}{-1}=-\dfrac{1}{4}\\\\r=\dfrac{-\frac{1}{6}}{\frac{1}{4}}=-\dfrac{1}{6}\cdot\dfrac{4}{1}=-\dfrac{2}{3}\neq-\dfrac{1}{4}

<h2>It's not a geometric sequence.</h2>

If a_3=-\dfrac{1}{16} then the common ratio is r=\dfrac{-\frac{1}{16}}{\frac{1}{4}}=-\dfrac{1}{16}\cdot\dfrac{4}{1}=-\dfrac{1}{4}

Put to the explicit formula:

a_n=-1\left(-\dfrac{1}{4}\right)^{n-1}

Put a_n=\dfrac{1}{1024} and solve for <em>n </em>:

-1\left(-\dfrac{1}{4}\right)^{n-1}=\dfrac{1}{1024}\qquad\text{use}\ a^n:a^m=a^{n-m}\\\\-\left(-\dfrac{1}{4}\right)^n:\left(-\dfrac{1}{4}\right)^1=\dfrac{1}{1024}\\\\-\left(-\dfrac{1}{4}\right)^n\cdot(-4)=\dfrac{1}{1024}\\\\(4)\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{1024}\qquad\text{divide both sides by 4}\ \text{/multiply both sides by}\ \dfrac{1}{4}/\\\\\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{4096}\\\\\dfrac{(-1)^n}{4^n}=\dfrac{1}{4^6}\qquad n\ \text{must be even number. Therefore}\ (-1)^n=1

\dfrac{1}{4^n}=\dfrac{1}{4^6}\iff n=6

5 0
3 years ago
How do you determine the number of zeros to annex in the product of 0.002 and 0.003?
Marat540 [252]
<span>0.002 x 0.003 = 0.000006

the zeroes are multiplied by the power of 10 which is 1/10 in particular.
For example.
The product of a whole number and a decimal number less than 1 will be greater than the whole number multiplied into. For this theorem to be proven. Let us state the mathematical expression into numbers such that </span><span><span>
1. </span> N x 0.1 = N/0.1 < N</span> <span><span>
2. </span> 1 x 0.5 = 0.5 </span><span><span>
3. </span> 2 x 0.1 = 0.2</span> <span><span>
4. </span> 100 x 0.55 = 55</span><span>  </span>

<span>These three examples and stances then suggest the claim that the product is not equal to the whole number used in the equation.<span>
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Al sumar fracciones con igual denominador, debemos sumar numerador con numerador y denominador con denominador. Verdadero o fals
Ahat [919]

Answer:

La respuesta es falso.  

Step-by-step explanation:

La respuesta es falso.  

Cuando se suman fraccciones con igual denominador, se suman los numeradores (numerador con numerador) y se deja el mismo denominador (el cual es común en ambos). Por ejemplo, la suma de 1/5 + 3/5 da como resultado:    

\frac{1}{5} + \frac{3}{5} = \frac{1 + 3}{5} = \frac{4}{5}

En el caso de fracciones con diferentes denominadores, tampoco se suma numerador con numerador y denominador con denominador. En ese caso se debe encontrar el mínimo común múltiplo.

 

Por lo tanto, la respuesta es falso.

Espero que te sea de utilidad!

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