-- 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
15 hours to fill the pool.

If the drain pipe had not been open, the filler pipe alone could have filled
the pool 2-1/2 times 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.

6 0

3 years ago

Solve the following equation; domain 0< x < 360; 3 cot x + sqrt 3 = 0

kondaur [170]

ANSWER

$x =120 \degree \: or \: 300 \degree$

EXPLANATION

We want to solve the trigonometric equation

$3 \cot(x) + \sqrt{3} = 0$

We group like terms to obtain,

$3\cot(x) = - \sqrt{3}$

Divide both sides of the equation by 3 to get,

$\cot(x) = - \frac{ \sqrt{3} }{3}$

We can reciprocate both sides to get,

$\tan(x) = - \frac{3}{ \sqrt{3} }$

We simplify the right hand side to get,

$\tan(x) = - \sqrt{3}$

Since the tangent ratio is negative, it implies that it is either in the second quadrant or fourth quadrant.

In the second quadrant,

$x = 180 \degree - arctan( \sqrt{3})$

$x = 180 \degree - 60 \degree = 120 \degree$

In the fourth quadrant,

$x = 360 \degree - arctan( \sqrt{3})$

$x = 360 \degree - 60 \degree$

$x = 300 \degree$