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san4es73 [151]
3 years ago
12

For which intervals is the function decreasing?

Mathematics
1 answer:
lilavasa [31]3 years ago
5 0

Answer:

The function is decreasing in the following intervals

A. (0, 1)

C. (2, pi)

Step-by-step explanation:

To answer this question, imagine that you draw lines of slope m parallel to the function shown at each point.

-If the slope of this line parallel to the function is negative for those points then the function is decreasing.

-If the slope of this line parallel to the function is positive for those points then the function is increasing.

Observe in the lines drawn in the attached image. You can see that they have slope less than zero in the following interval:

(0, 1) U (2, pi)

Therefore the correct option is:

A. (0, 1)

C. (2, pi)

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The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate (\dot V), measured in liters per hour, is directly proportional to draining time (t), measured in hours. That is:

\dot V \propto \frac{1}{t}

\dot V = \frac{k}{t} (Eq. 1)

Where k is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) <em>Large and small drains are opened</em>

\dot V_{s}+\dot V_{l} = \frac{k}{2} (Eq. 2)

\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}

(ii) <em>Only the small drain is opened</em>

\dot V_{s} = \frac{k}{t_{l}+3} (Eq. 3)

\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}

(iii) <em>Only the big drain is opened</em>

\dot V_{l} = \frac{k}{t_{l}} (Eq. 4)

\frac{\dot V_{l}}{k}  = \frac{1}{t_{l}}

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}

\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}

2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}

t_{l}^{2}-t_{l}-3 = 0 (Eq. 5)

Whose roots are determined by the Quadratic Formula:

t_{l,1}\approx 2.303\,h and t_{l,2} \approx -1.302\,h

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

t_{s} = 2.303\,h+3\,h

t_{s} = 5.303\,h

The draining time when only the small drain is opened is 5.303 hours.

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