Which of the following options best describes the function graphed below?

Mathematics

1 answer:

hodyreva [135]2 years ago

6 0

C.
The graph looks like it’s linear and is trending in an increasing way.

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One of the smallest pools on the ship has about 45,000 gallons of water in it.

gtnhenbr [62]

Amount of water in the pool at the end of the day is 5457798.7 gallons

<u>Explanation:</u>

Given:

Initial amount of water in the pool = 45,000 gallons

Increase in amount = 0.75 in per minute

Time, t = 1 day

t = 24 X 60 min

t = 1440 min

So,

Increase in amount of water in 1 day = 0.75 in X 1440

= 1080 in

Volume of 1080 in of water = (1080 in)³

Volume from cubic inch to gallon = 5453298.7 gallon

Amount of water in the pool at the end of the day = 45000 + 5453298.7 gallon

= 5457798.7 gallon

5 0

3 years ago

Find dy/dx for y= x^3 ln (cot x)

ICE Princess25 [194]

<h3>Answer</h3>

$\dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)$

<h3>Explanation</h3>

By the product rule $(d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)$ , we have

$\begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}$

By the chain rule:

$\begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}$

By the power rule:

$(x^3)' = 3x^2$

thus

$\begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}$