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3 years ago

HELP ASAP

Mathematics

1 answer:

almond37 [142]3 years ago

6 0

Answer:

Step-by-step explanation:

let the number of dimes=x

and number of quarters=y

x=2y

0.10 x+0.25 y=2.25

or 10x+25y=225

divide by 5

2x+5y=45

2 ×2y+5y=45

4y+5y=45

9y=45

y=45/9=5

x=2y=2×5=10

Hence dimes=10

quarters=5

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<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:

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By the power rule:

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thus

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

Nothing to do to simplify any further, other than factoring out $x^2$ .

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