14:2= 20: wat is the equivalent ratio

Mathematics

2 answers:

lina2011 [118]2 years ago

6 0

By cross multiplying, the answer should be 2.9

sukhopar [10]2 years ago

6 0

Answer:

$\large\boxed{14:2=20:\dfrac{20}{7}}$

Step-by-step explanation:

$14:2=20:x\\\\\dfrac{14\!\!\!\!\!\diagup^7}{2\!\!\!\!\diagup_1}=\dfrac{20}{x}\\\\\dfrac{7}{1}=\dfrac{20}{x}\qquad\text{cross multiply}\\\\7x=(20)(1)\\\\7x=20\qquad\text{divide both sides by 7}\\\\x=\dfrac{20}{7}$

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ExtremeBDS [4]

Answer:5 min I think

Step-by-step explanation: 5 min I think

6 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

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