3 years ago

Which describes how to find the solution 5k >22

Mathematics

1 answer:

inysia [295]3 years ago

8 0

Answer:

Step-by-step explanation:

$5k>22\qquad\text{divide both sides by 5}\\\\\dfrac{5k}{5}>\dfrac{22}{5}\\\\k>4\dfrac{2}{5}\to k>4.4\to k\in(4.4,\ \infty)$

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Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%20lnx%20%7D%20" id="TexFormula1" title=" \sqrt[4]{ lnx } " alt=" \sqrt[4]

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Use the power rule for differentiation:

$\dfrac{\text{d}}{\text{d}x} (f(x))^k = k(f(x))^{k-1}f'(x)$

You can use this formula if you remember that a root is just a rational exponential:

$\sqrt[4]\ln(x) = (\ln(x))^{\frac{1}{4}}$

So, remembering that the derivative of the logarithm is 1/x, you have

$\dfrac{\text{d}}{\text{d}x} (\ln(x))^{\frac{1}{4}} = \frac{1}{4}(\ln(x))^{\frac{1}{4}-1}\dfrac{1}{x}$

Which you can rewrite as

$\dfrac{1}{4}(\ln(x))^{\frac{1}{4}-1}\dfrac{1}{x} =\dfrac{1}{4}(\ln(x))^{\frac{-3}{4}}\dfrac{1}{x} =\dfrac{1}{4}\dfrac{1}{\sqrt[4]{\ln(x))^3}}\dfrac{1}{x} = \dfrac{1}{4x\sqrt[4]{\ln(x))^3}}$

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2 years ago