(X^-5y/y^3)^-1= A)x^5/y^2 B)y^2/x^5 C)x^5 y^2 D)x^5 y^-3

Mathematics

1 answer:

Nezavi [6.7K]1 year ago

3 0

$\left(\frac{x^{-5}y}{y^3} \right)^{-1} \\ \\ =\frac{y^3}{x^{-5}y} \\ \\ =\frac{y^2}{x^{-5}} \\ \\ =\boxed{x^5 y^2}$

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Find the general solution of <img src="https://tex.z-dn.net/?f=2cos%282x%20%2B%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%29%20%3D%20%5Cs

Aleks04 [339]

Answer:

$\large\boxed{x=-\dfrac{\pi}{4}+k\pi\ \text{or}\ x=k\pi\ \text{for}\ k\in\mathbb{Z}}$

Step-by-step explanation:

$2\cos\left(2x+\dfrac{\pi}{4}\right)=\sqrt2\qquad\text{divide both sides by 2}\\\\\cos\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt2}{2}\to2x+\dfrac{\pi}{4}=\pm\dfrac{\pi}{4}+2k\pi\\\\2x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+2k\pi\ \vee\ 2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+2k\pi\qquad\text{subtract}\ \dfrac{\pi}{4}\ \text{from all sides}\\\\2x=-\dfrac{2\pi}{4}+2k\pi\ \vee\ 2x=2k\pi\qquad\text{divide all sides by 2}\\\\x=-\dfrac{\pi}{4}+k\pi\ \vee\ x=k\pi\ \text{for}\ k\in\mathbb{Z}$