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Fynjy0 [20]
3 years ago
10

Plz help me!!!!!!!!!

Mathematics
1 answer:
Over [174]3 years ago
7 0

Answer:  \bold{\dfrac{4\pm \sqrt{6}}{2}}

<u>Step-by-step explanation:</u>

\dfrac{3}{y-2}-2=\dfrac{1}{y-1}\\\\\\\text{Multiply by the LCD (y-2)(y-1) to clear the denominator:}\\\\\dfrac{3}{y-2}(y-2)(y-1)-2(y-2)(y-1)=\dfrac{1}{y-1}(y-2)(y-1)\\\\\\3(y-1)-2(y-2)(y-1)=1(y-2)\\\\3y-3-2(y^2-3y+2)=y-2\\\\3y-3-2y^2+6y-4=y-2\\\\-2y^2+9y-7=y-2\\\\0=2y^2-8y+5\quad \rightarrow \quad a=2,\ b=-8,\ c=5\\\\\\\text{Quadratic formula is: }x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-8)\pm \sqrt{(-8)^2-4(2)(5)}}{2(2)}\\\\\\.\ =\dfrac{8\pm \sqrt{64-40}}{2(2)}

.\ =\dfrac{8\pm \sqrt{24}}{2(2)}\\\\\\.\ =\dfrac{8\pm 2\sqrt{6}}{2(2)}\\\\\\.\ =\dfrac{4\pm \sqrt{6}}{2}

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Answer:

<h2>FALSE</h2>

Step-by-step explanation:

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\text{Method 2}\\\\\text{Solve the equation:}\\\\3\sqrt2\cos\theta+2=-1\qquad\text{subtract 2 from both sides}\\\\3\sqrt2\cos\theta=-3\qquad\text{divide both sides by}\ 3\sqrt2\\\\\cos\theta=-\dfrac{3}{3\sqrt2}\\\\\cos\theta=-\dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}\\\\\cos\theta=-\dfrac{\sqrt2}{2}\to\theta=\dfrac{3\pi}{4}+2k\pi\ \vee\ \theta=-\dfrac{3\pi}{4}+2k\pi\ \text{for}\ k\in\mathbb{Z}\\\\\text{It's not equal to}\ \dfrac{\pi}{4}\ \text{for any value of }\ k.

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