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Svet_ta [14]
3 years ago
14

What is the solution to 4/a+5/3a<3

Mathematics
1 answer:
scoray [572]3 years ago
3 0
\dfrac{4}{a}+\dfrac{5}{3a} < 3\ \ \ |-3\ \ \ assume\ a\neq0\\\\\dfrac{3\cdot4}{3a}+\dfrac{5}{3a}-\dfrac{3a\cdot3}{3a} < 0\\\\\dfrac{12+5-9a}{3a} < 0\\\\\dfrac{-9a+17}{3a} < 0\iff 3a(-9a+17) < 0\\\\3a=0\to a=0\\\\-9a+17=0\to a=\dfrac{17}{9}
Look at the picture.
Answer:\ a\in\left(0;\ \dfrac{17}{9}\right)

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Given:

There are given that the cos function:

cos210^{\circ}=-\frac{\sqrt{3}}{2}

Explanation:

To find the value, first, we need to use the half-angle formula:

So,

From the half-angle formula:

cos(\frac{\theta}{2})=\pm\sqrt{\frac{1+cos\theta}{2}}

Then,

Since 105 degrees is the 2nd quadrant so cosine is negative

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By the formula:

\begin{gathered} cos(105^{\circ})=cos(\frac{210^{\circ}}{2}) \\ =-\sqrt{\frac{1+cos(210)}{2}} \end{gathered}

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