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prohojiy [21]
3 years ago
11

Solve the inequality 2x - 3 < x + 2 ≤ 3x + 5. Show your work.

Mathematics
1 answer:
enyata [817]3 years ago
6 0
2x - 3 < x + 2 ≤ 3x + 5
<u>     - 2        - 2          - 2</u>
2x - 5 < x ≤ 3x + 3
The solution set is (2x - 5, 3x + 3}.
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Can anybody show me how to do this half angle identity problem STEP BY STEP? I always seem to be missing something
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Answer:   \bold{-\dfrac{7\sqrt2}{10}}

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

It is given that θ is between 270° and 360°, which means that θ is located in Quadrant IV ⇒ (x > 0, y < 0).  Furthermore, the half-angle will be between 135° and 180°, which means the half-angle is in Quadrant II ⇒ cos\ \dfrac{\theta}{2}

It is given that sin θ = -\dfrac{7}{25}  ⇒  y = -7 & hyp = 25

Use Pythagorean Theorem to find "x":

x² + y² = hyp²

x² + (-7)² = 25²

x² + 49 = 625

x²         = 576

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Use the "x" and "hyp" values to find cos θ:

cos\ \theta=\dfrac{x}{hyp}=\dfrac{24}{25}    


Lastly, input cos θ into the half angle formula:

cos\bigg(\dfrac{\theta}{2}\bigg)=\pm \sqrt{\dfrac{1+cos\ \theta}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{1+\dfrac{24}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{\dfrac{25}{25}+\dfrac{24}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{\dfrac{49}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{49}{50}}\\\\\\.\qquad \quad =\pm \dfrac{7}{5\sqrt2}}\\\\\\.\qquad \quad =\pm \dfrac{7}{5\sqrt2}}\bigg(\dfrac{\sqrt2}{\sqrt2}\bigg)\\\\\\.\qquad \quad =\pm \dfrac{7\sqrt2}{10}

Reminder: We previously determined that the half-angle will be negative.

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