Helppp with 2 and 4 !!!!!

A polynomial function has -5+√3t as a root. Which of the following must also be a root of the function? a) -5-√3t b) -5+√3t c) 5

juin [17]

Answer:

a

Step-by-step explanation:

Radical roots occur as conjugate pairs

Given - 5 + $\sqrt{3}$ t is a root , then

- 5 - $\sqrt{3}$ t is also a root → a

4 0

3 years ago

Can anybody show me how to do this half angle identity problem STEP BY STEP? I always seem to be missing something

klio [65]

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

x = 24

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:

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