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Leviafan [203]
3 years ago
13

Type the correct answer for each.

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0
If point \left(x,\ \frac{\sqrt{7}}{3}\right) is on the unit ircle, then:

x^2+\left( \frac{ \sqrt{7} }{3} \right)^2=1 \\  \\ \Rightarrow x^2+ \frac{7}{9} =1 \\  \\ \Rightarrow x^2=1- \frac{7}{9} = \frac{2}{9}  \\  \\ \Rightarrow x= \sqrt{ \frac{2}{9} } = \frac{ \sqrt{2} }{3}

Since, the point is in the second quadrant, x is negative.

Thus, (x,\ y)=\left(x,\ \frac{\sqrt{7}}{3}\right)=\left(-\frac{ \sqrt{2} }{3},\ \frac{\sqrt{7}}{3}\right)

Part A:

3 \sqrt{7} =6\left( \frac{ \sqrt{7} }{2} \right) \\  \\ =6\left( \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \right) \\  \\ =6\left( \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } \right)=-6\left( \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } \right) \\  \\ =-6\left( \frac{y}{x} \right)=-6\tan\theta

Therefore, -6\tan\theta=3\sqrt{7}.



Part B:

- \frac{ \sqrt{7} }{2} =- \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} }  \\ \\ =- \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } = \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } = \frac{y}{x}  \\  \\ =\tan\theta

Therefore, \tan\theta=- \frac{ \sqrt{7} }{2}
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It is important because while some quadratics are factorable and can be solved not all are. The formula will solve all quadratic equations and can also give both real and imaginary solutions.  Using the formula will require less work than finding the factors if factorable. We will substitute a=9, b=-54 and c=-19.

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We will now solve for the plus and the minus.

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x=\frac{54+\sqrt{3600}}{18}\\x=\frac{54+60}{18}\\x=\frac{114}{18}\\x=6.3

and the minus...

x=\frac{54-\sqrt{3600}}{18}\\x=\frac{54-60}{18}\\x=\frac{-6}{18}\\x=-0.333...

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