Answer:
119.05°
Step-by-step explanation:
In general, the angle is given by ...
θ = arctan(y/x)
Here, that becomes ...
θ = arctan(9/-5) ≈ 119.05°
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<em>Comment on using a calculator</em>
If you use the ATAN2( ) function of a graphing calculator or spreadsheet, it will give you the angle in the proper quadrant. If you use the arctangent function (tan⁻¹) of a typical scientific calculator, it will give you a 4th-quadrant angle when the ratio is negative. You must recognize that the desired 2nd-quadrant angle is 180° more than that.
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It may help you to consider looking at the "reference angle." In this geometry, it is the angle between the vector v and the -x axis. The coordinates tell you the lengths of the sides of the triangle vector v forms with the -x axis and a vertical line from that axis to the tip of the vector. Then the trig ratio you're interested in is ...
Tan = Opposite/Adjacent = |y|/|x|
This is the tangent of the reference angle, which will be ...
θ = arctan(|y| / |x|) = arctan(9/5) ≈ 60.95°
You can see from your diagram that the angle CCW from the +x axis will be the supplement of this value, 180° -60.95° = 119.05°.
The geometric sequence is given by:
an=ar^(n-1)
where:
a=first term
r=common ratio
n is the nth term
given that a=4, and second term is -12, then
r=-12/4=-3
hence the formula for this case will be:
an=4(-3)^(n-1)
where n≥1
![\displaystyle\\Answer:\ x=\sqrt{2}\ \ \ \ \ x=-\sqrt{2}\ \ \ \ \ x=\frac{ln5}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5C%5CAnswer%3A%5C%20x%3D%5Csqrt%7B2%7D%5C%20%5C%20%5C%20%5C%20%5C%20x%3D-%5Csqrt%7B2%7D%5C%20%5C%20%5C%20%5C%20%5C%20x%3D%5Cfrac%7Bln5%7D%7B2%7D)
Step-by-step explanation:
![5x^2-x^2e^{2x}+2e^{2x}=10](https://tex.z-dn.net/?f=5x%5E2-x%5E2e%5E%7B2x%7D%2B2e%5E%7B2x%7D%3D10)
![Let:\ x^2=t \ \ \ \ \ e^{2x}=v\\Hence,\\5t-tv+2v^2=10\\5t-tv+2v^2-10=10-10\\5t-tv-10+2v^2=0\\t(5-v)-2(5-v)=0\\(5-v)(t-2)=0\\5-v=0\\5-v+v=0+v\\5=v\\5=e^{2x}](https://tex.z-dn.net/?f=Let%3A%5C%20x%5E2%3Dt%20%20%5C%20%5C%20%5C%20%5C%20%5C%20e%5E%7B2x%7D%3Dv%5C%5CHence%2C%5C%5C5t-tv%2B2v%5E2%3D10%5C%5C5t-tv%2B2v%5E2-10%3D10-10%5C%5C5t-tv-10%2B2v%5E2%3D0%5C%5Ct%285-v%29-2%285-v%29%3D0%5C%5C%285-v%29%28t-2%29%3D0%5C%5C5-v%3D0%5C%5C5-v%2Bv%3D0%2Bv%5C%5C5%3Dv%5C%5C5%3De%5E%7B2x%7D)
<em>Prologarithmize both parts of the equation:</em>
![ln5=lne^{2x}\\ln5=2x*lne\\ln5=2x*1\\ln5=2x\\](https://tex.z-dn.net/?f=ln5%3Dlne%5E%7B2x%7D%5C%5Cln5%3D2x%2Alne%5C%5Cln5%3D2x%2A1%5C%5Cln5%3D2x%5C%5C)
<u><em> Divide both parts of the equation by 2:</em></u>
![\displaystyle\\x=\frac{ln5}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5C%5Cx%3D%5Cfrac%7Bln5%7D%7B2%7D)
![t-2=0\\t-2+2=0+2\\t=2\\x^2=2\\x=\sqrt{2} \\x=-\sqrt{2}](https://tex.z-dn.net/?f=t-2%3D0%5C%5Ct-2%2B2%3D0%2B2%5C%5Ct%3D2%5C%5Cx%5E2%3D2%5C%5Cx%3D%5Csqrt%7B2%7D%20%5C%5Cx%3D-%5Csqrt%7B2%7D)
Step-by-step explanation:
hope this will help you......
Sorry this took so long! I managed to do everything except for the parts involving standard deviation. (I tried but I couldn't quite figure it out.)
~Hope this helps!~