The formula for A = Ao (1/2) ^ t/h
So sub in.
100 = 800 (1/2) ^ 639,000/?
That's all i can do lol
P(x)=2x^2-4x
q(x)=x-3
(p*q)(x)
(2x^2-4x)(x-3)
2x^3-10x^2+12x
D is a line, K is a point, beta is a plane and PY is a line segment.
I think the answer be C? Sorry if it's wrong
![x](https://tex.z-dn.net/?f=x)
is in quadrant I, so
![0](https://tex.z-dn.net/?f=0%3Cx%3C%5Cdfrac%5Cpi2)
, which means
![0](https://tex.z-dn.net/?f=0%3C%5Cdfrac%20x2%3C%5Cdfrac%5Cpi4)
, so
![\dfrac x2](https://tex.z-dn.net/?f=%5Cdfrac%20x2)
belongs to the same quadrant.
Now,
![\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}](https://tex.z-dn.net/?f=%5Ctan%5E2%5Cdfrac%20x2%3D%5Cdfrac%7B%5Csin%5E2%5Cfrac%20x2%7D%7B%5Ccos%5E2%5Cfrac%20x2%7D%3D%5Cdfrac%7B%5Cfrac%7B1-%5Ccos%20x%7D2%7D%7B%5Cfrac%7B1%2B%5Ccos%20x%7D2%7D%3D%5Cdfrac%7B1-%5Ccos%20x%7D%7B1%2B%5Ccos%20x%7D)
Since
![\sin x=\dfrac5{13}](https://tex.z-dn.net/?f=%5Csin%20x%3D%5Cdfrac5%7B13%7D)
, it follows that
![\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}](https://tex.z-dn.net/?f=%5Ccos%5E2x%3D1-%5Csin%5E2x%5Cimplies%20%5Ccos%20x%3D%5Cpm%5Csqrt%7B1-%5Cleft%28%5Cdfrac5%7B13%7D%5Cright%29%5E2%7D%3D%5Cpm%5Cdfrac%7B12%7D%7B13%7D)
Since
![x](https://tex.z-dn.net/?f=x)
belongs to the first quadrant, you take the positive root (
![\cos x>0](https://tex.z-dn.net/?f=%5Ccos%20x%3E0)
for
![x](https://tex.z-dn.net/?f=x)
in quadrant I). Then
![\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}](https://tex.z-dn.net/?f=%5Ctan%5Cdfrac%20x2%3D%5Cpm%5Csqrt%7B%5Cdfrac%7B1-%5Cfrac%7B12%7D%7B13%7D%7D%7B1%2B%5Cfrac%7B12%7D%7B13%7D%7D%7D)
![\tan x](https://tex.z-dn.net/?f=%5Ctan%20x)
is also positive for
![x](https://tex.z-dn.net/?f=x)
in quadrant I, so you take the positive root again. You're left with