![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
Answer:
The answer I got was 15 3/4 ft 2
Answer:
Step-by-step explanation:
height h = 11 cm
slant height s = 11.180339887499 cm
side length a = 4 cm
lateral edge length e = 11.357816691601 cm
1/2 side length r = 2 cm
volume V = 58.666666666667 cm3
lateral surface area L = 89.442719099992 cm2
base surface area B = 16 cm2
total surface area A = 105.44271909999 cm2
Answer:
Step-by-step explanation:
Key to this method is remembering that
![(a+b)^2 = a^2 + 2ab + b^2](https://tex.z-dn.net/?f=%28a%2Bb%29%5E2%20%3D%20a%5E2%20%2B%202ab%20%2B%20b%5E2)
So, if you divide the coefficient of x by 2 and square it, that will complete the square. With that in mind, things are easy.
6/2 = 3, so add ![3^2 = 9](https://tex.z-dn.net/?f=3%5E2%20%3D%209)
![x^2+6x+9 = (x+3)^2](https://tex.z-dn.net/?f=x%5E2%2B6x%2B9%20%3D%20%28x%2B3%29%5E2)
So, c=9
Similarly, for the other two,
![c = (\frac{-34}{2} )^2 = (-17)^2 = 289](https://tex.z-dn.net/?f=c%20%3D%20%28%5Cfrac%7B-34%7D%7B2%7D%20%29%5E2%20%3D%20%28-17%29%5E2%20%3D%20289)
You don't really have to worry about the sign, since squaring will always make it positive
![(\frac{25}{13}) ^2 = \frac{625}{169}](https://tex.z-dn.net/?f=%28%5Cfrac%7B25%7D%7B13%7D%29%20%5E2%20%3D%20%5Cfrac%7B625%7D%7B169%7D)