Answer:
19w^2-m
Step-by-step explanation:
4w^2-m+15w^2
19w^2-m
Find the two volumes separately:
Bale: V = 4(2)(2) ft^3 = 16 ft^3
Cylinder: V = pi*(2.5 ft)^2* (6 ft) = 117.8 ft^3
Divide 117.8 ft^3 by 16 ft^3:
117.8 ft^3
------------- = 7.36
16 ft^3
Thus, the cyl. bale could theoretically hold the total volume of 7.36 rect. bales. That could be rounded off either way: to 7 or to 8 bales; 7 is closer and thus is the better answer.
Answer:
See Below.
Step-by-step explanation:
We want to prove the trigonometric identity:
![\displaystyle \frac{\sec^2(\theta)-1}{\sin(\theta)}=\frac{\sin(\theta)}{1-\sin^2(\theta)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Csec%5E2%28%5Ctheta%29-1%7D%7B%5Csin%28%5Ctheta%29%7D%3D%5Cfrac%7B%5Csin%28%5Ctheta%29%7D%7B1-%5Csin%5E2%28%5Ctheta%29%7D)
To start, let's simplify the right side. Recall the Pythagorean Identity:
![\sin^2(\theta)+\cos^2(\theta)=1](https://tex.z-dn.net/?f=%5Csin%5E2%28%5Ctheta%29%2B%5Ccos%5E2%28%5Ctheta%29%3D1)
Therefore:
![\cos^2(\theta)=1-\sin^2(\theta)](https://tex.z-dn.net/?f=%5Ccos%5E2%28%5Ctheta%29%3D1-%5Csin%5E2%28%5Ctheta%29)
Substitute:
![\displaystyle \frac{\sin(\theta)}{1-\sin^2(\theta)}=\frac{\sin(\theta)}{\cos^2(\theta)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Csin%28%5Ctheta%29%7D%7B1-%5Csin%5E2%28%5Ctheta%29%7D%3D%5Cfrac%7B%5Csin%28%5Ctheta%29%7D%7B%5Ccos%5E2%28%5Ctheta%29%7D)
Split:
![\displaystyle =\frac{\sin(\theta)}{\cos(\theta)}\left(\frac{1}{\cos(\theta)}\right)=\tan(\theta)\sec(\theta)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%5Cfrac%7B%5Csin%28%5Ctheta%29%7D%7B%5Ccos%28%5Ctheta%29%7D%5Cleft%28%5Cfrac%7B1%7D%7B%5Ccos%28%5Ctheta%29%7D%5Cright%29%3D%5Ctan%28%5Ctheta%29%5Csec%28%5Ctheta%29)
Therefore, our equation becomes:
![\displaystyle \frac{\sec^2(\theta)-1}{\sin(\theta)}=\tan(\theta)\sec(\theta)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Csec%5E2%28%5Ctheta%29-1%7D%7B%5Csin%28%5Ctheta%29%7D%3D%5Ctan%28%5Ctheta%29%5Csec%28%5Ctheta%29)
From the Pythagorean Identity, we can divide both sides by cos²(θ). This yields:
![\displaystyle \tan^2(\theta)+1=\sec^2(\theta)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctan%5E2%28%5Ctheta%29%2B1%3D%5Csec%5E2%28%5Ctheta%29)
So:
![\tan^2(\theta)=\sec^2(\theta)-1](https://tex.z-dn.net/?f=%5Ctan%5E2%28%5Ctheta%29%3D%5Csec%5E2%28%5Ctheta%29-1)
Substitute:
![\displaystyle \frac{\tan^2(\theta)}{\sin(\theta)}=\tan(\theta)\sec(\theta)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Ctan%5E2%28%5Ctheta%29%7D%7B%5Csin%28%5Ctheta%29%7D%3D%5Ctan%28%5Ctheta%29%5Csec%28%5Ctheta%29)
Rewrite:
![\displaystyle (\tan(\theta))^2\left(\frac{1}{\sin(\theta)}\right)=\tan(\theta)\sec(\theta)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%28%5Ctan%28%5Ctheta%29%29%5E2%5Cleft%28%5Cfrac%7B1%7D%7B%5Csin%28%5Ctheta%29%7D%5Cright%29%3D%5Ctan%28%5Ctheta%29%5Csec%28%5Ctheta%29)
Recall that tan(θ) = sin(θ)/cos(θ). So:
![\displaystyle \frac{\sin^2(\theta)}{\cos^2(\theta)}\left(\frac{1}{\sin(\theta)}\right)=\tan(\theta)\sec(\theta)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Csin%5E2%28%5Ctheta%29%7D%7B%5Ccos%5E2%28%5Ctheta%29%7D%5Cleft%28%5Cfrac%7B1%7D%7B%5Csin%28%5Ctheta%29%7D%5Cright%29%3D%5Ctan%28%5Ctheta%29%5Csec%28%5Ctheta%29)
Simplify:
![\displaystyle \frac{\sin(\theta)}{\cos^2(\theta)}=\tan(\theta)\sec(\theta)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Csin%28%5Ctheta%29%7D%7B%5Ccos%5E2%28%5Ctheta%29%7D%3D%5Ctan%28%5Ctheta%29%5Csec%28%5Ctheta%29)
Simplify:
![\tan(\theta)\sec(\theta)=\tan(\theta)\sec(\theta)}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%5Csec%28%5Ctheta%29%3D%5Ctan%28%5Ctheta%29%5Csec%28%5Ctheta%29%7D)
Hence proven.
Answer:
-√32 < 59/11 < √30 < 5.49
2/5 Is the answer! Am pretty sure Hope this helped