The answer is Net A with a surface area of 390 square inches
From this diagram, you can see that a square is a quadrilateral, a parallelogram, a rectangle, and a rhombus!
Answer:
a) ![\cos(\theta) = \frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta%29%20%3D%20%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
b) ![\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B6%7D%29%5Cfrac%7B-3%5Csqrt%5B%5D%7B11%7D%2B4%7D%7B14%7D)
c) ![\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta-%5Cpi%29%3D%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
d)![\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B4%7D%29%20%3D%20%5Cfrac%7B%5Cfrac%7B-4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%2B1%7D%7B1%2B%5Cfrac%7B4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%7D)
Step-by-step explanation:
We will use the following trigonometric identities
.
Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that
which implies that ![x=-\sqrt[]{49-16} = -\sqrt[]{33}](https://tex.z-dn.net/?f=x%3D-%5Csqrt%5B%5D%7B49-16%7D%20%3D%20-%5Csqrt%5B%5D%7B33%7D)
. Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then
![\cos(\theta) = \frac{-\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta%29%20%3D%20%5Cfrac%7B-%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
b)Recall that ![\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}](https://tex.z-dn.net/?f=%5Csin%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%5Cfrac%7B1%7D%7B2%7D%20%2C%20%5Ccos%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D)
, then using the identity from above, we have that
![\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%20%2B%20%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Csin%28%5Ctheta%29%5Ccos%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%2B%5Ccos%28%5Calpha%29%5Csin%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Cfrac%7B4%7D%7B7%7D%5Cfrac%7B1%7D%7B2%7D-%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%20%3D%20%5Cfrac%7B-3%5Csqrt%5B%5D%7B11%7D%2B4%7D%7B14%7D)
c) Recall that 
. Then,
![\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta-%5Cpi%29%3D%5Ccos%28%5Ctheta%29%5Ccos%28%5Cpi%29%2B%5Csin%28%5Ctheta%29%5Csin%28%5Cpi%29%20%3D%20%5Cfrac%7B-%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D%5Ccdot%28-1%29%20%2B%200%20%3D%20%5Cfrac%7B%5Csqrt%5B%5D%7B33%7D%7D%7B7%7D)
d) Recall that 
and ![\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%20%3D%20%5Cfrac%7B%5Csin%28%5Ctheta%29%7D%7B%5Ccos%28%5Ctheta%29%7D%3D%5Cfrac%7B-4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D)
. Then
![\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%2B%5Cfrac%7B%5Cpi%7D%7B4%7D%29%20%3D%20%5Cfrac%7B%5Ctan%28%5Ctheta%29%2B%5Ctan%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%7D%7B1-%5Ctan%28%5Ctheta%29%5Ctan%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B-4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%2B1%7D%7B1%2B%5Cfrac%7B4%7D%7B%5Csqrt%5B%5D%7B33%7D%7D%7D)
Answer:
Can you show the picture that's a pdf
Answer:
12 Hershey bars
Step-by-step explanation:
if its 10 to 8 and you have 15 m&m then you would have to do the 10 which you would have to put 8 then 5 is half of 10 so you would just half 8 which is 4 so 8+4+12. 12 Hershey bars needed to restock the shelf.
