Y=1/2x+2
y=-x+5
subsitute since they both equal y
1/2x+2=y=-x+5
1/2x+2=-x+5
subtract 2 from both sides
1/2x=-x+3
multiply both sides by 2
x=-2x+6
add 2x to both sides
3x=6
divide both sides by 3
x=2
subsitue
y=-x=5
y=-2+5
y=3
x=2 and y=3
solution in (x,y) form is (2,3)
Answer:
20 people
Step-by-step explanation:
1/8 serves one so 1/4 does 2 . 1/2 does 3 and 1 does 4 so 4 *5 is 20
Answer:
1, 3, 19, 57, 361, and 1083.
Step-by-step explanation:
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
. 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
, 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
. 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)