Answer:
![\cos(\alpha +\beta)=\frac{2}{3}(1-\frac{\sqrt{5}}{5})](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%20%2B%5Cbeta%29%3D%5Cfrac%7B2%7D%7B3%7D%281-%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B5%7D%29)
Step-by-step explanation:
Let the hypotenuse of the smaller triangle be h units.
Then; from the Pythagoras Theorem.
![h^2=4^2+2^2](https://tex.z-dn.net/?f=h%5E2%3D4%5E2%2B2%5E2)
![h^2=16+4](https://tex.z-dn.net/?f=h%5E2%3D16%2B4)
![h^2=20](https://tex.z-dn.net/?f=h%5E2%3D20)
![h=\sqrt{20}](https://tex.z-dn.net/?f=h%3D%5Csqrt%7B20%7D)
![h=2\sqrt{5}](https://tex.z-dn.net/?f=h%3D2%5Csqrt%7B5%7D)
From the smaller triangle;
and ![\sin(\alpha)=\frac{2}{2\sqrt{5} }=\frac{1}{\sqrt{5} }](https://tex.z-dn.net/?f=%5Csin%28%5Calpha%29%3D%5Cfrac%7B2%7D%7B2%5Csqrt%7B5%7D%20%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B5%7D%20%7D)
From the second triangle, let the other other shorter leg of the second triangle be s units.
Then;
![s^2+4^2=6^2](https://tex.z-dn.net/?f=s%5E2%2B4%5E2%3D6%5E2)
![s^2+16=36](https://tex.z-dn.net/?f=s%5E2%2B16%3D36)
![s^2=36-16](https://tex.z-dn.net/?f=s%5E2%3D36-16)
![s^2=20](https://tex.z-dn.net/?f=s%5E2%3D20)
![s=\sqrt{20}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B20%7D)
![s=2\sqrt{5}](https://tex.z-dn.net/?f=s%3D2%5Csqrt%7B5%7D)
![\cos(\beta)=\frac{2\sqrt{5} }{6}=\frac{\sqrt{5} }{3}](https://tex.z-dn.net/?f=%5Ccos%28%5Cbeta%29%3D%5Cfrac%7B2%5Csqrt%7B5%7D%20%7D%7B6%7D%3D%5Cfrac%7B%5Csqrt%7B5%7D%20%7D%7B3%7D)
and
![\sin(\beta)=\frac{4}{6}=\frac{2}{3}](https://tex.z-dn.net/?f=%5Csin%28%5Cbeta%29%3D%5Cfrac%7B4%7D%7B6%7D%3D%5Cfrac%7B2%7D%7B3%7D)
We now use the double angle property;
![\cos(\alpha +\beta)=\cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta)](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%20%2B%5Cbeta%29%3D%5Ccos%28%5Calpha%29%5Ccos%28%5Cbeta%29%20-%5Csin%28%5Calpha%29%5Csin%28%5Cbeta%29)
we plug in the values to obtain;
![\cos(\alpha +\beta)=\frac{2}{\sqrt{5} }\times \frac{\sqrt{5} }{3}-\frac{1}{\sqrt{5} }\times \frac{2}{3}](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%20%2B%5Cbeta%29%3D%5Cfrac%7B2%7D%7B%5Csqrt%7B5%7D%20%7D%5Ctimes%20%5Cfrac%7B%5Csqrt%7B5%7D%20%7D%7B3%7D-%5Cfrac%7B1%7D%7B%5Csqrt%7B5%7D%20%7D%5Ctimes%20%5Cfrac%7B2%7D%7B3%7D)
![\cos(\alpha +\beta)=\frac{2}{3}(1-\frac{\sqrt{5}}{5})](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%20%2B%5Cbeta%29%3D%5Cfrac%7B2%7D%7B3%7D%281-%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B5%7D%29)
The answer is B
Have a good day!
questions
1) through: (2, 3), parallel to y = 4x - 1
2) through: (5, -4), parallel to y = -
3/5x +1
3)through: (1, 2), parallel to y = 6x + 3
4)through: (-1, -1), parallel to y = -4x + 1
answers=
1)y = 4x - 5
2)y = -
3/5-1
3) y = 6x - 4
4)y = -4x - 5
$30 is your constant
Months would be your independent variable
Total cost would be dependent
Therefore, your total cost would “depend” on the number of months “independent” times your constant of 30