Starting from the fundamental trigonometric equation, we have
![\cos^2(\alpha)+\sin^2(\alpha)=1 \iff \sin(\alpha)=\pm\sqrt{1-\cos^2(\alpha)}](https://tex.z-dn.net/?f=%5Ccos%5E2%28%5Calpha%29%2B%5Csin%5E2%28%5Calpha%29%3D1%20%5Ciff%20%5Csin%28%5Calpha%29%3D%5Cpm%5Csqrt%7B1-%5Ccos%5E2%28%5Calpha%29%7D)
Since
, we know that the angle lies in the third quadrant, where both sine and cosine are negative. So, in this specific case, we have
![\sin(\alpha)=-\sqrt{1-\cos^2(\alpha)}](https://tex.z-dn.net/?f=%5Csin%28%5Calpha%29%3D-%5Csqrt%7B1-%5Ccos%5E2%28%5Calpha%29%7D)
Plugging the numbers, we have
![\sin(\alpha)=-\sqrt{1-\dfrac{64}{289}}=-\sqrt{\dfrac{225}{289}}=-\dfrac{15}{17}](https://tex.z-dn.net/?f=%5Csin%28%5Calpha%29%3D-%5Csqrt%7B1-%5Cdfrac%7B64%7D%7B289%7D%7D%3D-%5Csqrt%7B%5Cdfrac%7B225%7D%7B289%7D%7D%3D-%5Cdfrac%7B15%7D%7B17%7D)
Now, just recall that
![\sin(-\alpha)=-\sin(\alpha)](https://tex.z-dn.net/?f=%5Csin%28-%5Calpha%29%3D-%5Csin%28%5Calpha%29)
to deduce
![\sin(-\alpha)=-\sin(\alpha)=-\left(-\dfrac{15}{17}\right)=\dfrac{15}{17}](https://tex.z-dn.net/?f=%5Csin%28-%5Calpha%29%3D-%5Csin%28%5Calpha%29%3D-%5Cleft%28-%5Cdfrac%7B15%7D%7B17%7D%5Cright%29%3D%5Cdfrac%7B15%7D%7B17%7D)
We will call:
M=Mari
J=Jen
Initial equations:
M=2J
M+J=480
(the fact that they work 20 hours a week is not needed in this problem information)
Plug in:
2J+J=480
3J=480
J=160
Jen makes $160 a week.
Therefore:
M=2J
M=2(160)
M=320
Mari makes $320 per week
Jen makes $160 per week
The slope is 6.cuse the slope also means sometimes 6
- Your answer is no solution.
- Because if we transpose 10x to the left hand side with a change in the symbol, then it will become 0.
- 10x - 1 = 10x +4
- or, 10x - 10x = 4 + 5
- or, 0 = 9
- Hence, the equation has <em><u>no solution</u></em>.
Hope you could get an idea from here.
Doubt clarification - use comment section.