Answer:
it will be approximately 10 weeks
Step-by-step explanation:
Answer:

Step-by-step explanation:
First, simplify each term:

Then given expression is equivalent to
![\cos ^3\alpha+(-\sin \alpha)^3-(-\sin \alpha)+(-\cos \alpha)\\ \\=\cos ^3\alpha-\sin^3 \alpha+\sin \alpha-\cos \alpha\\ \\=(\cos\alpha-\sin\alpha)(\cos^2\alpha+\cos\alpha\sin\alpha+\sin^2\alpha)-(\cos\alpha-\sin\alpha)\\ \\=(\cos\alpha-\sin\alpha)(1+\cos\alpha\sin\alpha-1)\ \ [\cos^2\alpha+\sin^2\alpha=1]\\ \\=\cos\alpha\sin\alpha(\cos\alpha-\sin\alpha)](https://tex.z-dn.net/?f=%5Ccos%20%5E3%5Calpha%2B%28-%5Csin%20%5Calpha%29%5E3-%28-%5Csin%20%5Calpha%29%2B%28-%5Ccos%20%5Calpha%29%5C%5C%20%5C%5C%3D%5Ccos%20%5E3%5Calpha-%5Csin%5E3%20%5Calpha%2B%5Csin%20%5Calpha-%5Ccos%20%5Calpha%5C%5C%20%5C%5C%3D%28%5Ccos%5Calpha-%5Csin%5Calpha%29%28%5Ccos%5E2%5Calpha%2B%5Ccos%5Calpha%5Csin%5Calpha%2B%5Csin%5E2%5Calpha%29-%28%5Ccos%5Calpha-%5Csin%5Calpha%29%5C%5C%20%5C%5C%3D%28%5Ccos%5Calpha-%5Csin%5Calpha%29%281%2B%5Ccos%5Calpha%5Csin%5Calpha-1%29%5C%20%5C%20%5B%5Ccos%5E2%5Calpha%2B%5Csin%5E2%5Calpha%3D1%5D%5C%5C%20%5C%5C%3D%5Ccos%5Calpha%5Csin%5Calpha%28%5Ccos%5Calpha-%5Csin%5Calpha%29)
Answer: Ann = 67 Josh = 60 Claire = 30
Step-by-step explanation: You get the equation 5x+7 = 157. You get the equation because for example, Claire = x then josh = 2x and Ann = 2x + 7. You do all the math and boom! You get the answer :)
14(10) + 2, 14x10 is 140 plus 2 = 142
2(7(10) + 1) 7x10 = 70 + 1 = 2(71) = 142
the coordinates of the vertices after the given transformation are:
- H' (4, -1)
- I' (4, 3)
- G' (1, 0)
<h3>
How to get the coordinates of the vertices after the transformation?</h3>
First, we can identify the original vertices, which are:
- H (4, 1)
- I (4, -3)
- G (1, 0)
Now we apply a reflection across the x-axis, it will only change the sign of the y-component of each of the above points, then the coordinates of the vertices after the given transformation are:
- H' (4, -1)
- I' (4, 3)
- G' (1, 0)
If you want to learn more about transformations:
brainly.com/question/4289712
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