Answer:



Step-by-step explanation:


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

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
Answer:
1/3
Step-by-step explanation:
sin=p/h
p/h= 3/9
3/9=1/3
8x + 3 = -39
8x + 3 -3 = -39 - 3
8x = -42
8x/8 = -42/8
x = -5.25
Equations:
Y=-50x+14,040
Y=20x+12,500
Solution X=22, Y=12,950
They will be at the same spot when the graphs intersect and the poi is (22 , 12,940)
So they will meet up at 22, 12,940