What is the greatest common divisor of 1665,4554,1683, and 4050?
1 answer:
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Vertex at the origin and opening down → y=ax^2
Width: w=16
x=w/2→x=16/2→x=8
x=8, y=-16→y=ax^2→-16=a(8)^2→-16=a(64)→-16/64=a(64)/64→-1/4=a→a=-1/4
y=ax^2→y=-(1/4)x^2
7 m from the edge of the tunnel → x=w/2-7=8 m-7 m→x=1 m
x=1→y=-(1/4)x^2=-(1/4)(1)^2=-(1/4)(1)→y=-1/4
Vertical clearance: 16-1/4=16-0.25→Vertical clearance=15.75 m
Please, see the attached file.
Answer: Third option 15.75 m
Width: w=16
x=w/2→x=16/2→x=8
x=8, y=-16→y=ax^2→-16=a(8)^2→-16=a(64)→-16/64=a(64)/64→-1/4=a→a=-1/4
y=ax^2→y=-(1/4)x^2
7 m from the edge of the tunnel → x=w/2-7=8 m-7 m→x=1 m
x=1→y=-(1/4)x^2=-(1/4)(1)^2=-(1/4)(1)→y=-1/4
Vertical clearance: 16-1/4=16-0.25→Vertical clearance=15.75 m
Please, see the attached file.
Answer: Third option 15.75 m