anonymous
  • anonymous
Will fan and medal. What is the equation of the line that represents the initial climb?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
@pooja195
MrNood
  • MrNood
the initial line appears to be vertical -hence the X value doesn't change, therefore the equation is x='constant' . You can get the value of the constant from the graph (NOTE - if this is intended to be the flight of a aircraft then it is evidently impossible to climb like that - ie it gains altitiude in 0 time - an infinite rate of climb. Note 2 - it is not the whole line that is valid for the climb - it is a 'line segment' and is bounded by y=0 and y = 9

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MrNood
  • MrNood
also - the numbers on your graph don't seem to make sense: you have -10 at both ends of the x axis you have what looks like 1/5 halfway along the negative axis the path hits the ground twice where did this drawing originate?
anonymous
  • anonymous
@MrNood I didn't mean to have multiple negative 10's. This is a roller coaster project im working on.
MrNood
  • MrNood
OK so the x axis is not time - it is distance? and y axis is height? it does mean that you have a truly vertical lift at the beginning - most rides climb up a sloping track.
anonymous
  • anonymous
@MrNood Yes. But it is possible to have one going straight up
MrNood
  • MrNood
yes ok
anonymous
  • anonymous
@MrNood will you help me with the equation?

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