If two angles of an isosceles triangle measure 35° and the base of the triangle is 18 feet, find the perimeter of the triangle.
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As the fellow above said, the particle does make a U-turn at the vertex, and we can find that out by getting the derivative of s(t) and zeroing it out to get the point of the horizontal tangent.
Bearing in mind that ds/dt is really v(t) or the velocity equation, so when we zero out the ds/dt, is the same as saying the velocity went to 0, since it got zeroed out, and then the particle goes over the vertex and changes direction.
the negative acceleration value, simply means a decreasing rate of change, so, it means the particle is slowing down and possibly coming to a stop before changing direction again.
Bearing in mind that ds/dt is really v(t) or the velocity equation, so when we zero out the ds/dt, is the same as saying the velocity went to 0, since it got zeroed out, and then the particle goes over the vertex and changes direction.
the negative acceleration value, simply means a decreasing rate of change, so, it means the particle is slowing down and possibly coming to a stop before changing direction again.
The value of csc theta is
It is important to note that the ratio for cot is given as;
Cot theta = adjacent/opposite
cosec theta = hypotenuse/opposite
If cot theta = 3/2, then
Adjacent = 3
Opposite = 2
Using Pythagoras theorem, we have
Hypotenuse square = opposite + adjacent square
Cosec theta =
Thus, the value of csc theta is
Learn more about trigonometric ratio here:
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