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1 answer:
Answer:
Tangent line states that a line in the plane of a circle that intersect the circle in exactly one point.
Common external tangent states that a common tangent that does not intersects the line segment joining the centers of circle.
Common internal tangent states that a common tangent that intersects the line segment joining the centers of circle.
Circumscribe polygon states that a polygon with all sides tangent to a circle contained within the polygon.
Therefore:
A polygon with all sides tangent to a circle contained within the polygon = Circumscribe polygon
A common tangent that intersects the line segment joining the centers of circle = Common internal tangent
A common tangent that does not intersects the line segment joining the centers of circle = Common external tangent
a line in the plane of a circle that intersect the circle in exactly one point = Tangent line
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Therefore, we must perform an algebraic manipulation in order to get rid of the indeterminacy.
We can solve this limit in two ways.
<h3>Way 1:</h3>By comparison of infinities:
We first expand the binomial squared, so we get
Note that in the numerator we get x⁴ while in the denominator we get x³ as the highest degree terms. Therefore, the degree of the numerator is greater and the limit will be \infty. Recall that when the degree of the numerator is greater, then the limit is \infty if the terms of greater degree have the same sign.
<h3>Way 2</h3>Dividing numerator and denominator by the term of highest degree:
Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.
Answer: On the 29th day
Step-by-step explanation:
According to this problem, no lilypad dies and the lilypads always reproduce, so we can apply the following reasoning.
On the first day there is only 1 lilypad in the pond. On the second day, the lilypad from the first reproduces, so there are 2 lilypads. On day 3, the 2 lilypads from the second day reproduce, so there are 2×2=4 lilypads. Similarly, on day 4 there are 8 lilypads. Following this pattern, on day 30 there are 2×N lilypads, where N is the number of lilypads on day 29.
The pond is full on the 30th day, when there are 2×N lilypads, so it is half-full when it has N lilypads, that is, on the 29th day. Actually, there are lilypads on the 30th, and
lilypads on the 29th. This can be deduced multiplying succesively by 2.