What is 11/12 of a full rotation
1 answer:
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First of all, you have to understand 
<span> is a square-root function.
</span>Square-root functions are continuous across their entire domain, and their domain is all real x-<span>values for which the expression within the square-root is non-negative.
</span>
In other words, for any square-root function
and any input 
in the domain of (except for its endpoint), we know that this equality holds: 
Let's take
<span>as an example.
</span>
The domain of
is all real numbers such that 
. Since 
is the endpoint of the domain, the two-sided limit at that point doesn't exist (you can't approach 
<span>from the left).
</span>
<span>However, continuity at an endpoint only demands that the one-sided limit is equal to the function's value:
</span>
In conclusion, the equality holds for any square-root function and any real number in the domain of e<span>xcept for its endpoint, where the two-sided limit should be replaced with a one-sided limit. </span>
The input
, is within the domain of <span>.
</span>
Therefore, in order to find
we can simply evaluate at 
<span>.
</span>
</span>Square-root functions are continuous across their entire domain, and their domain is all real x-<span>values for which the expression within the square-root is non-negative.
</span>
In other words, for any square-root function
Let's take
</span>
The domain of
</span>
<span>However, continuity at an endpoint only demands that the one-sided limit is equal to the function's value:
</span>
In conclusion, the equality
The input
</span>
Therefore, in order to find
</span>
The sector is of the circle so the area is