1030303030303030 divided by 220202020
2 answers:
<h2>Answer:</h2><h2>103030303030303/22020202</h2><h2>____________________________________</h2><h3>Honey, all you need to do is reduce the expression, if possible, by cancelling the common factors!</h3><h3>Then you'll get your answers:</h3><h3>Exact Form:</h3><h3>103030303030303/22020202</h3><h3>or</h3><h3>
</h3><h3 /><h3>Decimal Form:</h3><h3>4678899.08686137</h3><h3>Mixed Number Form:</h3><h3>4678899 1912705/22020202</h3><h3>or</h3><h3>
</h3><h2>____________________________________</h2><h2>I know this is probably just a troll but your welcome!</h2><h2>Hope you have a good day, Loves~ <3</h2>
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Answer:
The correct option is;
No, the distance from (0, 0) to (2, √6) is not 3 units
Step-by-step explanation:
The given parameters of the question are as follows;
Circle center (h, k) = (0, 0)
Point on the circle (x, y) = (3, 0)
We are required to verify whether point (2, √6) lie on the circle
We note that the radius of the circle is given by the equation of the circle as follows;
Distance² = (x - h)² + (y - k)² = r² which gives;
(3 - 0)² + (0 - 0)² = 3²
Hence r² = 3² and r = 3 units
We check the distance of the point (2, √6) from the center of the circle (0, 0) as follows;
Therefore;
(2 - 0)² + (√6 - 0)² = 2² + √6² = 4 + 6 = 10 = √10²
Which gives the distance of the point (2, √6) from the center of the circle (0, 0) = √10
Hence the distance from the circle center (0, 0) to (2, √6) is not √10 which s more than 3 units hence the point (2, √6), does not lie on the circle.
Step-by-step explanation:
Domain of a rational function is everywhere except where we set vertical asymptotes. or removable discontinues
Here, we have
First, notice we have x in both the numerator and denomiator so we have a removable discounties at x.
Since, we don't want x to be 0,
We have a removable discontinuity at x=0
Now, we have
We don't want the denomiator be zero because we can't divide by zero.
so
So our domain is
All Real Numbers except-2 and 0.
The vertical asymptors is x=-2.
To find the horinzontal asymptote, notice how the numerator and denomator have the same degree. So this mean we will have a horinzontal asymptoe of
The leading coeffixent of the numerator/ the leading coefficent of the denomiator.
So that becomes
So we have a horinzontal asymptofe of 2