10 points.............
2 answers:
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Answer:
An equation of the circle with centre (-2,1) and radius 3 is
Option D is correct.
Step-by-step explanation:
Looking at the figure we get centre of circle C (-2,1) and radius of circle r = 3
The equation of circle is of form: where (h,k) is centre and r is radius.
We have centre C (-2,1) so, we have h = -2 and k = 1
We have radius = 3 so, r = 3
Putting values in the equation and finding the required equation:
So, an equation of the circle with centre (-2,1) and radius 3 is
Option D is correct.
Answer:
x = 5
Step-by-step explanation:
x + 5 = {Midpoint theorem}
2(x+ 5) = x² - x
2x + 10 = x² - x
x² - x - 2x - 10 = 0
x² - 3x - 10 = 0
x² - 5x + 2x - (5*2) = 0
x(x - 5) + 2(x - 5) = 0
(x -5)(x + 2) = 0
{x + 2 is ignored because measurement could not be in negative value}
x - 5 = 0
x = 5
bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.
le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.