What is the greatest common factor of 6d² and 18d
1 answer:
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Check the picture below.
there are a few angles shaded, I'll call that angle at O ∡O, and that angle at "x" ∡x and the green angle at A, ∡A, and the angle at B, ∡B, just so you know which angle we're referring to.
well, first off let's notice that B and C are points of tangency, meaning we get right-angles as you see in the picture, if we run an angle bisector from ∡x, towards point O, we get two congruent triangles, lemme shorten that up some, 90 + 90 + ∡O +∡x = 360, which means that ∡O + ∡x = 180, meaning that ∡O = 180 - ∡x, if that's so, the angle at point O across the shade, is 360 - ∡O or 360 - (180 - ∡x).
alrighty, by the inscribed angle theorem, ∡A is half of ∡O.
let's now focus on the triangle AOB, we'll be using half of ∡A and half of ∡O and ∡B on that triangle, let's proceed.
now, let's multiply both sides by the LCD of all denominators, hmmmm in this case that'll be 4, so multiplying both sides by the LCD will do away with the denominators, so let's do so.
