One geometry question i'm failing and having a mental breakdown as we speak
me i wanna graduate i got 40 something daysmy mom will disown me if i don't the woman already wants to kick me out i need just this one i ha ve a 37 in geometry this is the only class im failing

So this circle has two lines, called tangent lines, coming off of it. How do we know they're tangent lines? They each just touch the circle at a single point (and make a right angle with a radius at that point), R for one, T for the other. And these two lines make an angle. We are asked to find the measure of this angle.

Once you learn this rule, you can apply it easily. The theorem states that if two secant lines, tangent lines, or one of each form an angle outside of the circle, then that angle is half the major arc minus the minor arc . That sounds confusing, I know, but basically we take the major arc TR, which has to be 360° – 134° = 226° (remember, the total has to be 360°, and one piece is 134°), and subtract the minor arc TR, which is 134°, so we get 92°. Remember, we want half of the difference, so half of 92°, which is 46°.

I really hope you are able to pass. Reach out to your teacher, use online resources like Khan Academy, ask classmates, and so on. Best of luck on your gradation!

Best answer that i found on yahoo ok   set :   x= length of 2nd piece in cm  x- 3= length of 1st piece in cm  x-4= length of 3rd piece in cm  then since you know they all came from the 20 cm ribbon you add them together and set them equal to 20x+ x-3 +x-4 = 20  3x - 7= 20  3x= 27  x= 9  the shortest piece is the third piece which is x- 4  x-4  9-4 = 5  the length of the smallest piece is 5 cm.  the other lengths are  9 cm  6 cm

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12