Let u = < 3, -1> , v = < -6, -6> . find 9u + 2v.

9u means you're multiplying 9 into that vector, both components.  Same with the 2v.  9*3 = 27 and 9*-1 = -9, so your new vector u is <27, -9>.  Now let's do v.  2* -6 (twice) = -12, so your new v vector is <-12, -12>.  Add those together now, first components of each and second components of each.  27 + (-12) = 15; -9+(-12)=-21.  So the addition of those gives us a final vector with a displacement of <15, -21>

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