1. ab is tangent to circle o at b. the diagram is not drawn to scale. if ab = 6 and ao = 11.4, what is the length of the radius (r)? round the answer to the nearest tenth. (1 point) 8.9 5.7 13.1 10.0 2. jk, kl, and lj are all tangent to circle o. the diagram is not drawn to scale. if ja = 12, al = 15, and ck = 5, what is the perimeter of δjkl? (1 point) 34 64 32 54 3. wz and xr are diameters of circle c. the diagram is not drawn to scale. (1 point) what is the measure of zwx 355º 5º 292º 243º 4. the radius of circle o is 32, and oc = 13. the diagram is not drawn to scale. what is the length of ab? round the answer to the nearest tenth. (1 point) 58.5 29.2 34.5 69.0

1.) The radius which is OB also is 10.0.
   It is a right triangle so c^2 = a^2 + b^2 : In this case, AO^2 = AB^2 + OB^2
   11.4^2 = 6^2 + OB^2
   129.96 = 36 + OB^2
   OB^2 = 93.96
   OB = 9.69

   Note: I think your given is wrong since there's no 9.69 in the choices. But a similar problem is found on the internet and the given are  AB = 6 and AO = 11.7. So If these are the given data, the answer would be 10.0, Letter D.

2.) The perimeter is 64.
   In an incircle, the distances between a vertex and the two nearest tangent points are equal to each other.
   JA = JB
   LA = LC
   KC = KB

   2 × (12 + 15 + 5) = 64

3.) The measure of ZWX is 243º.
   A diameter forms an arc of 180°, so ZRW would equal to 180°. To get ZWX, we'll add ZRW and WX.
   180° + 63° = 243°

4.) The length of AB is 58.5.
   We'll get the length of CB first, by using OB^2 = OC^2 + CB^2, substitute and transpose to get CB.
   CB = sqrt(32^2 - 13^2)
   CB = sqrt(1024 - 169)
   CB = sqrt(855)
   CB = 29.24

   AB is equal to 2CB.
   AB = 2(29.24) = 58.5

Answer to your question: a. the fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. the roots are ​ x=1±i7√2 .number of roots=degree of polynomialhere, degree of polynomial is 2 so number of roots will be 2.hope it *_*