50 pls with attached and show your work

∠1 = 39°∠3 = 51°

Step-by-step explanation:

∠2 is a right angle, so ∠3 is complementary to the 39° angle shown. It is ...

  ∠3 = 90° -39° = 51°

ΔDAB is isosceles, so ∠DBA ≅ ∠3 and ∠1 ≅ 39°.

If you're starting completely from scratch, you might go to the trouble to show ΔABC ≅ ΔADC, hence AC is a bisector of angles A and C. If X is where the diagonals cross, then you can show ΔABX ≅ ΔADX and ∠2 is 90°.

I prefer to start with the fact that the diagonals of a (symmetrical) kite cross at right angles. That makes all the triangles right triangles, so the angles are easily found, given any one of them. (Acute angles of a right triangle are complementary.) Of course, the triangles on one side of the line of symmetry (AC) are congruent to those on the other side.

Ineed show the photo because i need more information

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 *_*