Answer:
PΔJKL=66
Step-by-step explanation:
so we are given the line segments JK, KL, and LJ which are tangent to k(O), and also that JA=9, AL=10, and CK=14
JL=JA+AL (parts whole postulate)
JL=9+10=19 (substitution, algebra)
JA=JB=9 (tangent segments from the same point are congruent)
CK=KB=14 (tangent segments from the same point are congruent)
JK=JB+KB (parts whole postulate)
JK=9+14=23 (substitution, algebra)
LA=LC=10 (tangent segments from the same point are congruent)
LK=LC+CK (parts whole postulate)
LK=10+14=24 (substitution, algebra)
Perimeter of ΔJKL=LK+KL+LJ (perimeter formula for triangles)
Perimeter of ΔJKL=23+24+19=66 (substitution, algebra)
Answer. The standard form of a line is in the form Ax + By = C where A is a positive integer, and B, and C are integers. Hope this helps, let me know if it’s correct so others can use it :)
Good luck.
Option b is the better deal
Answer:
there is none prime factorization of 3. Since it cant be multiplied by anything to get 3 like 2 and 5 and 7
Step-by-step explanation:
