Answer:
A
Step-by-step explanation:
it is going from low to high
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: x= -7
Step-by-step explanation:
3x/5 + 5 = 4/5
Put the fractions in 3x/5 +5 over a common denomintator.
Answer:
2x²y⁴
Step-by-step explanation:
Area = W × L
W = A / L = 
= 2x²y⁴
