ΔJKL, sin(JLK)=JK/KL=3/6=0.5
<JKL=30deg.
cos(JLK)=JL/KL
JL=KL*cos(JLK)=6cos(30deg.)
<JLK + <JLM = <MLK
30 + <JLM = 61
<JLM=31deg
in ΔMJL
tan(JLM)=JM/JL
JM=JL*tan(JLM)
=6cos(30deg.)*tan(31deg)
=3.12
Answer:
b
Step-by-step explanation:
Step-by-step explanation:
step 1. The x values refer to the domain of the function.
step 2. if you set y = 0 and solve the quadratic equation the x values refer to the point the graph crosses the x axis.