T1 = 3
t2 = 2*t1 = 6
t3 = 2*t2 = 12
t4 = 24
t5 = 48
Yes I got the third one too
Answer:
6
Step-by-step explanation:
1. re-write the given system:
cx+3y=c-3; => y= -cx/3 +(c-3)/3
12x+cy=c; => y= -12x/c+c/12
2. according the condition the rule for the parallel graphs is:
-c/3= -12/c
3. to calculate the unknown 'c':
c²=36; ⇔c=±6
