Answer:
Step-by-step explanation:
As you can see from the graph I attached you, the possible solutions in the interval from 0 to 2π are approximately:
So, it's useful to solve the equation too, in order to verify the result:
Taking the inverse sine of both sides:
Using this result we can conclude the solutions in the interval from 0 to 2π are approximately:
You can count from 6-10 and you would get c) 5/10
Answer:
i think its -3
Step-by-step explanation:
Answer:
D 7
Step-by-step explanation:
Total no. of bulb= 5
no. of defected bulb= 3
no. of not defected bulb=2
Total no. of bulb combination = 5C2
=5!/2!(5-2)!
= 5!/2!3!
= 5×4×3×2×1/2×1×3×2×1
=120/12
=10
( since a room can be lighted with one bulb also)
total no. of bulb combination when room shall not light = 3C2
3!/2!(3-2)!
= 3!/2! 1!
= 3×2×1/2×1×1
= 6/2
=3
Now,
Total no. of trial when room shall light
=10-3
=7
Hence, number of trial when the room shall be lighted is 7 which is option d
