I need to evaluate the limit and I also need to show the work for it. Please help me out if you can.
1 answer:
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Answer:
Jogging 6th time.
Step-by-step explanation:
We have been given that Vicki started jogging the first time she ran she ran 3/16 mile the second time she ran 3/8 mile and the third time she ran 9/16 mile.
We can see that the distance Vicki covers each time forms a arithmetic sequence, where 1st term is 3/16.
We know that an arithmetic sequence is in form , where,
= nth term of sequence,
= 1st term of sequence,
n = Number of terms in sequence,
d = Common difference.
Let us find common difference of our given sequence as:
Since Vicki needs to cover more than 1 mile, so we nth term of sequence should be greater than 1.
Let us solve for n.
We can also write next terms of our sequence as:
Therefore, Vicki will run more than 1 mile when she is jogging for 6th time.
3 sin x = 2 ( 1 - sin² x )
3 sin x = 2 - 2 sin² x
2 sin² x + 3 sin x - 2 = 0
Substitution: t = sin x
2 t² + 3 t - 2 = 0
t 1/2 =
t 1 = 1/2
t 2 = - 2 ( this solution is not acceptable )
sin x = 1/2
Answer:
x 1 = π / 6 + 2 kπ, x 2 = 5 π / 6 + 2 k π, k ∈ Z
3 sin x = 2 - 2 sin² x
2 sin² x + 3 sin x - 2 = 0
Substitution: t = sin x
2 t² + 3 t - 2 = 0
t 1/2 =
t 1 = 1/2
t 2 = - 2 ( this solution is not acceptable )
sin x = 1/2
Answer:
x 1 = π / 6 + 2 kπ, x 2 = 5 π / 6 + 2 k π, k ∈ Z