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shane kepley

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  • JAG legal advice for National Guard?

    I'm in the Florida National Guard. I used to be active in the 82nd and I know that if a service member had a legal issue (noncriminal), they could go to JAG to get legal advice. Now I'm in the National Guard and I just ended a lease. Just prior to ending the least the landlord sold the duplex, and the new landlord is refusing to give me my security deposit back (she actually is claiming I did not pay the entire deposit and is demanding I owe her money).

    What I need to know is if there is a guard version of JAG that can help me determine how I need to proceed with a civil action to get my money back. We left the apt in immaculate condition with no damages and have receipts and proof that we paid the entire security deposit to the old land lord when we moved in. I'm just not sure where I need to go or what paperwork I need to file to get this taken to small claims court. Any help or phone numbers would be appreciated.

    4 AnswersMilitary10 years ago

  • limit of the sequence...?

    n-(sqrt(n^2-3n))

    the answer is 3/2. I keep coming up with n - n = 0

    n - sqrt(n^2-3n)). If I factor the radical I get

    n - sqrt (n)*sqrt (n-3). Since sqrt (n-3) is slightly less than sqrt (n), it should behave roughly like sqrt(n)*sqrt(n) =n.

    Or if I factor out n^2 I get

    n - sqrt (n^2)*sqrt(1-3/n) = n - n*sqrt(1) = n-n.

    We just took our calc 2 final and I got this question horribly wrong. Our semester was cut short a few days and we spent very little time on infinite sequence/series so I'm not sure how to get 3/2 from this.

    1 AnswerMathematics1 decade ago

  • Integral of tan^4 x sec x dx?

    I've tried integration by parts to get tangent into an odd power, or secant into an even power so I can use sec^2 x = 1-tan^2 x. No luck. I've also tried integration by parts attempting to use tan^3 x as u, with sec x tan x as dv. Doing this gives me du of 3 tan^2 x * sec^2 x and v = sec x. Applying the product rule I get tan^3 x * sec x - integral 3tan^2 x sec^3 x.

    My thoughts from here are since trig identities failed, to somehow pick a u and dv such that the next integral of (udv) is equal to my original integral of tan^4 x sec x dx and then subtract from both sides similar to how you would use integration by parts to solve a product of 2 non-reducing factors. I just can't seem to figure out any combination that allows me to get back to my original integral.

    What am I missing?

    2 AnswersMathematics1 decade ago