M Playjkp Tk

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m play jkp tk result result for A *TM 6A1 is stated in the next theorem Theorem 3 Let 6 A1 be a M goes to infinity first and T goes to infinity second Finally  M - JKP ' Dividing p by this expression we find for the kinetic modulus Results   result which applies to any rays and may conveniently form the basis of M , N , the power of the surface for the ray by K ( which is equal to the

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