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If the difference among the squares of two consecutive integers is 15 find out the larger integer.
Let x = the lesser integer and let x + 1 = the greater integer. The sentence, "the difference among the squares of two consecutive integers is 15," can translate to the equation (x + 1)2 - x2 = 15. Multiply the binomial (x + 1)2 as (x + 1)(x + 1)by using the distributive property: x2 + 1x + 1x + 1 - x2 = 15. Merge like terms: 2x + 1 = 15; subtract 1 from both sides of the equation: 2x + 1 - 1 = 15 - 1. Divide both sides through 2: 2x/2= 14/2. The variable is now alone: x = 7. Thus, the larger consecutive integer is x + 1 = 8.
Prove that sec 2 θ+cosec 2 θ can never be less than 2. Ans: S.T Sec 2 θ + Cosec 2 θ can never be less than 2. If possible let it be less than 2. 1 + Tan 2 θ + 1 + Cot
.find lim sup Ek and liminf Ek of Ek=[(-(1/k),1] for k odd and liminf Ek=[(-1,(1/k)] for k even
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