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How many integers satisfy the inequality |10(x+1)/x^2+2x+3|=1? Solution) first thing thats not an inequality,and second thing its very easy if thats the question.the LHS = |10/x^2 + 10/x + 2x + 3| = 110/x^2 will also be an integer only if x^2<10.....i.e. x<3.hence x= 0,+1,-1,+2,-2,+3,-3.but the only value that holds the equality is -1.hence the no. of INTEGERS is only 1.
ABC is a triangle right angled at c. let BC=a, CA=b, AB=c and lrt p be the length of the perpendicular from C on AB. prove that cp=ab and 1/p2=1/a2+1/b2
-x^3+6x-7
Find the series solution of2x2y”+xy’+(x2-3)Y=0 about regular singular point
info about right triangles
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