At the end of the Year 1 the population will be = 110,000 + 4% of 110,000 =P1At the end of the Year 2 the population will be=P1+ 4% of P1= 110,000 + 4% of 110,000 + 4% of (110,000 + 4% of 110,000) = 110,000 + 4% of (110,000+ 110,000 + 4% of 110,000) = 110,000 + 2*4% of 110,000 + (4%)2 of 110,000= P2At the end of the Year 3 the population will be= P2+ 4% of P2= 110,000 +2* 4% of 110,000 + (4%)2 of 110,000 + 4% of [ 110,000 +2* 4% of 110,000 + (4%)2 of 110,000] = 110,000 +2* 4% of 110,000 + (4%)2 of 110,000 + 4% of 110,000 + 2* (4%)2 of 110,000 + (4%)3 of 110,000 =110,000 +3* 4% of 110,000 +3* (4%)2 of 110,000 + (4%)3 of 110,000 =P3At the end of the Year 4 the population will be= P3+ 4% of P3=110,000 +3* 4% of 110,000 +3* (4%)2 of 110,000 + (4%)3 of 110,000 +4% of [110,000 +3* 4% of 110,000 +3* (4%)2 of 110,000 + (4%)3 of 110,000] =110,000 +4* 4% of 110,000 + 6*(4%)2 of 110,000+4* (4%)3 of 110,000+ (4%)4 of 110,000 Now if we substitute n for 110,000 and r for 4% yearly rate of increase, we can rewrite,Before Year 1, at Year 0, the population, P0= n = n(1+r)0At the end of the Year 1 the population , P1= n+rn= x(1+r)1At the end of the Year 2 the population , P2= n+2rn+r2n =n(1+2r+r2)=n(1+r)2At the end of the Year 3 the population , P3= n+3rn+2r2n+r3n=n(1+3r+3r2+r3)=x(1+r)3At the end of the Year 4 the population , P4= n+4rn+6r2n+4r3n+r4n=n(1+r)4. . . . . . At the end of the Year n the population , Px=n+nrx+(x-1)r2n+(x-2)r3n+...........+(x-x+2)r(x-1)x+(x-x+1)rxn=n(1+r)x.....At the end of the Year 16 the population , P16= n+16rn+15r2n+14r3n+.............+3r14n+2r15n+r16n=n(1+r)16 Thus under given condition of rate of growth, the Population P(x) at xth year will be P(x)=n(1+r)x Therefore, a population of 110,000 growing at 4% per year, in 16 years will be= 110,000(1+4/100)16= 110,000(1.04)16=206027.93702999115428132473599427≈206028
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In geometry, a hypotenuse is the longest side of a right-angled triangle, the side opposite the right angle. The length of the hypotenuse can be found using the Pythagorean theorem, which states that the square of the length of the hypotenuse equals the sum of the squares of the lengths of the other two sides.
