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Help solve an algebra problem see inside please? and got a better answer
Response from Дима[+++]
Response from 0[+++++]
Invested 10000 r first increase m is 10000+10000*m =10000+100*m. the second increase 10000+100*m + 0.05* 10000+100*m and it equals 11550 10000+100m+500+5m=11550 105m=11550-10000-500 105m=1050 m = 10 Answer 10
Invested 10000 r first increase m is 10000+10000*m =10000+100*m. the second increase 10000+100*m + 0.05* 10000+100*m and it equals 11550 10000+100m+500+5m=11550 105m=11550-10000-500 105m=1050 m = 10 Answer 10
Response from 0[+++]
k is the interest at the beginning of 1 year 10000 + 10000*k = amount after 1 year 10000 + 10000*k + 10000 + 10000*k * k+5 = 11550 amount after 2 years now solve the quadratic equation
k is the interest at the beginning of 1 year 10000 + 10000*k = amount after 1 year 10000 + 10000*k + 10000 + 10000*k * k+5 = 11550 amount after 2 years now solve the quadratic equation
Response from 0[+++++]
By convention 10000* 1+x * 1+2*x=11550 where x*100 is the bank's primary annual interest x^2+2.05*x-0.105=0=0 x=-2.1 or 0.05 The former does not fit Answer: 0.05*100 =5 annual
By convention 10000* 1+x * 1+2*x=11550 where x*100 is the bank's primary annual interest x^2+2.05*x-0.105=0=0 x=-2.1 or 0.05 The former does not fit Answer: 0.05*100 =5 annual