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
I hope this is helpful. I found the solution on the internet. Hope you have a good day and bye.
Using the revenue model for most company sales agents y = 150x - x^2, and the weekly revenue of $10,000 assured by the applicant, you get:
150x - x^2 = 10,000
From which you can solve for x:
x^2 - 150x + 10,000 = 0
When you use the quadratic equation you find that the equation does not have real solution driving to the conclusion that the agent is not tellling the true.
The constant should be added to form a perfect square trinomial will be 1/4. Then the correct option is D.
<h3>What is a quadratic equation?</h3>
It's a polynomial with a value of zero. There exist polynomials of variable power 2, 1, and 0 terms. A quadratic equation is an equation with one statement in which the degree of the parameter is a maximum of 2.
The expression is x² + x.
Then the constant should be added to form a perfect square trinomial.
Then the constant will be
The square of the half of the coefficient of the variable x is to be added to make a perfect square.
Then the constant will be 1/4.
Then the perfect square will be
More about the quadratic equation link is given below.
