Question

mr. ken's class is collecting empty water bottles. they started with 17 water bottles on day 1. on day 3, they had 30 water bottles. on day 5, they had 47 water bottles. how many water bottles did they have in day 2 and on day 4? how many water bottles will there be on day 7?

Answer 1

They had 23 and 38 water bottles on day 2 and day 4 respectively. Also, there will be 68 water bottles on day 7.

Explanation

Suppose, the equation representing the number of bottles is  $y= ax^2 +bx+c$ , where $x$ is the number of days.

There are 17 water bottles on day 1,  30 water bottles on day 3 and  47 water bottles on day 5

So, the three points in the form of (x, y) are:   (1, 17) , (3, 30) and (5, 47)

Plugging these three points into the above equation, we will get .....

$17=a(1)^2 + b(1) +c\\ a+b+c=17 ............................ (1)\\ \\ 30=a(3)^2+b(3)+c\\ 9a+3b+c=30 ......................... (2) \\ \\ 47=a(5)^2+b(5)+c\\ 25a+5b+c=47 ......................... (3)$

Subtracting equation (1) from equation (2) , we will get .....

$8a+2b=13 ..................... (4)$

Subtracting equation (2) from equation (3) , we will get .....

$16a+2b=17 ...................... (5)$

Now, subtracting equation (4) from equation (5) , we will get ......

$16a -8a = 17-13\\ \\ 8a= 4\\ \\ a=\frac{4}{8}= 0.5$

Substituting this $a= 0.5$ into equation (4) ........

$8(0.5)+2b=13\\ \\ 4+2b=13\\ \\ 2b=9\\ \\ b=\frac{9}{2}=4.5$

Again, substituting $a=0.5$ and $b=4.5$ into equation (1) , we will get ......

$0.5+4.5+c=17\\ \\ 5+c=17\\ \\ c=17-5 =12$

So, the equation will be now:  $y= 0.5x^2 +4.5x+12$

For finding the number of bottles on day 2 , day 4 and day 7 , we need to plug $x=2 , 4, 7$ respectively into the above equation.

For $x= 2$ ,  $y= 0.5(2)^2 +4.5(2)+12 = 2+9+12= 23$

For $x= 4$ ,  $y= 0.5(4)^2 +4.5(4)+12 = 8+18+12= 38$

For $x= 7$ ,  $y= 0.5(7)^2 +4.5(7)+12 = 24.5+31.5+12= 68$

So, they had 23 water bottles on day 2 and 38 water bottles on day 4

Also, there will be 68 water bottles on day 7