contestada

Alex starts with a population of 1,000 amoeba that quadruples every hour for a number of hours, h. He writes the expression 1,000(4h) to find the number of amoeba after h hours. Emma starts with a population of 1 amoeba that increases 40% every hour for a number of hours, h. She writes the expression (1 + 0.4)h to find the number of amoeba after h hours. Use the drop-down menus to explain what each part of Alex's and Emma's expressions mean.

Respuesta :

4: growth factor for each hour
h: number of hours
1000: initial population
4^h: growth factor after h hours

Emma
0.4: percent increase
h: number of hours
1+0.4: growth factor for each hour

Answer:

In Alex expression,

[tex]1000(4^h)[/tex] shows the number of Amoeba after h hours.

Where, [tex]4^h[/tex] is the growth factor for h hours.

Let,

[tex]f(h) = 1000(4^h)[/tex]

Initially, h = 0 ⇒ f(0) = 1000

Hence, 1000 shows the initial number of amoeba,

Now, after 1 hours, h = 1,

⇒ Number of amoeba, f(1) = 1000 × 4

For h = 2, number of amoeba, f(2) = 1000 × 4 × 4

For h = 3,  number of amoeba, f(3) = 1000 × 4 × 4 × 4

..........so on...

⇒ Number of amoeba is increasing with the growth factor 4,

4 is the growth factor by with number of amoeba is increasing.

Now, in Emma expression,

[tex](1+0.4)^h[/tex] shows the number of amoeba after h hours,

Let,

[tex]H(h)=(1+0.4)^h[/tex]

Initially, h = 0 ⇒ H(0) = 1

Initial number of amoeba = 1,

Now, after h = 1,

H(1) = (1+0.4)

For h = 2, H(2) = [tex](1+0.4)^2[/tex] = 1 (1+0.4)(1+0.4)

For h = 3, H(3) = [tex](1+0.4)^3[/tex] = 1 (1+0.4)(1+0.4)(1+0.4)

...... so on,...

Hence, the number of amoeba is increasing with the rate of  0.4 and with the growth factor of (1+0.4).

0.4 is the growth rate.

And, (1+0.4) is the growth for each hour.

Otras preguntas