danielaguardado
contestada

Suppose that a population of fish in a pond doubled every 6 months. There were originally 8 fish in the pond. How many fish were in the pond after 10 years?
A. 5,120
B. 8,192
C. 4,194,304
D. 8,388,608

Respuesta :

Answer is D. Just use 8x2^20
To solve this is pure multiplication.

1 year is made up of 12 months.
So 12 divided by 2 is 6.

So that means the fish doubled in the pond twice a year.
For 10 years

2 x 10 = 20

In the 10 years the fish doubled 20 times.

Starting with the original 8
8 x 2 = 16     1
16 x 2 =32      2
32 x 2 = 64        3
64 x 2 = 128         4
128 x 2 = 256         5
256 x 2 = 512          6
512 x 2 = 1024          7
1024 x 2 = 2048        8
2048 x 2 = 4096        9
4096 x 2 = 8192       10
8192 x 2 = 16384      11
16384 x 2 = 32768    12
32768 x 2 = 65536    13
65536 x 2 = 131072   14
131072 x 2 = 262144  15
262144 x 2 = 524288  16
524288 x 2 = 1048576 17
1048576 x 2 = 2097152  18
2097152 x 2 = 4194304   19
4194304 x 2 = 8388608    20

D being your answer :)

Hope this helps.
Brainliest is always appreciated if you feel its deserved.

Otras preguntas