Question

In 2016, a school started its own aquarium with x fish. In 2019, there were x3 fish in the same aquarium. There were 512 fish in 2019. How many fish were there in 2016? Show your work.

Answer 1

Let's use algebra to solve the problem. We can start by using the information given to set up an equation relating the number of fish in 2016 (x) to the number of fish in 2019 (x3):

x3 = 512

To solve for x, we can isolate x by dividing both sides of the equation by 3:

x = 512/3

x = 170.67 (rounded to two decimal places)

Therefore, there were approximately 170.67 fish in the aquarium in 2016. However, since we cannot have a fraction of a fish, we can round up to the nearest whole number to get the final answer:

x ≈ 171

So there were 171 fish in the aquarium in 2016.

Explanation:

Answer 2

Explanation:

We are given that in 2016, the aquarium had x fish, and in 2019, there were x3 fish. We also know that in 2019, there were 512 fish. So we can set up the equation:

x3 = 512

To solve for x, we need to take the cube root of both sides of the equation:

x = ∛512

Using a calculator or simplifying by prime factorization, we can find that:

512 = 2^9

So,

∛512 = ∛2^9 = 2^3 = 8

Therefore, there were 8 fish in the aquarium in 2016.