Jim makes the following conjecture. other than 1, there are no numbers less than 100 that are both perfect squares and perfect cubes. what is a counter example that proves his conjecture false

64  is a counter example that proves Jim's conjecture as false.

Explanation

The number should be less than 100 and need to be both perfect square and perfect cubes.

So, 64 is a perfect squared number.

Now,

So, 64 is also a perfect cubed number.

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!