50 points and Brainliest to whoever solves this problem I invented it so u guys can't cheat
Also if you just put in some random answer, I will report you!

So, if you have a balance and 8 coins
Amongst the 8 coins there is one that is heavier
you can only use the balance twice
how do you find the heavier one?
I have the answer if u want to know how to solve it!
Good luxk contestants :))

in order to maximize efficiency of this balancer you put four of the coins on the balancer and then you do it again and record both of how much each set of four coins weigh and you see that eg B's set of coins weighs more so then you divide As weigh by 4 and then you see that (whatever number it is) should be what the answer is then you divide Bs by four (with the odd one out) depending on how much heavier it is you would just take away from the weigh until it Raquel's what A equals then you would add that then whatever B divide by four equals you would add however many of the remainder of B and ad dit to the answer of B divided by four... done that's how you do it

Split the coins into 3 groups – 2 groups with 3 coins each and 1 group with 2 coins.

Now put the 2 groups of 3 coins on the scale.

Scenario #1: Both groups weigh the same. This means that the counterfeit coin is in the group with 2 coins. So take the 2 coins from that group and use the scale to determine the counterfeit coin.

Scenario #2: One group weighs less than the other. Take any 2 coins from this group of 3 coins. If they weigh the same, then the 3rd coin is the counterfeit coin. If they don’t weigh the same, then the heavier one is obviously the counterfeit coin.

Step-by-step explanation:

c. (0.6,-1.4)

step-by-step explanation:

hope this

can you put the picture up?

step-by-step explanation: